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Quantum tunneling of the magnetic moment in a free nanoparticle
Journal of Magnetism and Magnetic Materials 324 (2012) 2871–2878
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Quantum tunneling of the magnetic moment in a free nanoparticle M.F. O’Keeffe, E.M. Chudnovsky n, D.A. Garanin Physics Department, Lehman College, City University of New York, 250 Bedford Park Boulevard West, Bronx, New York, 10468-1589, USA
a r t i c l e i n f o
abstract
Article history: Received 11 January 2012 Received in revised form 30 March 2012 Available online 7 May 2012
We study tunneling of the magnetic moment in a particle that has full rotational freedom. Exact energy levels are obtained and the ground-state magnetic moment is computed for a symmetric rotor. The effect of mechanical freedom on spin tunneling manifests itself in a strong dependence of the magnetic moment on the moments of inertia of the rotor. The energy of the particle exhibits quantum phase transitions between states with different values of the magnetic moment. Particles of various shapes are investigated and the quantum phase diagram is obtained. & 2012 Elsevier B.V. All rights reserved.
Keywords: Spin Rotator Tunneling
1. Introduction There has been much recent interest in the quantum mechanics of nanoscopic magnets that possess mechanical freedom. Experimental work in this area focused on free magnetic clusters [1–3], magnetic particles that are free to move inside solid nanocavities [4], magnetic microresonators [5,6], and magnetic molecules bridged between conducting leads [7–9]. Theoretical research on free magnetic particles has been scarce. The generic problem is that of a rigid quantum rotor with a spin. Without a spin this problem is tractable by analytical methods only for a symmetric rotor [10]. Complications resulting from spin degrees of freedom make even symmetric cases significantly more difficult [11]. The first attempt to understand how mechanical freedom of a small magnetic particle affects tunneling of the magnetic moment was made in Ref. [12], where it was noticed that tunneling of a macrospin in a free particle must be entangled with mechanical rotations in order to conserve the total angular momentum (spinþorbital). A similar situation occurs for tunneling of a superconducting current between clockwise and counterclockwise directions in a SQUID [13–17]. Recently, it was demonstrated [18] that the problem of a rigid rotor with a spin can be solved exactly in the laboratory frame when mechanical rotation is allowed only about a fixed axis and the spin states are reduced to spin-up and spin-down due to strong magnetic anisotropy. The latter is typical for magnetic molecules and small ferromagnetic clusters [19–21]. The reduction to two spin states in a system rotating about a fixed axis also allows one to obtain a simple solution of the problems of a magnetic molecule embedded in a
n
Corresponding author. E-mail address: [email protected] (E.M. Chudnovsky).
0304-8853/$ - see front matter & 2012 Elsevier B.V. All rights reserved. http://dx.doi.org/10.1016/j.jmmm.2012.04.032
microcantilever [22], a magnetic molecule vibrating between conducting leads [23], and a macrospin tunneling inside a torsional resonator [24,25]. However, the problem for arbitrary rotations of a two-state spin system, which is relevant to free magnetic nanoparticles, remained unsolved until now. In this paper we show that the problem of a two-state macrospin inside a symmetric rigid rotor has a rigorous solution for arbitrary rotations in the coordinate frame that is rigidly coupled to the rotor. The magnetic moment of electrically neutral rotor is entirely due to spin. It depends on the relative contribution of the up and down spin states. When a nanoparticle is embedded in a solid, tunneling of the spin results in a zero ground-state magnetic moment. This situation changes for a free particle due to a complex interplay between spin and mechanical angular momentum that conserves the total angular momentum. We show that the energy of the particle exhibits first- or secondorder quantum phase transition between states with different values of the total angular momentum. The order of the transition depends on the shape of the particle. The ground-state magnetic moment of a free particle with a total spin S can be anything between zero and g mB S, depending on the principal moments of inertia (with mB being the Bohr magneton and g being the gyromagnetic factor associated with the spin). The structure of the article is as follows. Quantum theory of a rigid rotator is briefly reviewed in Section 2. Theory of a tunneling macrospin is reviewed in Section 3. Quantum states of a rigid rotator containing a tunneling macrospin are constructed in Section 4. Ground state of a symmetric rotor with a spin is analyzed in Section 5. Ground-state magnetic moment is studied in Section 6. Our conclusions are presented in Section 7. Mathematics of the problem studied in this article is based upon anomalous commutation relations of the operator of angular momentum in the coordinate frame that is rigidly coupled with
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the rotating particle. To simplify the reading and to make the paper self-contained, some old results that are difficult to find in the literature are derived in the Appendix.
antisymmetric combinations of 9c 7 S S, respectively [21], 1
C þ ¼ pffiffiffi ð9cS S þ9cS SÞ, 2
1
C ¼ pffiffiffi ð9cS S9cS SÞ,
2. Quantization of rigid body rotations
2
Consider first the problem of a rigid body rotating without a spin. We choose the coordinate frame that is rigidly coupled with the rotating body and direct the axes of that frame x, y, and z, along the principle axes of the tensor of moments of inertia of the body. In such coordinate frame the Hamiltonian of mechanical rotations is given by [10] ! 2 2 2 L2 ^ R ¼ _ Lx þ y þ Lz : H ð1Þ 2 Ix Iy Iz Here Ix , Iy , Iz are the principal moments of inertia and Lx , Ly , Lz are projections of the operator of the mechanical angular momentum (defined in the laboratory frame) onto the body axes x, y, z. Such a choice of coordinates and operators results in the anomalous commutation relations [26], ½Li ,Lj ¼ iEijk Lk (notice the minus sign in the right-hand side), but does not affect the relations L2 ¼ LðLþ 1Þ, ½L2 ,Lz ¼ 0. For a symmetric rotor two of the moments of inertia are the same, Ix ¼Iy, and the Hamiltonian can be written as 2 2 2 2 ^ R ¼ _ L þ _ Lz 1 1 : H ð2Þ Iz Ix 2Ix 2 The corresponding eigenstates are characterized by three quantum numbers L, K, and M, 2
L 9LKMS ¼ LðLþ 1Þ9LKMS, Lz 9LKMS ¼ K9LKMS, LZ 9JKMS ¼ M9LKMS,
L ¼ 0; 1,2, . . . ,
which satisfy ^ SC 7 ¼ E 8 C 7 , H
ð7Þ
where E þ E D:
ð8Þ
The tunnel splitting D is generally many orders of magnitude smaller than the distance to other spin energy levels, which makes the two-state approximation very accurate at low energies. For example, ^ S ¼ DS2 þ dS2 H z y
ð9Þ
with d 5D describes the biaxial anisotropy of spin-10 molecular nanomagnet Fe-8, where the tunnel splitting in the limit of large S is given by [27] S 8S3=2 d D ¼ 1=2 D: ð10Þ 4D p The distance to the next excited spin level is ð2S1ÞD, which is large compared to D. It is convenient to describe these lowest energy spin states C 7 with a pseudospin-1/2. Components of the corresponding Pauli operator r are
sx ¼ 9cS S/cS 9þ 9cS S/cS 9, sy ¼ i9cS S/cS 9i9cS S/cS 9,
K ¼ L,L þ1, . . . ,L1,L,
sz ¼ 9cS S/cS 99cS S/cS 9:
ð11Þ
M ¼ L,Lþ 1 . . . ,L1,L,
^ S onto 9c S states is The projection of H 7S X ^ ^ Hs ¼ /m9H S 9nS9mS/n9:
ð12Þ
ð3Þ
where LZ is the angular momentum operator defined with respect to the laboratory coordinate frame (X,Y,Z). The eigenvalues of (2) are degenerate on K: _2 LðLþ 1Þ _2 K 2 1 1 þ ELK ¼ : ð4Þ 2Ix Iz I x 2 The general form for the energy levels of a rotating asymmetric rigid body, Ix a Iy aIz , does not exist, although it is possible to calculate matrix elements of the Hamiltonian for a given L.
3. Tunneling of a large spin Let S be a fixed-length spin embedded in a stationary body. Naturally, the magnetic anisotropy is defined with respect to the body axes. The general form of the crystal field Hamiltonian is ^ J þH ^ ?, ^S ¼H H
ð6Þ
ð5Þ
^ ? is a perturbation that does ^ J commutes with Sz and H where H not commute with Sz. The states 9 7SS are degenerate ground ^ J , where S is the total spin of the nanomagnet. H ^? states of H slightly perturbs these states, adding to them small contributions from other 9mS S states. We will call these degenerate perturbed states 9c 7 S S. Physically they describe the magnetic moment aligned in one of the two directions along the anisotropy axis. Full ^ S properturbation theory with account of the degeneracy of H vides quantum tunneling between the 9c 7 S S states for integer S. The ground state and first excited state are symmetric and
m,n ¼ c 7 S
Expressing 9c 7 S S in terms of C 7 one obtains ^ S 9c S ¼ 0, /c 7 S 9H 7S
^ S 9c S ¼ D , /c 7 S 9H 8S 2
ð13Þ
which gives the two-state Hamiltonian ^ s ¼ D sx H 2
ð14Þ
having eigenvalues 7 D=2. In the absence of tunneling a classical magnetic moment is localized in the up or down state. It is clear that delocalization of the magnetic moment due to spin tunneling reduces the energy by D=2. In a free particle, however, tunneling of the spin must be accompanied by mechanical rotations in order to conserve the total angular momentum. Such rotations cost energy, so it is not a priori clear whether the tunneling will survive in a free particle and what the ground state is going to be. This problem is addressed in the following Section.
