Following light cone quantisation of the bosonic string, in the last post we briefly reviewed the free open string spectrum. Now let’s take a look at the closed string spectrum.

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For the closed string spectrum, let’s review from the start. The method of light cone quantisation is essentially the same. We impose the same gauge conditions for the closed-string as we did for the open string (p.25). The one nuance here, as Polchinski highlights, is that in the open string case the gauge was determined completely by the conditions imposed. This is not true for the closed string, as there is extra coordinate freedom. We may state this as,

    \[ \sigma^{\prime} = \sigma + s(\tau)mod \ l \]

Further review should be sought on p.25. Moving forward, one of the main differences between the open string case and the closed string case to be highlighted here concerns how we’re now working with the periodicity condition: \sigma \sim \sigma + 2\pi.

As the closed string forms, in a sense, a closed loop, we must also take into account our left and right-movers. Consider, for instance,

    \[X^{\mu}(\tau, \sigma) = X_{L}^{\mu}(\tau, \sigma) + X_{R}^{\mu}(\tau - \sigma) \]

Where u = \tau + \sigma and v = \tau - \sigma. We may also use the following shorthand notation,

    \[ X^{\mu}(\tau, \sigma) = X_{L}^{\mu}(u) + X_{R}^{\nu}(v) \]

With the periodicity condition noted, the point remains to be stated explicitly that the left and right-movers are equivalent, which is to also say

    \[X^{\mu}(\tau, \sigma + 2\pi) = X^{\mu}(\tau, \sigma) \]

This makes complete sense if you consider,

    \[X^{\mu}(\tau, \sigma + 2\pi) = X_{L}(u + 2\pi) + X_{R}(v - 2\pi) = X_{L}(u) + X_{L}(v) \]

    \[\implies X_{L}(u + 2\pi) - X_{L}(u) = X_{R}(v) - X_{R}(v - 2\pi) \ (1) \]

Moreover, a key lesson or insight here is that the closed string is periodic. What is interesting, and what perhaps may have already ‘jumped out’ to the engaged reader, this statement about periodicity also implies that failure of the left-side to be periodic is equal to the right-side failing to be periodic. So, curiously, both sides are in a sense sharing information.

Now, before proceeding, we need to look at what it means to take the derivative of both u and v:

    \[ \frac{\partial}{\partial u} \ : X_{L}^{\mu \prime}(u + 2\pi) - X_{L}^{\mu \prime}(u) = 0 \]

    \[ \frac{\partial}{\partial v} \ : X_{R}^{\mu \prime}(v) - X_{R}^{\mu \prime}(v - 2\pi) = 0 \]

It follows that for our primed left and right-movers, we have:

    \[ X_{L}^{\mu \prime}(u) = \sqrt{\frac{\alpha^{\prime}}{2}}\sum_{n \in Z} \tilde{\alpha}_{n}^{\mu}e^{-inu} \]

    \[ X_{R}^{\mu \prime}(v) = \sqrt{\frac{\alpha^{\prime}}{2}}\sum_{n \in Z} \alpha_{n}^{\mu}e^{inv} \]

Notice that what we have is representative of a set of independent oscillators. Indeed, and moreover, notice that in the above we have essentially two copies of a product of an open string. When we integrate, we obtain what follows

    \[ X_{L}^{\mu}(u) = x_{0}^{\mu} + \sqrt{\frac{\alpha^{\prime}}{2}}\tilde{\alpha}_{0}^{\mu}u + i\sqrt{\frac{\alpha^{\prime}}{2}}\sum_{n \neq 0} \frac{1}{n}\tilde{\alpha}_{n}^{\mu}e^{-inu} \]

    \[ X_{R}^{\mu}(v) = x_{0}^{\mu} + \sqrt{\frac{\alpha^{\prime}}{2}}\alpha_{0}^{\mu}v + i\sqrt{\frac{\alpha^{\prime}}{2}}\sum_{n \neq 0} \frac{1}{n}\alpha_{n}^{\mu}e^{-inv} \]

This should look familiar, where we have our spacetime coordinate, momentum and on the far right-hand side the excitations. From (1) also notice,

    \[ \sqrt{\frac{\alpha^{\prime}}{2}}\tilde{\alpha}_{0}^{\mu}(2\pi) = \sqrt{\frac{\alpha^{\prime}}{2}}\alpha_{0}^{\mu}(2\pi) \]

    \[\rightarrow \tilde{\alpha}_{0}^{\mu} = \alpha_{0}^{\mu} \]

Which is to say that the periodicity condition that describes the waves moving left and right along the string, it basically constrains the system in such a way to show that the momenta on both sides are equal. Or, more accurately, there is only one momentum operator for the closed quantum string. This is rather neat.

