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leastaction
leastaction
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Consider the closed bosonic string in 26 dimensions, of which one physical dimension, say $X^{24}$ is compactified into a circle of radius $R$. The lowest lying massless states are the graviton, the antisymmetric $B_{\mu\nu}$ field, the dilaton $\phi$, and two ``vectors'', the so called graviphoton and B-vector. They are described by

$\alpha^{i}_{-1} \tilde{\alpha}^{24}_{-1}|p\rangle = 0$ and $\alpha^{24}_{-1}\tilde{\alpha}^{i}_{-1}|p\rangle$, with $p^2 = 0$.

What is the difference between these two states? I see that they are defined differently, but the only difference is in which (antiholomorphic) creation operator appears first.

Is it correct to say that the symmetrized version is the graviphoton and the antisymmetrized version is the B vector field?

Consider the closed bosonic string in 26 dimensions, of which one physical dimension, say $X^{24}$ is compactified into a circle of radius $R$. The lowest lying massless states are the graviton, the antisymmetric $B_{\mu\nu}$ field, the dilaton $\phi$, and two ``vectors'', the so called graviphoton and B-vector. They are described by

$\alpha^{i}_{-1} \tilde{\alpha}^{24}_{-1}|p\rangle = 0$ and $\alpha^{24}_{-1}\tilde{\alpha}^{i}_{-1}|p\rangle$, with $p^2 = 0$.

What is the difference between these two states? I see that they are defined differently, but the only difference is in which (antiholomorphic) creation operator appears first.

Consider the closed bosonic string in 26 dimensions, of which one physical dimension, say $X^{24}$ is compactified into a circle of radius $R$. The lowest lying massless states are the graviton, the antisymmetric $B_{\mu\nu}$ field, the dilaton $\phi$, and two ``vectors'', the so called graviphoton and B-vector. They are described by

$\alpha^{i}_{-1} \tilde{\alpha}^{24}_{-1}|p\rangle = 0$ and $\alpha^{24}_{-1}\tilde{\alpha}^{i}_{-1}|p\rangle$, with $p^2 = 0$.

What is the difference between these two states? I see that they are defined differently, but the only difference is in which (antiholomorphic) creation operator appears first.

Is it correct to say that the symmetrized version is the graviphoton and the antisymmetrized version is the B vector field?

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leastaction
leastaction
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T-Duality for closed bosonic string: what is the difference between a graviphoton and a B-vector?

Consider the closed bosonic string in 26 dimensions, of which one physical dimension, say $X^{24}$ is compactified into a circle of radius $R$. The lowest lying massless states are the graviton, the antisymmetric $B_{\mu\nu}$ field, the dilaton $\phi$, and two ``vectors'', the so called graviphoton and B-vector. They are described by

$\alpha^{i}_{-1} \tilde{\alpha}^{24}_{-1}|p\rangle = 0$ and $\alpha^{24}_{-1}\tilde{\alpha}^{i}_{-1}|p\rangle$, with $p^2 = 0$.

What is the difference between these two states? I see that they are defined differently, but the only difference is in which (antiholomorphic) creation operator appears first.