4. Rigid rotor containing tunneling macrospin Consider now a tunneling macrospin embedded in a free particle having the body z-axis as the magnetic anisotropy direction. When the magnetic anisotropy is dominated by magnetostatic interactions, this choice is justified by the fact that the principal axes of the demagnetizing tensor would always coincide with the principal axes of the rotator [28]. For a symmetric
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rotator, macroscopic magnetostatic interactions alone do not provide the transverse term in the Hamiltonian that generates tunneling of the spin. In this case one has to assume that the tunneling is provided by the transverse component of the crystal field due to microscopic structure of the particle, see Eqs. (5) and (9). The free particle is characterized by the total angular momentum, J ¼ L þS. In the body frame this operator may appear unconventional due to the different sign of commutation relations for L and S (see Appendix). However, this problem can be easily fixed [11] by the transformation S-S~ ¼ S that changes the sign of the commutation relation for S. Such a transformation does not change the results of the previous section because the crystal field Hamiltonian contains only even powers of S, so that ^S ¼H ^ ~ . It is interesting to notice that while in the laboratory H S frame ½Ji ,Sj ¼ iEijk Sk , components of the operators J and S~ defined in the body frame commute with each other [11]. In addition, the operator J2 is the same in the body and laboratory frames [26]. This permits description of quantum states of the particle in terms of quantum numbers associated independently with the total angular momentum and spin. The full Hamiltonian is given by the sum of the rotational energy and magnetic anisotropy energy
easy-axis magnetic anisotropy and the tunnel splitting can decrease or increase, depending on the ratio Ix =Iz . Projection of Eq. (17) on the two spin states along the lines of the previous section gives 2 2 2 2 2 ^ ¼ _ J þ _ J z 1 1 D sx _ S J sz , H ð20Þ Iz Ix Iz z 2Ix 2 2
2 2 _2 L2y _2 L2z ^ ¼ _ Lx þ ^ ~: H þ þH S 2Ix 2Iy 2Iz
where EJK is provided by Eq. (4) with L replaced by J. The upper (lower) sign in Eq. (24) corresponds to the lower (upper) sign in Eq. (22). For K a 0 each state is degenerate with respect to the sign of K. For K ¼ 0; 1,2, . . . the coefficients in Eq. (22) are given by rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi C 7 ¼ 17 aK= S2 þ ðaKÞ2 , ð25Þ
ð15Þ
Note that the mechanical part and the spin part of this Hamiltonian are not independent because in the body frame the operators L and S~ do not commute, ½Li , S~ j ¼ iEijk S~ k . It, therefore, makes sense to express the above Hamiltonian in terms of the commuting body-frame operators of the total angular momentum J and ~ reversed spin S: 0 1 ! 2 2 2 J2y J2z S~ y S~ z _2 J 2x _2 @S~ x ^ H¼ þ þ þ þ A þ 2 Ix I y Iz 2 Ix Iy Iz ! J S~ x J y S~ y J z S~ z ^ ~: þ _2 x þ þ ð16Þ þH S Ix Iy Iz For a symmetric rigid rotor with Ix ¼Iy this Hamiltonian reduces to 2 2 2 2 ^ ¼ _ J þ _ Jz 1 1 H I z Ix 2Ix 2 ! ~ ~ J x S x þJ y S y J z S~ z ^0, þ ð17Þ þH þ _2 S~ Ix Iz where 2
^~þ_ ^0 ¼H H S S~ 2
2 1 1 ~ 2 _2 S~ : Sz þ Iz Ix 2Ix
ð18Þ
0
^ ~ is an unessential constant, _2 SðS þ 1Þ=ð2Ix Þ. The The last term in H S second term provides renormalization of the crystal field in a freely rotating particle. For, e.g., the biaxial spin Hamiltonian given by Eq. (9) it leads to _2 1 1 D-D : ð19Þ 2 Iz Ix This, in turn, renormalizes the tunnel splitting given by Eq. (10). For a particle that is allowed to rotate about the Z-axis only (that is, in the limit of Ix -1) these results coincide with the results obtained by the instanton method in Ref. [29], where it was shown that, in practice, the renormalization of the magnetic anisotropy and spin tunnel splitting by mechanical rotations is small. Eq. (19) provides a generalization of this effect for arbitrary rotations of a symmetric rotator with a spin. According to this equation and Eq. (9), when rotations are allowed the effective
where we have used /c 7 S 9Sz 9c 7 S S ¼ 7 S,
/c 7 S 9Sx,y 9c 7 S S ¼ 0:
ð21Þ
We construct eigenstates of this Hamiltonian according to 1 9CJK S ¼ pffiffiffi ðC 7 S 9cS S7 C 8 S 9cS SÞ9JKS, 2
ð22Þ
where J2 9JKS ¼ JðJ þ1Þ9JKS, Jz 9JKS ¼ K9JKS,
J ¼ 0; 1,2, . . . ,
K ¼ J, . . . ,J:
^ CJK S ¼ E9CJK S gives energy levels as Solution of H9 vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi !2ffi u 2 u D _2 KS ð7Þ t þ EJK ¼ EJK 7 , Iz 2
ð23Þ
ð24Þ
where a is a dimensionless magneto-mechanical ratio,
a¼
2ð_SÞ2 : Iz D
ð26Þ
The energy levels in Eq. (24) can be given a simple semiclassical interpretation. Indeed, the last term in this equation is the tunnel splitting of the levels in the effective magnetic field that appears in the body reference frame due to rotation about the spin quantization axis at the angular velocity _K=Iz . When S¼0 (which also means D ¼ 0) Eq. (24) with J¼ L gives the energy of the quantum symmetric rigid rotor without a spin, Eq. (4). In the case of a heavy body (large moments of inertia) the ground state and the first excited state correspond to J¼K ¼0, and we recover the tunnel-split spin states in a non-rotating macroscopic body, E00 7 ¼ 7 D=2. In the general case, spin states of the rotator are entangled with mechanical rotations. Eqs. (22)–(25) are our main analytical results for the lowenergy states of a free magnetic particle. In general, numerical analysis is needed to find the ground state of the particle. Special cases of the aspect ratio that will be analyzed below include a needle of vanishing diameter (which is equivalent to the problem of the rotation about a fixed axis treated previously in the laboratory frame by two of the authors [18]), a finite-diameter needle, a sphere, and a disk.
5. Ground state Minimization of the energy in Eq. (24) on J with the account of the fact that J cannot be smaller than K immediately yields J¼ K, that is, the ground state always corresponds to the maximal projection of the total angular momentum onto the spin quantization axis. In semiclassical terms this means that the minimal energy states in the presence of spin tunneling always correspond to mechanical rotations about the magnetic anisotropy axis. This is easy to understand by noticing that the sole reason for mechanical rotation is the necessity
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to conserve the total angular momentum while allowing spin tunneling to lower the energy. To accomplish this the particle needs to oscillate between clockwise and counterclockwise rotations about the spin quantization axis in unison with the tunneling spin. If such mechanical oscillation costs more energy than the energy gain from spin tunneling, then both spin tunneling and mechanical motion must be frozen in the ground state as, indeed, happens in very light particles (see below). Rotations about axes other than the spin quantization axis can only increase the energy and, thus, should be absent in the ground state. For further analysis it is convenient to write Eq. (24) in the dimensionless form, sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi " # EðJK7 Þ a JðJ þ 1ÞK 2 K2 1 K2 ¼ lþ 2 7 ð27Þ 1 þ 2 a2 , 2 2 D 4 S S S in terms of dimensionless parameters a and the aspect ratio for the moments of inertia
l ¼ Iz =Ix :
ð28Þ
The range of l for a symmetric rotator is 0 r l r2. For, e.g., a symmetric ellipsoid with semiaxes a ¼ b ac, one has l ¼ 2a2 = ða2 þ c2 Þ. The dependence of the energy levels (24) on J at K ¼J is shown in Figs. 1 and 2. It exhibits a quantum phase transition on the parameter a between states with different values of J. Only for a needle of infinite length, Iz =Ix -0, which corresponds to c-1 in the case of an ellipsoid, is the transition second order, see Fig. 1. It occurs at a ¼ ½11=ð2SÞ2 1 . This case is equivalent to the rotation about a fixed axis studied in Ref. [18]. For any finite ratio Iz =Ix the transition is first order if J is treated as a continuous variable, see Fig. 2. It occurs at the value of a that depends on Iz =Ix . The origin of the transfer from a second-order transition at l ¼ 0 to the firstorder transition at l a 0 can be traced to the term ½JðJ þ 1ÞK 2 S2 l in Eq. (27). We should notice that for a finite-size nanomagnet the analogy with first- and second-order phase transition is, of course, just an analogy. To talk about real phase transitions one has to take the limit of S-1, Ix,z -1 when the distances between quantum levels go to zero. For a given l, as a increases the ground-state switches from J¼0 to higher J when ðÞ 0 0 EðÞ 00 ðaJ ðlÞÞ ¼ EJJ ðaJ ðlÞÞ:
ð29Þ
Fig. 2. Dependence of energy on J at K¼ J and Iz =Ix ¼ 2 for different values of a. The plot shows first-order quantum phase transition on a.