Regarding the matter of coordinates, for the full string closed string coordinate we may write,

    \[X^{\mu}(\tau, \sigma) = \frac{1}{2}(x_{0}^{L \mu} + x_{0}^{R\mu}) + \sqrt{2\alpha^{\prime}}\alpha_{0}^{\mu}\tau + i\sqrt{\frac{\alpha^{\prime}}{2}} \sum_{n \neq 0}^{\infty} \frac{e^{-in\tau}}{n}(\alpha_{n}^{\mu}e^{in\sigma} + \tilde{\alpha}^{\mu}_{n}e^{-in\sigma}) \]

Recall, also, that the momentum density of the string is always,

    \[P^{\tau \mu} = \frac{1}{2\pi \alpha^{\prime}}\frac{\partial X^{\mu}}{\partial \tau} \]

And so,

    \[P^{\mu} = \int_{0}^{\pi} P^{\tau \mu} d\sigma = \sqrt{\frac{2}{\alpha^{\prime}}}\alpha_{0}^{\mu} \]

Where the \alpha term can be seen to be proportional to the spacetime momentum of the string.

As a next step in working our way toward the closed string spectrum, it may be helpful to record the \tau and \sigma derivatives of the coordinate, recalling from before X^{\mu} = X_{L}^{\mu}(\tau + \sigma) + X_{R}^{\mu}(\tau - \sigma). And so,

    \[\dot{X}^{\mu} = X_{L}^{\prime \mu}(\tau + \sigma) + X_{R}^{\prime \mu}(\tau - \sigma) \]

    \[X^{\prime \mu} = X_{L}^{\prime \mu}(\tau + \sigma) - X_{R}^{\prime \mu}(\tau - \sigma) \]

When we add and subtract these we find “that the barred oscillators do not mix with the unbarred oscillators in these combinations of derivatives” (Zwiebach, p.284). Ultimately, as we have something analogous to the open string expansion, it follows that for the commutators of the closed string we don’t have to perform any new calculations (Polchinski, pp. 26-27).

As review, recall that we have the commutators,

    \[ [X^{I}(\tau, \sigma), P^{IJ}(\tau, \sigma^{\prime}] = i\delta(\sigma - \sigma^{\prime})\eta^{IJ} \]

    \[ [\alpha_{m}^{I}, \alpha_{n}^{J}] = [\tilde{\alpha}_{m}^{I}, \tilde{\alpha}_{n}^{J}] = m\delta_{m + n, 0}\delta^{IJ} \]

    \[ [\alpha_{m}^{I}, \tilde{\alpha}_{n}^{J}] = 0 \]

Following this, we may now turn our attention to the Virasoro operators for the closed string. For the open string, we have come to understand that for the LC coordinate X^{-} the transverse Virasoro operators are the modes \alpha_{n}^{-}. But, in the closed string case, we have both barred and unbarred modes (I have been using tildes to highlight this fact) using the X^{-} coordinates.

How do the X^{-} coordinates relate the transverse Virasoro operators in the closed string case? From the study of the LC equations of motion we have,

    \[ \dot{X}^{-} \pm X^{\prime -} = \frac{1}{\beta \alpha^{\prime}}\frac{1}{2p^{+}}(\dot{X}^{I} \pm X^{\prime I})^{2} \]

Except, in our present case we set \beta = 1. In short, we find that as X^{-}(\tau, \sigma) = X_{0}^{-}, P^{-} it follows that P^{-} \rightarrow \alpha_{0}^{-} or \tilde{\alpha}_{0}^{-} in which it also follows that we have L_{0}^{\perp} or \tilde{L}_{0}^{\perp}. And so, in full,

    \[ L_{0}^{\perp} = \frac{\alpha^{\prime}}{4}p^{I}p^{I} + \sum_{p=1}^{\infty} \alpha_{-p}^{I}\alpha_{p}^{I} = \frac{\alpha^{\prime}}{4}p^{I}p^{I} + N^{\perp} \]

    \[\tilde{L}_{0}^{\perp} = \frac{\alpha^{\prime}}{4}p^{I}p^{I} + \sum_{p=1}^{\infty} \tilde{\alpha}_{-p}^{I}\alpha_{p}^{I} = \frac{\alpha^{\prime}}{4}p^{I}p^{I} + \tilde{N}^{\perp} \]

Now, an interesting question about the above expressions concerns how there does not seem to be a way to set them equal. What is going on? Well, one way to handle this is to enforce the following constraint (though still quite complicated):

    \[ (L_{0}^{\perp} - \tilde{L}_{0}^{\perp}) |\psi > = 0 \]

    \[ (N_{\perp} - \tilde{N}_{\perp}) | \psi > = 0 \]

For the first, we have the closed string state. For the second, this kills the states.

