Corrigendum to “Spinning strings in AdS3×S3 with NS–NS flux” [Nucl. Phys. B 888 (2014) 236–247]

Corrigendum to “Spinning strings in AdS3×S3 with NS–NS flux” [Nucl. Phys. B 888 (2014) 236–247]

Available online at www.sciencedirect.com ScienceDirect Nuclear Physics B 895 (2015) 303–304 www.elsevier.com/locate/nuclphysb Corrigendum Corrigen...

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Available online at www.sciencedirect.com

ScienceDirect Nuclear Physics B 895 (2015) 303–304 www.elsevier.com/locate/nuclphysb

Corrigendum

Corrigendum to “Spinning strings in AdS3 × S 3 with NS–NS flux” [Nucl. Phys. B 888 (2014) 236–247] Rafael Hernández ∗ , Juan Miguel Nieto Departamento de Física Teórica I, Universidad Complutense de Madrid, 28040 Madrid, Spain Received 8 April 2015; accepted 11 April 2015 Available online 20 April 2015 Editor: Stephan Stieberger

In this corrigendum we want to address three typos in our article. The first typo is a missing factor of 2 in Eq. (2.33), which should read √ √   λ λ J 2 ω1 = √ + ¯1 +m ¯ 2 )(1 − q ) 1 − 2 m ¯ 1 (m qm ¯ 2 + ··· . 2J J λ The second typo is a sign in the definition of the energy. Eq. (3.11) should read 2π √  dσ 2 E= λ (z κ − qz12 β1 ) . 2π 0 0

This change implies a correction in Eq. (3.19), which should read E=



λκ + 

¯ S(κ − q k) . ¯ κ 2 + k¯ 2 − 2q kκ

This modifies the final expression of the energy as a function of the angular momenta and the spin to be  ¯  √ kS λ E = J + S − λq m ¯1 +2 ¯ 21 J1 + m ¯ 22 J2 + k¯ 2 S)(1 − q 2 ) + 2 (m J 2J λ ¯ + ··· ¯ m − 3 2q 2 kS( ¯ 1 J + kS) J DOI of original article: http://dx.doi.org/10.1016/j.nuclphysb.2014.10.001. * Corresponding author.

E-mail addresses: [email protected] (R. Hernández), [email protected] (J.M. Nieto). http://dx.doi.org/10.1016/j.nuclphysb.2015.04.011 0550-3213/© 2015 The Authors. Published by Elsevier B.V.

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R. Hernández, J.M. Nieto / Nuclear Physics B 895 (2015) 303–304

The third typo appears in Eq. (3.22), which should be replaced by   √ ¯ (ω1 + q m ¯ 2 )2 − (m ¯ 21 − m ¯ 22 )(1 − q 2 ) λ(m ¯1 −m ¯ 2 )ω12 + 2 λ(m ¯ 2 J + kS)ω 1   ¯ J − (m ¯ 2 (1 − q 2 ) = 0 . − (m ¯1 +m ¯ 2 )J + 2kS ¯1 +m ¯ 2 )(m ¯ 1 J + kS)