Fig. 3. Dependence of the ground-state energy on a for a spherical particle. Inset shows the discontinuity of the derivative of the ground-state energy on a.
Solution of this equation for a0J ðlÞ gives
a0J ¼
ð2SÞ2 ðJ þ lÞ J½ð2SÞ2 ðJ þ lÞ2
:
ð30Þ
This first transition occurs for the smallest value of a0J ðlÞ and the transition is from J¼0 to the corresponding critical value, Jc (Fig. 3). For a o a0Jc the ground state corresponds to J¼0 and C 7 S ¼ 1. After the first transition from J¼0 to J ¼ Jc , the ground state switches to sequentially higher J at values of a which satisfy ðÞ EðÞ J1J1 ðaJ ðlÞÞ ¼ EJJ ðaJ ðlÞÞ:
ð31Þ
Solution of this equation for aJ ðlÞ gives ð2SÞ2 TðJ, lÞ
aJ ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiqffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi , 2
2
ð2SÞ ð2J1Þ TðJ, lÞ2
ð32Þ
ð2SÞ2 TðJ, lÞ2
with TðJ, lÞ ¼ 2J1þ l:
Fig. 1. Dependence of energy on J at K¼J and Iz =Ix ¼ 0 for different values of a. The plot shows second-order quantum phase transition on a.
ð33Þ
The critical aJ has poles at l ¼ 2ðSJÞ þ 1. For l Z 1 there is no longer a ground-state transition to J¼S, even for very large values of a. Needle of infinite length: The case of a particle that can only rotate about its anisotropy axis [18] is equivalent in our model to
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a needle of infinite length (c-1 for an ellipsoid), having l ¼ 0. It is also equivalent to the problem of tunneling of the angular momentum of a superconducting current in a flux qubit coupled to a torsional resonator. In this limit we reproduce results of Ref. [18]. The quantum number K determines the ground state, as the energy, Eq. (27), no longer formally depends on J. However, the values of aJ at l ¼ 0, for which ground-state transitions occur, are ðÞ the same as those for which EðÞ JK1 ¼ EJK , and we will use J to describe the ground state of the axial rotor as well. The first ground-state transition occurs from J ¼0 to J¼1 at aðlÞ ¼ a1 ð0Þ ¼ a01 ð0Þ, because Jc ¼1 for l t 0:01. At a ¼ a2 ð0Þ the ground state switches from J ¼1 to J¼2, and so on. The final transition is to a completely localized spin state J¼S in which spin tunneling is frozen for all a 4 aS ð0Þ. For example, when S ¼10, a1 ð0Þ ¼ a01 ð0Þ ¼ 1:0025 and a10 ¼ 3:2066. Needle of finite diameter: The ground state of a needle of finite diameter (a 5 c for an ellipsoid) with l ¼ 0:1, that is free to rotate about any axis, shows qualitatively different behavior. As a increases, the ground-state changes from J¼0 to Jc ¼3 at a ¼ a03 ð0:1Þ, as the smallest value of a0J ð0:1Þ for 1 r J rS occurs for Jc ¼3. The J¼1, 2 states never become the ground state. After this, transitions occur to successively higher J, beginning with J¼4 at a ¼ a4 ð0:1Þ, and eventually localizing the spin with J¼S for a 4 aS ð0:1Þ. For S¼10, a03 ð0:1Þ ¼ 1:0588 and a10 ð0:1Þ ¼ 3:3935. Sphere: As l increases towards unity, the particle becomes more symmetric with the moment of inertia having (prolate) ellipsoidal symmetry, until it reaches spherical symmetry at l ¼ 1. The first ground-state transition occurs from J¼0 to J ¼ Jc ¼ 5 at a ¼ a05 ð1Þ, and subsequent transitions occur at a ¼ aJ ð1Þ. However, the spin never localizes in the J ¼S state even for very large a, as aJ ð1Þ has a pole at J¼S, so the last transition occurs to the J ¼ S1 state at a ¼ aS1 ð1Þ. For S ¼10, a05 ð1Þ ¼ 1:3187 and a9 ð1Þ ¼ 2:4325. Disk: With l increasing from unity, the symmetry of the body becomes that of an oblate ellipsoid, and begins to flatten in the plane perpendicular to the anisotropy axis. It is easy to check from Eq. (27) that for 1 o l r2 the state with J ¼S always has higher energy than the state with J ¼ S1, even in the limit of a-1. This means that for an oblate particle some spin tunneling (accompanied by mechanical rotations) survives in the ground state no matter how light the particle is. This purely quantummechanical result has no semi-classical analogy. In the case of a disk of vanishing thickness, l ¼ 2, the first ground-state transition occurs from J¼0 to J ¼ Jc ¼ 6 at a ¼ a06 ð2Þ, and subsequent transitions occur at a ¼ aJ ð2Þ up through J ¼ S1. For S ¼10, a06 ð2Þ ¼ 1:5873 and a9 ð2Þ ¼ 3:5849.
6. Ground-state magnetic moment As has been already mentioned, the magnetic moment is due entirely to the spin of the particle, as Lz represents mechanical motion of the particle as a whole, and not electronic orbital angular momentum. Thus,
aK
m ¼ g mB /CJK 9Sz 9CJK S ¼ g mB S qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi : S2 þ ðaKÞ2
ð34Þ
Here g is the spin gyromagnetic factor, and the minus sign reflects the negative gyromagnetic ratio g ¼ g mB =_. The ground state always corresponds to J¼K, so these are used interchangeably in descriptions of the ground state. The dependence of the magnetic moment on a for different aspect ratios of the particle is shown in Fig. 4. For a o aJc ðlÞ the ground state corresponds to J¼K¼0, so the spin-up and spin-down states are in an equal superposition which produces zero magnetic moment. At greater values of a the spin states contribute in unequal
Fig. 4. Ground-state magnetic moment for a needle of vanishing diameter ðl ¼ 0Þ, finite-diameter needle ðl ¼ 0:1Þ, sphere ðl ¼ 1Þ, and a disk of vanishing thickness ðl ¼ 2Þ.
amounts which leads to a non-zero magnetic moment. As a becomes large, the magnetic moment approaches its maximal value 9mmax 9 ¼ g mB S. Note that the magnetic moment approaches its maximum value even for values of l that do not admit transitions to J¼S states. Because the ground state is completely determined by the parameters a and l, we can depict the ground-state behavior in a quantum phase diagram shown in Fig. 5. The curves separate areas in the (a, l) plane that correspond to different values of J and different values of the magnetic moment. Notice the fine structure of the diagram (lower picture in Fig. 5) near the first critical a. This very rich behavior of the ground state on parameters must have significant implications for magnetism of rigid atomic clusters.
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They may also apply to free magnetic molecules if one can justify the condition of rigidity. Direct comparison between theory and experiment may be possible for atomic clusters (molecules) in magnetic traps. To see that the quantum problem studied in this paper may, indeed, be relevant to the quantum states of free nanomagnets, consider, e.g., a spherical atomic cluster of radius R and average mass density r having spin S¼10 that, when embedded in a large body, can tunnel between up and down at a frequency of a few GHz, thus providing D 0:1 K. Significant changes in the magnetic moment of such a cluster would occur at a 1, which, according to Eq. (26), corresponds to I ¼ 8prR5 =15 1042 kg m2 and R 1 nm. For a magnetic molecule like, e.g., Mn12, the moments of inertia would also be in the ballpark of 1042 kg m2 . However, the natural spin tunnel splitting in Mn12 is very small, thus, providing a very large a. The same is true for Fe8 magnetic molecules. In this case the spin tunneling in a free molecule must be completely frozen. Even if the molecule cannot be considered as entirely rigid, such effect, if observed, would receive natural interpretation within the framework of our theory.
Acknowledgements This work has been supported by the U.S. Department of Energy through Grant no. DE-FG02-93ER45487.
Appendix A. Coupling of spin and rotational angular momenta
Fig. 5. Quantum phase diagram for the ground-state magnetic moment and the total angular momentum.