After some computation and lengthy consideration (Zwiebach, pp. 287-288), particularly concerning the constant ambiguities due to the ordering of our operators, we arrive at the following equations for n = 0,

    \[\sqrt{2\alpha^{\prime}}\bar{\alpha}_{0}^{-} = \frac{2}{p^{+}}(\tilde{L}_{0}^{\perp} - 1), \ \sqrt{2\alpha^{\prime}}\alpha_{0}^{-} = \frac{2}{p^{+}}(L_{0}^{\perp} - 1) \]

Which, in terms of the closed string Hamiltonian we may write compactly as,

    \[ H = L_{0}^{\perp} + \tilde{L}_{0}^{\perp} - 2 \]

We can also calculate the mass squared,

    \[M^{2} = \frac{2}{\alpha^{\prime}} (N^{\perp} + \tilde{N}^{\perp} - 2) \]

At this point we should note that, if not already clear, L_{0}^{\perp} + \tilde{L}_{0}^{\perp} generate \tau translations. So, it can also be said L_{0}^{\perp} + \tilde{L}_{0}^{\perp} \sim H generates \tau evolution. For sigma translations we have, [L_{0}^{\perp} - \tilde{L}_{0}^{\perp}, X^{I}(\tau, \sigma)] = i\frac{\partial X^{I}}{\partial \sigma}.

Moreover, in producing \sigma translations the states are also killed. That is, the statement L_{0}^{\perp} - \tilde{L}_{0}^{\perp} generates \sigma translations but kills all states, what this says in other words is that all physical states are invariant under \sigma translations. Thus, too, we are led to the fact that no one has been able to figure out how to select a special point in a closed string (at least not that I know of, and was also highlighted in a past lecture series by Zwiebach). That is, at what point might one determine the position of \sigma = 0 for the closed string? It’s quite interesting to think about and explore.

With all of that out of the way, we can now focus on the state space for the closed string. For the ground states we have |p^{+},p_{\tau}> in which,

    \[ | \lambda, \tilde{\lambda}> \equiv [\prod_{n=1}^{\infty} \prod_{I=2}^{D-1}(\alpha_{n}^{I})^{\dagger \lambda n, I}][\prod_{m=1}^{\infty} \prod_{J=2}^{D-1}(\tilde{\alpha}_{m}^{J})^{\dagger \tilde{\lambda} n, J}] | p^{+},p_{\tau}> \]

We may also note that,

    \[N^{\perp} = \sum_{n, I} n\lambda_{n,I} \]

    \[\tilde{N}^{\perp} = \sum_{n, I}n\tilde{\lambda}_{n,I} \]

And so,

    \[ N^{\perp} = \tilde{N}^{\perp} \]

Now, for N^{\perp} = \tilde{N}^{\perp} = 0 and M^{2} = -\frac{4}{\alpha^{\prime}}, we have a scalar field – a tachyon.

A few words were already shared about the tachyon in the study of the open string. It is a subject that deserves an entire paper in itself. But, for the sake of maintaining focus, let’s consider the case where N^{\perp} = \tilde{N}^{\perp} = 1 and M^{2} = 0. We also have R_{IJ} = S_{IJ} + A_{IJ} = \tilde{S}_{IJ} + S^{\prime}\delta_{IJ} + A_{IJ}, in which \tilde{S}_{IJ} is symmetric traceless. It follows, after computation,

    \[R_{IJ} (\alpha_{1}^{I})^{\dagger} - (\tilde{\alpha}_{1}^{J})^{\dagger} | p^{+},p_{\tau}> \]

    \[\tilde{S}_{IJ} (\alpha_{1}^{I})^{\dagger} - (\tilde{\alpha}_{1}^{J})^{\dagger} |p^{+},p_{\tau}> \ graviton \ states \]

    \[A_{IJ} (\alpha_{1}^{I})^{\dagger} - (\tilde{\alpha}_{1}^{J})^{\dagger} | p^{+},p_{\tau}> \ Kalb-Ramond states \]

    \[S^{\prime}_{IJ} (\alpha_{1}^{I})^{\dagger} - (\tilde{\alpha}_{1}^{J})^{\dagger} | p^{+},p_{\tau}> \ dilaton \ scalar \]

Note: the KR field has to do with charge in ST. The dilaton is an important field that we will come into contact with again in the future, as it controls the string coupling.

References

Joseph Polchinski. (2005). “String Theory: An Introduction to the Bosonic String“, Vol. 1.

Kevin Wray. (2009). “An Introduction to String Theory”.

Barton Zwiebach. (2009). “A First Course in String Theory”, 2nd ed.