7. Conclusions We have studied the problem of a quantum rotator containing a tunneling spin. This problem is relevant to quantum mechanics of free magnetic nanoparticles. It also provides an interesting insight into quantum mechanics of molecules studied from the macroscopic end. The answer obtained for the energy levels of a symmetric rotator, Eq. (24), is non-perturbative and highly nontrivial. It is difficult to imagine how it could be obtained from first principles without the reduction to two spin states. Indeed, for spin S the tunnel splitting itself generally appears in the S-th order of perturbation theory, see Eq. (10), so the path from the full crystal-field Hamiltonian like, e.g., Eq. (9) to Eq. (24) must be very long. Eqs. (22) and (24) represent, therefore, a unique exact solution of the quantum-mechanical problem of a mechanical rotator with a spin. A striking feature of this solution is the presence of first- and second-order quantum phase transitions between states with different values of the magnetic moment. Our results provide the framework for comparison between theory and experiment on very small free magnetic clusters. Our main conclusion for experiment is that rotational states and magnetic moments of such clusters depend crucially and in a predictable way on size and aspect ratio. This dependence results in a complex phase diagram that separates regions in the parameter space, corresponding to different values of the magnetic moment. A broad distribution of the magnetic moments that does not simply scale with the volume, has, in fact, been reported in beams of free atomic clusters of ferromagnetic materials [1–3]. Our results may shed some additional light on these experiments.
In this Appendix, we derive the so-called anomalous commutation relations of angular momentum in a rotating frame. This treatment applies to the general case of mechanical rotations of a quantum system with internal angular momentum degrees of freedom (see Ref. [11] and references therein). In our case this internal degree of freedom is the spin of the particle, S. Starting with the usual commutation relations for the components of S, components of the angular momentum of mechanical rotations L, and total angular momentum J ¼ S þL in the fixed frame, we derive commutation relations in the rotating (body) frame. We show that in the body frame all components of J and S commute, and therefore the corresponding quantum numbers provide good description of the quantum states of the particle. The nanomagnet is a rigid rotator around its center of mass, which we define as the common origin O of two right-handed coordinate systems, shown in Fig. A1. The X, Y, Z axes make up the lab frame that is fixed in space. The x, y, z axes define the coordinate system that is rigidly coupled to the body, and are directed along the principle
Z z
y
O
X
N
Y
x
Fig. A1. Geometry considered in this paper. The Euler angles f, y, c specify the orientation of the rotating body xyz with respect to the fixed laboratory XYZ frame.
M.F. O’Keeffe et al. / Journal of Magnetism and Magnetic Materials 324 (2012) 2871–2878
moments of inertia of the rotator with z chosen along the symmetry axis of the rotator. Initially, these two frames coincide. At any other instant the orientation of the rotating xyz frame relative to the fixed XYZ frame is specified by the Euler angles f, y, c, which are defined using the zyz-convention [30]. The spherical polar angles represent the orientation of the symmetry axis z of the moving frame relative to the fixed frame. A vector R with components RA (we use uppercase Roman letters to indicate fixed frame components) as measured in the fixed XYZ frame can be projected onto the rotating xyz coordinate frame. In the rotating frame, this vector is described by r and has components r a (lowercase Greek indices denote components in the rotating frame). The transformation r ¼ CR is given by the rotation matrix C, 0 1
lxX Bl C ¼ @ yX lzX
lxY lyY lzY
lxZ lyZ C A, lzZ
ðA:1Þ
where the laA are direction cosines between fixed frame and rotating frame axes, and can be expressed in terms of Euler angles. The total angular momentum J obeys the usual commutation relations in the fixed XYZ frame, ½JA ,J B ¼ iEABC JC ,
ðA:2Þ
where EABC is the fully antisymmetric Levi-Civita tensor (summation over repeated indices is implicit throughout this Appendix). In the rotating xyz frame, the total angular momentum has components J a ¼ laA J A
ðA:3Þ
½Ja ,Jb ¼ iEabg Jg :
ðA:4Þ
The components of the mechanical angular momentum can be resolved in either frame, or in terms of the Euler angles [31] f, y, c which describe precession of the symmetry axis z about the fixed Z axis, rotation about the axis of symmetry z, nutation of the z axis with respect to the Z axis, and precession of the body about its axis of symmetry, respectively. The operator forms of the corresponding angular momenta are pf ¼ i_
@ , @f
py ¼ i_
@ , @y
pc ¼ i_
@ @c
LX ¼ cot y cos fpf sin fpy þcscy cos fpc , LY ¼ cot y sin fpf þ cos fpy þ cscy sin fpc , ðA:6Þ
or the body frame coordinate system, Lx ¼ cscy cos cpf þsin cpy þ cot y cos cpc , Ly ¼ cscy sin cpf þ cos cpy coty sin cpc , Lz ¼ pc :
ðA:7Þ
The commutation relations can be obtained by direct calculation, with the fixed frame components satisfying ½LA ,LB ¼ iEABC LC ,
ðA:8Þ
while the rotating frame components obey ½La ,Lb ¼ iEabg Lg :
½SA ,SB ¼ iEABC SC :
ðA:10Þ
Using the fact that ½SA ,LB ¼ 0, it is easy to see that the components of J ¼ L þ S in the fixed frame satisfy the commutation relation given by Eq. (A.2). Projecting this spin onto the rotating axes, Sa ¼ laA SA , and noticing ½SA , lbB ¼ 0, we obtain ½Sa ,Sb ¼ laA lbB ½SA ,SB ¼ iEABC laA lbB SC ¼ iEabg lgC SC ¼ iEabg Sg :
ðA:11Þ
The relation
EABC laA lbB ¼ Eabg lgC
ðA:12Þ
follows from the fact that for a special orthogonal matrix any element is equal to its cofactor. In order to obtain the same sign for all angular momenta in the rotating frame, we define reversed spin S-S~ ¼ S, giving ½S~ a , S~ b ¼ iEabg S~ g :
ðA:13Þ
~ and the components of J satisfy the Now we may write J ¼ LS, anomalous commutation relations, Eq. (A.4). Alternatively, the commutation relation for the total angular momentum in the lab frame can be calculated directly. Using the fact that the direction cosines transform according to ½laA ,J B ¼ ½laA ,LB ¼ iEABC laC ,
ðA:14Þ
and that the spin and rotational angular momentum do not commute in the rotating frame ðA:15Þ
gives ½J a ,Jb ¼ ½La ,Lb ½La , S~ b ½S~ a ,Lb þ ½S~ a , S~ b ¼ iEabg ðLg þ S~ g Þ ¼ iEabg J g ,
ðA:16Þ
where the middle two terms cancel. Similarly, we can show that ½J a , S~ b ¼ ½La , S~ b ½S~ a , S~ b ¼ 0,
ðA:17Þ
allowing us to simultaneously choose quantum numbers corre~ sponding to both J and S.
ðA:5Þ
which mutually commute. The rotational angular momentum operators can be projected onto the laboratory frame coordinate system,
LZ ¼ pf ,
The spin obeys the same regular commutation relations in either frame. To show this, we define the spin components in the laboratory frame where
½S~ a ,Lb ¼ lbB ½laA ,LB SA ¼ iEabg S~ g
and the sign of i in the commutation relation is reversed,
2877
ðA:9Þ
References [1] I.M.L. Billas, A. Chˆatelain, W.A. de Heer, Science 265 (1994) 1682. [2] X.S. Xu, S. Yin, W.A. de Heer, Physical Review Letters 95 (2005) 237209. [3] F.W. Payne, W. Jiang, J.W. Emmert, J. Deng, L.A. Bloomfield, Physical Review B 75 (2007) 094431. [4] J. Tejada, R.D. Zysler, E. Molins, E.M. Chudnovsky, Physical Review Letters 104 (2010) 027202. [5] T.M. Wallis, J. Moreland, P. Kabos, Applied Physics Letters 89 (2006) 122502. [6] J.P. Davis, D. Vick, D.C. Fortin, J.A.J. Burgess, W.K. Hiebert, M.R. Freeman, Applied Physics Letters 96 (2010) 072513. [7] H.B. Heersche, Z. de Groot, J.A. Folk, H.S.J. van der Zant, C. Romeike, M.R. Wegewijs, L. Zobbi, D. Barreca, E. Tondello, A. Cornia, Physical Review Letters 96 (2006) 206801. [8] J.J. Henderson, C.M. Ramsey, E. del Barco, A. Mishra, G. Christou, Journal of Applied Physics 101 (2007) 09E102. [9] S. Voss, M. Fonin, U. Rudiger, M. Burgert, U. Groth, Physical Review B 78 (2008) 155403. [10] A.R. Edmonds, Angular Momentum in Quantum Mechanics, Princeton University Press, Princeton, New Jersey, 1957. [11] J.H. Van Vleck, Reviews of Modern Physics 23 (1951) 213. [12] E.M. Chudnovsky, Physical Review Letters 72 (1994) 3433. [13] J.R. Friedman, V. Patel, W. Chen, S.K. Tolpygo, J.E. Lukens, Nature—London 406 (2000) 43. [14] C.H. van der Wal, A.C.J. ter Haar, F.K. Wilhelm, R.N. Schouten, C.J.P.M. Harmans, T.P. Orlando, S. Lloyd, J.E. Mooij, Science 290 (2000) 773. [15] J. Clarke, F.K. Wilhelm, Nature—London 453 (2008) 1031. [16] M.D. LaHaye, J. Suh, P.M. Echternach, K.C. Schwab, M.L. Roukes, Nature 459 (2009) 960.
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[17] A.D. O’Connell, M. Hofheinz, M. Ansmann, R.C. Bialczak, M. Lenander, E. Lucero, M. Neeley, D. Sank, H. Wang, M. Weides, J. Wenner, J.M. Martinis, A.N. Cleland, Nature 464 (2010) 697. [18] E.M. Chudnovsky, D.A. Garanin, Physical Review B 81 (2010) 214423. [19] E.M. Chudnovsky, Soviet Physics JETP 50 (1979) 1035. [JETP 50 (1979) 1035]. [20] E.M. Chudnovsky, L. Gunther, Physical Review Letters 60 (1988) 661. [21] E.M. Chudnovsky, J. Tejada, Macroscopic Quantum Tunneling of the Magnetic Moment, Cambridge University Press, Cambridge, UK, 1998. [22] R. Jaafar, E.M. Chudnovsky, Physical Review Letters 102 (2009) 227202. [23] R. Jaafar, E.M. Chudnovsky, D.A. Garanin, Europhysics Letters 89 (2009) 27001.
[24] A.A. Kovalev, L.X. Hayden, G.E.W. Bauer, Y. Tserkovnyak, Physical Review Letters 106 (2011) 147203. [25] D.A. Garanin, E.M. Chudnovsky, Physical Review X 1 (2011) 011005. ¨ Physik 58 (1929) 530. [26] O. Klein, Zeitschrift fur [27] D.A. Garanin, Journal of Physics A 24 (1991) L61. [28] E.M. Chudnovsky, J. Tejada, Lectures on Magnetism, Rinton Press, Princeton, NJ, 2006. [29] M.F. O’Keeffe, E.M. Chudnovsky, Physical Review B 83 (2011) 092402. [30] H. Goldstein, C.P. Poole, J.L. Safko, Classical Mechanics, Addison Wesley, 2001. [31] W.H. Shaffer, Journal of Molecular Spectroscopy 1 (1957) 69.
Contents lists available at SciVerse ScienceDirect
Journal of Magnetism and Magnetic Materials journal homepage: www.elsevier.com/locate/jmmm
Quantum tunneling of the magnetic moment in a free nanoparticle M.F. O’Keeffe, E.M. Chudnovsky n, D.A. Garanin Physics Department, Lehman College, City University of New York, 250 Bedford Park Boulevard West, Bronx, New York, 10468-1589, USA
a r t i c l e i n f o
abstract
Article history: Received 11 January 2012 Received in revised form 30 March 2012 Available online 7 May 2012
We study tunneling of the magnetic moment in a particle that has full rotational freedom. Exact energy levels are obtained and the ground-state magnetic moment is computed for a symmetric rotor. The effect of mechanical freedom on spin tunneling manifests itself in a strong dependence of the magnetic moment on the moments of inertia of the rotor. The energy of the particle exhibits quantum phase transitions between states with different values of the magnetic moment. Particles of various shapes are investigated and the quantum phase diagram is obtained. & 2012 Elsevier B.V. All rights reserved.
Keywords: Spin Rotator Tunneling
1. Introduction There has been much recent interest in the quantum mechanics of nanoscopic magnets that possess mechanical freedom. Experimental work in this area focused on free magnetic clusters [1–3], magnetic particles that are free to move inside solid nanocavities [4], magnetic microresonators [5,6], and magnetic molecules bridged between conducting leads [7–9]. Theoretical research on free magnetic particles has been scarce. The generic problem is that of a rigid quantum rotor with a spin. Without a spin this problem is tractable by analytical methods only for a symmetric rotor [10]. Complications resulting from spin degrees of freedom make even symmetric cases significantly more difficult [11]. The first attempt to understand how mechanical freedom of a small magnetic particle affects tunneling of the magnetic moment was made in Ref. [12], where it was noticed that tunneling of a macrospin in a free particle must be entangled with mechanical rotations in order to conserve the total angular momentum (spinþorbital). A similar situation occurs for tunneling of a superconducting current between clockwise and counterclockwise directions in a SQUID [13–17]. Recently, it was demonstrated [18] that the problem of a rigid rotor with a spin can be solved exactly in the laboratory frame when mechanical rotation is allowed only about a fixed axis and the spin states are reduced to spin-up and spin-down due to strong magnetic anisotropy. The latter is typical for magnetic molecules and small ferromagnetic clusters [19–21]. The reduction to two spin states in a system rotating about a fixed axis also allows one to obtain a simple solution of the problems of a magnetic molecule embedded in a
n
Corresponding author. E-mail address: [email protected] (E.M. Chudnovsky).
0304-8853/$ - see front matter & 2012 Elsevier B.V. All rights reserved. http://dx.doi.org/10.1016/j.jmmm.2012.04.032
microcantilever [22], a magnetic molecule vibrating between conducting leads [23], and a macrospin tunneling inside a torsional resonator [24,25]. However, the problem for arbitrary rotations of a two-state spin system, which is relevant to free magnetic nanoparticles, remained unsolved until now. In this paper we show that the problem of a two-state macrospin inside a symmetric rigid rotor has a rigorous solution for arbitrary rotations in the coordinate frame that is rigidly coupled to the rotor. The magnetic moment of electrically neutral rotor is entirely due to spin. It depends on the relative contribution of the up and down spin states. When a nanoparticle is embedded in a solid, tunneling of the spin results in a zero ground-state magnetic moment. This situation changes for a free particle due to a complex interplay between spin and mechanical angular momentum that conserves the total angular momentum. We show that the energy of the particle exhibits first- or secondorder quantum phase transition between states with different values of the total angular momentum. The order of the transition depends on the shape of the particle. The ground-state magnetic moment of a free particle with a total spin S can be anything between zero and g mB S, depending on the principal moments of inertia (with mB being the Bohr magneton and g being the gyromagnetic factor associated with the spin). The structure of the article is as follows. Quantum theory of a rigid rotator is briefly reviewed in Section 2. Theory of a tunneling macrospin is reviewed in Section 3. Quantum states of a rigid rotator containing a tunneling macrospin are constructed in Section 4. Ground state of a symmetric rotor with a spin is analyzed in Section 5. Ground-state magnetic moment is studied in Section 6. Our conclusions are presented in Section 7. Mathematics of the problem studied in this article is based upon anomalous commutation relations of the operator of angular momentum in the coordinate frame that is rigidly coupled with
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the rotating particle. To simplify the reading and to make the paper self-contained, some old results that are difficult to find in the literature are derived in the Appendix.
antisymmetric combinations of 9c 7 S S, respectively [21], 1
C þ ¼ pffiffiffi ð9cS S þ9cS SÞ, 2
1
C ¼ pffiffiffi ð9cS S9cS SÞ,
2. Quantization of rigid body rotations
2
Consider first the problem of a rigid body rotating without a spin. We choose the coordinate frame that is rigidly coupled with the rotating body and direct the axes of that frame x, y, and z, along the principle axes of the tensor of moments of inertia of the body. In such coordinate frame the Hamiltonian of mechanical rotations is given by [10] ! 2 2 2 L2 ^ R ¼ _ Lx þ y þ Lz : H ð1Þ 2 Ix Iy Iz Here Ix , Iy , Iz are the principal moments of inertia and Lx , Ly , Lz are projections of the operator of the mechanical angular momentum (defined in the laboratory frame) onto the body axes x, y, z. Such a choice of coordinates and operators results in the anomalous commutation relations [26], ½Li ,Lj ¼ iEijk Lk (notice the minus sign in the right-hand side), but does not affect the relations L2 ¼ LðLþ 1Þ, ½L2 ,Lz ¼ 0. For a symmetric rotor two of the moments of inertia are the same, Ix ¼Iy, and the Hamiltonian can be written as 2 2 2 2 ^ R ¼ _ L þ _ Lz 1 1 : H ð2Þ Iz Ix 2Ix 2 The corresponding eigenstates are characterized by three quantum numbers L, K, and M, 2
L 9LKMS ¼ LðLþ 1Þ9LKMS, Lz 9LKMS ¼ K9LKMS, LZ 9JKMS ¼ M9LKMS,
L ¼ 0; 1,2, . . . ,
which satisfy ^ SC 7 ¼ E 8 C 7 , H
ð7Þ
where E þ E D:
ð8Þ
The tunnel splitting D is generally many orders of magnitude smaller than the distance to other spin energy levels, which makes the two-state approximation very accurate at low energies. For example, ^ S ¼ DS2 þ dS2 H z y
ð9Þ
with d 5D describes the biaxial anisotropy of spin-10 molecular nanomagnet Fe-8, where the tunnel splitting in the limit of large S is given by [27] S 8S3=2 d D ¼ 1=2 D: ð10Þ 4D p The distance to the next excited spin level is ð2S1ÞD, which is large compared to D. It is convenient to describe these lowest energy spin states C 7 with a pseudospin-1/2. Components of the corresponding Pauli operator r are
sx ¼ 9cS S/cS 9þ 9cS S/cS 9, sy ¼ i9cS S/cS 9i9cS S/cS 9,
K ¼ L,L þ1, . . . ,L1,L,
sz ¼ 9cS S/cS 99cS S/cS 9:
ð11Þ
M ¼ L,Lþ 1 . . . ,L1,L,
^ S onto 9c S states is The projection of H 7S X ^ ^ Hs ¼ /m9H S 9nS9mS/n9:
ð12Þ
ð3Þ
where LZ is the angular momentum operator defined with respect to the laboratory coordinate frame (X,Y,Z). The eigenvalues of (2) are degenerate on K: _2 LðLþ 1Þ _2 K 2 1 1 þ ELK ¼ : ð4Þ 2Ix Iz I x 2 The general form for the energy levels of a rotating asymmetric rigid body, Ix a Iy aIz , does not exist, although it is possible to calculate matrix elements of the Hamiltonian for a given L.
3. Tunneling of a large spin Let S be a fixed-length spin embedded in a stationary body. Naturally, the magnetic anisotropy is defined with respect to the body axes. The general form of the crystal field Hamiltonian is ^ J þH ^ ?, ^S ¼H H
ð6Þ
ð5Þ
^ ? is a perturbation that does ^ J commutes with Sz and H where H not commute with Sz. The states 9 7SS are degenerate ground ^ J , where S is the total spin of the nanomagnet. H ^? states of H slightly perturbs these states, adding to them small contributions from other 9mS S states. We will call these degenerate perturbed states 9c 7 S S. Physically they describe the magnetic moment aligned in one of the two directions along the anisotropy axis. Full ^ S properturbation theory with account of the degeneracy of H vides quantum tunneling between the 9c 7 S S states for integer S. The ground state and first excited state are symmetric and
m,n ¼ c 7 S
Expressing 9c 7 S S in terms of C 7 one obtains ^ S 9c S ¼ 0, /c 7 S 9H 7S
^ S 9c S ¼ D , /c 7 S 9H 8S 2
ð13Þ
which gives the two-state Hamiltonian ^ s ¼ D sx H 2
ð14Þ
having eigenvalues 7 D=2. In the absence of tunneling a classical magnetic moment is localized in the up or down state. It is clear that delocalization of the magnetic moment due to spin tunneling reduces the energy by D=2. In a free particle, however, tunneling of the spin must be accompanied by mechanical rotations in order to conserve the total angular momentum. Such rotations cost energy, so it is not a priori clear whether the tunneling will survive in a free particle and what the ground state is going to be. This problem is addressed in the following Section.
4. Rigid rotor containing tunneling macrospin Consider now a tunneling macrospin embedded in a free particle having the body z-axis as the magnetic anisotropy direction. When the magnetic anisotropy is dominated by magnetostatic interactions, this choice is justified by the fact that the principal axes of the demagnetizing tensor would always coincide with the principal axes of the rotator [28]. For a symmetric
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rotator, macroscopic magnetostatic interactions alone do not provide the transverse term in the Hamiltonian that generates tunneling of the spin. In this case one has to assume that the tunneling is provided by the transverse component of the crystal field due to microscopic structure of the particle, see Eqs. (5) and (9). The free particle is characterized by the total angular momentum, J ¼ L þS. In the body frame this operator may appear unconventional due to the different sign of commutation relations for L and S (see Appendix). However, this problem can be easily fixed [11] by the transformation S-S~ ¼ S that changes the sign of the commutation relation for S. Such a transformation does not change the results of the previous section because the crystal field Hamiltonian contains only even powers of S, so that ^S ¼H ^ ~ . It is interesting to notice that while in the laboratory H S frame ½Ji ,Sj ¼ iEijk Sk , components of the operators J and S~ defined in the body frame commute with each other [11]. In addition, the operator J2 is the same in the body and laboratory frames [26]. This permits description of quantum states of the particle in terms of quantum numbers associated independently with the total angular momentum and spin. The full Hamiltonian is given by the sum of the rotational energy and magnetic anisotropy energy
easy-axis magnetic anisotropy and the tunnel splitting can decrease or increase, depending on the ratio Ix =Iz . Projection of Eq. (17) on the two spin states along the lines of the previous section gives 2 2 2 2 2 ^ ¼ _ J þ _ J z 1 1 D sx _ S J sz , H ð20Þ Iz Ix Iz z 2Ix 2 2
2 2 _2 L2y _2 L2z ^ ¼ _ Lx þ ^ ~: H þ þH S 2Ix 2Iy 2Iz
where EJK is provided by Eq. (4) with L replaced by J. The upper (lower) sign in Eq. (24) corresponds to the lower (upper) sign in Eq. (22). For K a 0 each state is degenerate with respect to the sign of K. For K ¼ 0; 1,2, . . . the coefficients in Eq. (22) are given by rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi C 7 ¼ 17 aK= S2 þ ðaKÞ2 , ð25Þ
ð15Þ
Note that the mechanical part and the spin part of this Hamiltonian are not independent because in the body frame the operators L and S~ do not commute, ½Li , S~ j ¼ iEijk S~ k . It, therefore, makes sense to express the above Hamiltonian in terms of the commuting body-frame operators of the total angular momentum J and ~ reversed spin S: 0 1 ! 2 2 2 J2y J2z S~ y S~ z _2 J 2x _2 @S~ x ^ H¼ þ þ þ þ A þ 2 Ix I y Iz 2 Ix Iy Iz ! J S~ x J y S~ y J z S~ z ^ ~: þ _2 x þ þ ð16Þ þH S Ix Iy Iz For a symmetric rigid rotor with Ix ¼Iy this Hamiltonian reduces to 2 2 2 2 ^ ¼ _ J þ _ Jz 1 1 H I z Ix 2Ix 2 ! ~ ~ J x S x þJ y S y J z S~ z ^0, þ ð17Þ þH þ _2 S~ Ix Iz where 2
^~þ_ ^0 ¼H H S S~ 2
2 1 1 ~ 2 _2 S~ : Sz þ Iz Ix 2Ix
ð18Þ
0
^ ~ is an unessential constant, _2 SðS þ 1Þ=ð2Ix Þ. The The last term in H S second term provides renormalization of the crystal field in a freely rotating particle. For, e.g., the biaxial spin Hamiltonian given by Eq. (9) it leads to _2 1 1 D-D : ð19Þ 2 Iz Ix This, in turn, renormalizes the tunnel splitting given by Eq. (10). For a particle that is allowed to rotate about the Z-axis only (that is, in the limit of Ix -1) these results coincide with the results obtained by the instanton method in Ref. [29], where it was shown that, in practice, the renormalization of the magnetic anisotropy and spin tunnel splitting by mechanical rotations is small. Eq. (19) provides a generalization of this effect for arbitrary rotations of a symmetric rotator with a spin. According to this equation and Eq. (9), when rotations are allowed the effective
where we have used /c 7 S 9Sz 9c 7 S S ¼ 7 S,
/c 7 S 9Sx,y 9c 7 S S ¼ 0:
ð21Þ
We construct eigenstates of this Hamiltonian according to 1 9CJK S ¼ pffiffiffi ðC 7 S 9cS S7 C 8 S 9cS SÞ9JKS, 2
ð22Þ
where J2 9JKS ¼ JðJ þ1Þ9JKS, Jz 9JKS ¼ K9JKS,
J ¼ 0; 1,2, . . . ,
K ¼ J, . . . ,J:
^ CJK S ¼ E9CJK S gives energy levels as Solution of H9 vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi !2ffi u 2 u D _2 KS ð7Þ t þ EJK ¼ EJK 7 , Iz 2
ð23Þ
ð24Þ
where a is a dimensionless magneto-mechanical ratio,
a¼
2ð_SÞ2 : Iz D
ð26Þ
The energy levels in Eq. (24) can be given a simple semiclassical interpretation. Indeed, the last term in this equation is the tunnel splitting of the levels in the effective magnetic field that appears in the body reference frame due to rotation about the spin quantization axis at the angular velocity _K=Iz . When S¼0 (which also means D ¼ 0) Eq. (24) with J¼ L gives the energy of the quantum symmetric rigid rotor without a spin, Eq. (4). In the case of a heavy body (large moments of inertia) the ground state and the first excited state correspond to J¼K ¼0, and we recover the tunnel-split spin states in a non-rotating macroscopic body, E00 7 ¼ 7 D=2. In the general case, spin states of the rotator are entangled with mechanical rotations. Eqs. (22)–(25) are our main analytical results for the lowenergy states of a free magnetic particle. In general, numerical analysis is needed to find the ground state of the particle. Special cases of the aspect ratio that will be analyzed below include a needle of vanishing diameter (which is equivalent to the problem of the rotation about a fixed axis treated previously in the laboratory frame by two of the authors [18]), a finite-diameter needle, a sphere, and a disk.
5. Ground state Minimization of the energy in Eq. (24) on J with the account of the fact that J cannot be smaller than K immediately yields J¼ K, that is, the ground state always corresponds to the maximal projection of the total angular momentum onto the spin quantization axis. In semiclassical terms this means that the minimal energy states in the presence of spin tunneling always correspond to mechanical rotations about the magnetic anisotropy axis. This is easy to understand by noticing that the sole reason for mechanical rotation is the necessity
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to conserve the total angular momentum while allowing spin tunneling to lower the energy. To accomplish this the particle needs to oscillate between clockwise and counterclockwise rotations about the spin quantization axis in unison with the tunneling spin. If such mechanical oscillation costs more energy than the energy gain from spin tunneling, then both spin tunneling and mechanical motion must be frozen in the ground state as, indeed, happens in very light particles (see below). Rotations about axes other than the spin quantization axis can only increase the energy and, thus, should be absent in the ground state. For further analysis it is convenient to write Eq. (24) in the dimensionless form, sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi " # EðJK7 Þ a JðJ þ 1ÞK 2 K2 1 K2 ¼ lþ 2 7 ð27Þ 1 þ 2 a2 , 2 2 D 4 S S S in terms of dimensionless parameters a and the aspect ratio for the moments of inertia
l ¼ Iz =Ix :
ð28Þ
The range of l for a symmetric rotator is 0 r l r2. For, e.g., a symmetric ellipsoid with semiaxes a ¼ b ac, one has l ¼ 2a2 = ða2 þ c2 Þ. The dependence of the energy levels (24) on J at K ¼J is shown in Figs. 1 and 2. It exhibits a quantum phase transition on the parameter a between states with different values of J. Only for a needle of infinite length, Iz =Ix -0, which corresponds to c-1 in the case of an ellipsoid, is the transition second order, see Fig. 1. It occurs at a ¼ ½11=ð2SÞ2 1 . This case is equivalent to the rotation about a fixed axis studied in Ref. [18]. For any finite ratio Iz =Ix the transition is first order if J is treated as a continuous variable, see Fig. 2. It occurs at the value of a that depends on Iz =Ix . The origin of the transfer from a second-order transition at l ¼ 0 to the firstorder transition at l a 0 can be traced to the term ½JðJ þ 1ÞK 2 S2 l in Eq. (27). We should notice that for a finite-size nanomagnet the analogy with first- and second-order phase transition is, of course, just an analogy. To talk about real phase transitions one has to take the limit of S-1, Ix,z -1 when the distances between quantum levels go to zero. For a given l, as a increases the ground-state switches from J¼0 to higher J when ðÞ 0 0 EðÞ 00 ðaJ ðlÞÞ ¼ EJJ ðaJ ðlÞÞ:
ð29Þ
Fig. 2. Dependence of energy on J at K¼ J and Iz =Ix ¼ 2 for different values of a. The plot shows first-order quantum phase transition on a.
Fig. 3. Dependence of the ground-state energy on a for a spherical particle. Inset shows the discontinuity of the derivative of the ground-state energy on a.
Solution of this equation for a0J ðlÞ gives
a0J ¼
ð2SÞ2 ðJ þ lÞ J½ð2SÞ2 ðJ þ lÞ2
:
ð30Þ
This first transition occurs for the smallest value of a0J ðlÞ and the transition is from J¼0 to the corresponding critical value, Jc (Fig. 3). For a o a0Jc the ground state corresponds to J¼0 and C 7 S ¼ 1. After the first transition from J¼0 to J ¼ Jc , the ground state switches to sequentially higher J at values of a which satisfy ðÞ EðÞ J1J1 ðaJ ðlÞÞ ¼ EJJ ðaJ ðlÞÞ:
ð31Þ
Solution of this equation for aJ ðlÞ gives ð2SÞ2 TðJ, lÞ
aJ ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiqffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi , 2
2
ð2SÞ ð2J1Þ TðJ, lÞ2
ð32Þ
ð2SÞ2 TðJ, lÞ2
with TðJ, lÞ ¼ 2J1þ l:
Fig. 1. Dependence of energy on J at K¼J and Iz =Ix ¼ 0 for different values of a. The plot shows second-order quantum phase transition on a.
ð33Þ
The critical aJ has poles at l ¼ 2ðSJÞ þ 1. For l Z 1 there is no longer a ground-state transition to J¼S, even for very large values of a. Needle of infinite length: The case of a particle that can only rotate about its anisotropy axis [18] is equivalent in our model to
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a needle of infinite length (c-1 for an ellipsoid), having l ¼ 0. It is also equivalent to the problem of tunneling of the angular momentum of a superconducting current in a flux qubit coupled to a torsional resonator. In this limit we reproduce results of Ref. [18]. The quantum number K determines the ground state, as the energy, Eq. (27), no longer formally depends on J. However, the values of aJ at l ¼ 0, for which ground-state transitions occur, are ðÞ the same as those for which EðÞ JK1 ¼ EJK , and we will use J to describe the ground state of the axial rotor as well. The first ground-state transition occurs from J ¼0 to J¼1 at aðlÞ ¼ a1 ð0Þ ¼ a01 ð0Þ, because Jc ¼1 for l t 0:01. At a ¼ a2 ð0Þ the ground state switches from J ¼1 to J¼2, and so on. The final transition is to a completely localized spin state J¼S in which spin tunneling is frozen for all a 4 aS ð0Þ. For example, when S ¼10, a1 ð0Þ ¼ a01 ð0Þ ¼ 1:0025 and a10 ¼ 3:2066. Needle of finite diameter: The ground state of a needle of finite diameter (a 5 c for an ellipsoid) with l ¼ 0:1, that is free to rotate about any axis, shows qualitatively different behavior. As a increases, the ground-state changes from J¼0 to Jc ¼3 at a ¼ a03 ð0:1Þ, as the smallest value of a0J ð0:1Þ for 1 r J rS occurs for Jc ¼3. The J¼1, 2 states never become the ground state. After this, transitions occur to successively higher J, beginning with J¼4 at a ¼ a4 ð0:1Þ, and eventually localizing the spin with J¼S for a 4 aS ð0:1Þ. For S¼10, a03 ð0:1Þ ¼ 1:0588 and a10 ð0:1Þ ¼ 3:3935. Sphere: As l increases towards unity, the particle becomes more symmetric with the moment of inertia having (prolate) ellipsoidal symmetry, until it reaches spherical symmetry at l ¼ 1. The first ground-state transition occurs from J¼0 to J ¼ Jc ¼ 5 at a ¼ a05 ð1Þ, and subsequent transitions occur at a ¼ aJ ð1Þ. However, the spin never localizes in the J ¼S state even for very large a, as aJ ð1Þ has a pole at J¼S, so the last transition occurs to the J ¼ S1 state at a ¼ aS1 ð1Þ. For S ¼10, a05 ð1Þ ¼ 1:3187 and a9 ð1Þ ¼ 2:4325. Disk: With l increasing from unity, the symmetry of the body becomes that of an oblate ellipsoid, and begins to flatten in the plane perpendicular to the anisotropy axis. It is easy to check from Eq. (27) that for 1 o l r2 the state with J ¼S always has higher energy than the state with J ¼ S1, even in the limit of a-1. This means that for an oblate particle some spin tunneling (accompanied by mechanical rotations) survives in the ground state no matter how light the particle is. This purely quantummechanical result has no semi-classical analogy. In the case of a disk of vanishing thickness, l ¼ 2, the first ground-state transition occurs from J¼0 to J ¼ Jc ¼ 6 at a ¼ a06 ð2Þ, and subsequent transitions occur at a ¼ aJ ð2Þ up through J ¼ S1. For S ¼10, a06 ð2Þ ¼ 1:5873 and a9 ð2Þ ¼ 3:5849.
6. Ground-state magnetic moment As has been already mentioned, the magnetic moment is due entirely to the spin of the particle, as Lz represents mechanical motion of the particle as a whole, and not electronic orbital angular momentum. Thus,
aK
m ¼ g mB /CJK 9Sz 9CJK S ¼ g mB S qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi : S2 þ ðaKÞ2
ð34Þ
Here g is the spin gyromagnetic factor, and the minus sign reflects the negative gyromagnetic ratio g ¼ g mB =_. The ground state always corresponds to J¼K, so these are used interchangeably in descriptions of the ground state. The dependence of the magnetic moment on a for different aspect ratios of the particle is shown in Fig. 4. For a o aJc ðlÞ the ground state corresponds to J¼K¼0, so the spin-up and spin-down states are in an equal superposition which produces zero magnetic moment. At greater values of a the spin states contribute in unequal
Fig. 4. Ground-state magnetic moment for a needle of vanishing diameter ðl ¼ 0Þ, finite-diameter needle ðl ¼ 0:1Þ, sphere ðl ¼ 1Þ, and a disk of vanishing thickness ðl ¼ 2Þ.
amounts which leads to a non-zero magnetic moment. As a becomes large, the magnetic moment approaches its maximal value 9mmax 9 ¼ g mB S. Note that the magnetic moment approaches its maximum value even for values of l that do not admit transitions to J¼S states. Because the ground state is completely determined by the parameters a and l, we can depict the ground-state behavior in a quantum phase diagram shown in Fig. 5. The curves separate areas in the (a, l) plane that correspond to different values of J and different values of the magnetic moment. Notice the fine structure of the diagram (lower picture in Fig. 5) near the first critical a. This very rich behavior of the ground state on parameters must have significant implications for magnetism of rigid atomic clusters.
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They may also apply to free magnetic molecules if one can justify the condition of rigidity. Direct comparison between theory and experiment may be possible for atomic clusters (molecules) in magnetic traps. To see that the quantum problem studied in this paper may, indeed, be relevant to the quantum states of free nanomagnets, consider, e.g., a spherical atomic cluster of radius R and average mass density r having spin S¼10 that, when embedded in a large body, can tunnel between up and down at a frequency of a few GHz, thus providing D 0:1 K. Significant changes in the magnetic moment of such a cluster would occur at a 1, which, according to Eq. (26), corresponds to I ¼ 8prR5 =15 1042 kg m2 and R 1 nm. For a magnetic molecule like, e.g., Mn12, the moments of inertia would also be in the ballpark of 1042 kg m2 . However, the natural spin tunnel splitting in Mn12 is very small, thus, providing a very large a. The same is true for Fe8 magnetic molecules. In this case the spin tunneling in a free molecule must be completely frozen. Even if the molecule cannot be considered as entirely rigid, such effect, if observed, would receive natural interpretation within the framework of our theory.
Acknowledgements This work has been supported by the U.S. Department of Energy through Grant no. DE-FG02-93ER45487.
Appendix A. Coupling of spin and rotational angular momenta
Fig. 5. Quantum phase diagram for the ground-state magnetic moment and the total angular momentum.
7. Conclusions We have studied the problem of a quantum rotator containing a tunneling spin. This problem is relevant to quantum mechanics of free magnetic nanoparticles. It also provides an interesting insight into quantum mechanics of molecules studied from the macroscopic end. The answer obtained for the energy levels of a symmetric rotator, Eq. (24), is non-perturbative and highly nontrivial. It is difficult to imagine how it could be obtained from first principles without the reduction to two spin states. Indeed, for spin S the tunnel splitting itself generally appears in the S-th order of perturbation theory, see Eq. (10), so the path from the full crystal-field Hamiltonian like, e.g., Eq. (9) to Eq. (24) must be very long. Eqs. (22) and (24) represent, therefore, a unique exact solution of the quantum-mechanical problem of a mechanical rotator with a spin. A striking feature of this solution is the presence of first- and second-order quantum phase transitions between states with different values of the magnetic moment. Our results provide the framework for comparison between theory and experiment on very small free magnetic clusters. Our main conclusion for experiment is that rotational states and magnetic moments of such clusters depend crucially and in a predictable way on size and aspect ratio. This dependence results in a complex phase diagram that separates regions in the parameter space, corresponding to different values of the magnetic moment. A broad distribution of the magnetic moments that does not simply scale with the volume, has, in fact, been reported in beams of free atomic clusters of ferromagnetic materials [1–3]. Our results may shed some additional light on these experiments.
In this Appendix, we derive the so-called anomalous commutation relations of angular momentum in a rotating frame. This treatment applies to the general case of mechanical rotations of a quantum system with internal angular momentum degrees of freedom (see Ref. [11] and references therein). In our case this internal degree of freedom is the spin of the particle, S. Starting with the usual commutation relations for the components of S, components of the angular momentum of mechanical rotations L, and total angular momentum J ¼ S þL in the fixed frame, we derive commutation relations in the rotating (body) frame. We show that in the body frame all components of J and S commute, and therefore the corresponding quantum numbers provide good description of the quantum states of the particle. The nanomagnet is a rigid rotator around its center of mass, which we define as the common origin O of two right-handed coordinate systems, shown in Fig. A1. The X, Y, Z axes make up the lab frame that is fixed in space. The x, y, z axes define the coordinate system that is rigidly coupled to the body, and are directed along the principle
Z z
y
O
X
N
Y
x
Fig. A1. Geometry considered in this paper. The Euler angles f, y, c specify the orientation of the rotating body xyz with respect to the fixed laboratory XYZ frame.
M.F. O’Keeffe et al. / Journal of Magnetism and Magnetic Materials 324 (2012) 2871–2878
moments of inertia of the rotator with z chosen along the symmetry axis of the rotator. Initially, these two frames coincide. At any other instant the orientation of the rotating xyz frame relative to the fixed XYZ frame is specified by the Euler angles f, y, c, which are defined using the zyz-convention [30]. The spherical polar angles represent the orientation of the symmetry axis z of the moving frame relative to the fixed frame. A vector R with components RA (we use uppercase Roman letters to indicate fixed frame components) as measured in the fixed XYZ frame can be projected onto the rotating xyz coordinate frame. In the rotating frame, this vector is described by r and has components r a (lowercase Greek indices denote components in the rotating frame). The transformation r ¼ CR is given by the rotation matrix C, 0 1
lxX Bl C ¼ @ yX lzX
lxY lyY lzY
lxZ lyZ C A, lzZ
ðA:1Þ
where the laA are direction cosines between fixed frame and rotating frame axes, and can be expressed in terms of Euler angles. The total angular momentum J obeys the usual commutation relations in the fixed XYZ frame, ½JA ,J B ¼ iEABC JC ,
ðA:2Þ
where EABC is the fully antisymmetric Levi-Civita tensor (summation over repeated indices is implicit throughout this Appendix). In the rotating xyz frame, the total angular momentum has components J a ¼ laA J A
ðA:3Þ
½Ja ,Jb ¼ iEabg Jg :
ðA:4Þ
The components of the mechanical angular momentum can be resolved in either frame, or in terms of the Euler angles [31] f, y, c which describe precession of the symmetry axis z about the fixed Z axis, rotation about the axis of symmetry z, nutation of the z axis with respect to the Z axis, and precession of the body about its axis of symmetry, respectively. The operator forms of the corresponding angular momenta are pf ¼ i_
@ , @f
py ¼ i_
@ , @y
pc ¼ i_
@ @c
LX ¼ cot y cos fpf sin fpy þcscy cos fpc , LY ¼ cot y sin fpf þ cos fpy þ cscy sin fpc , ðA:6Þ
or the body frame coordinate system, Lx ¼ cscy cos cpf þsin cpy þ cot y cos cpc , Ly ¼ cscy sin cpf þ cos cpy coty sin cpc , Lz ¼ pc :
ðA:7Þ
The commutation relations can be obtained by direct calculation, with the fixed frame components satisfying ½LA ,LB ¼ iEABC LC ,
ðA:8Þ
while the rotating frame components obey ½La ,Lb ¼ iEabg Lg :
½SA ,SB ¼ iEABC SC :
ðA:10Þ
Using the fact that ½SA ,LB ¼ 0, it is easy to see that the components of J ¼ L þ S in the fixed frame satisfy the commutation relation given by Eq. (A.2). Projecting this spin onto the rotating axes, Sa ¼ laA SA , and noticing ½SA , lbB ¼ 0, we obtain ½Sa ,Sb ¼ laA lbB ½SA ,SB ¼ iEABC laA lbB SC ¼ iEabg lgC SC ¼ iEabg Sg :
ðA:11Þ
The relation
EABC laA lbB ¼ Eabg lgC
ðA:12Þ
follows from the fact that for a special orthogonal matrix any element is equal to its cofactor. In order to obtain the same sign for all angular momenta in the rotating frame, we define reversed spin S-S~ ¼ S, giving ½S~ a , S~ b ¼ iEabg S~ g :
ðA:13Þ
~ and the components of J satisfy the Now we may write J ¼ LS, anomalous commutation relations, Eq. (A.4). Alternatively, the commutation relation for the total angular momentum in the lab frame can be calculated directly. Using the fact that the direction cosines transform according to ½laA ,J B ¼ ½laA ,LB ¼ iEABC laC ,
ðA:14Þ
and that the spin and rotational angular momentum do not commute in the rotating frame ðA:15Þ
gives ½J a ,Jb ¼ ½La ,Lb ½La , S~ b ½S~ a ,Lb þ ½S~ a , S~ b ¼ iEabg ðLg þ S~ g Þ ¼ iEabg J g ,
ðA:16Þ
where the middle two terms cancel. Similarly, we can show that ½J a , S~ b ¼ ½La , S~ b ½S~ a , S~ b ¼ 0,
ðA:17Þ
allowing us to simultaneously choose quantum numbers corre~ sponding to both J and S.
ðA:5Þ
which mutually commute. The rotational angular momentum operators can be projected onto the laboratory frame coordinate system,
LZ ¼ pf ,
The spin obeys the same regular commutation relations in either frame. To show this, we define the spin components in the laboratory frame where
½S~ a ,Lb ¼ lbB ½laA ,LB SA ¼ iEabg S~ g
and the sign of i in the commutation relation is reversed,
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ðA:9Þ
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