Covariant Quantisation and the Time Operator in String Theory

Question

Covariant quantisation in string theory is accomplished by giving the commutator relations

$$[X^\mu(\sigma,\tau),P^\nu(\sigma',\tau)] = i \eta^{\mu\nu} \delta(\sigma - \sigma').$$

Although string theory is not a field theory this approach is identical to the quantisation of fields $X^\mu$ in 1+1 dimensions. From this point of view it is also clear how to proceed from here (e.g. getting rid of the negative norm states due to the "wrong" sign of the zeroth component).

However, there are two issues that I do not quite understand. Physical states are supposed to have positive energy, i.e. $P^0 \geq 0$. Especially when supersymmetry comes into the game, the algebra so to say dictates positive energies. But when the energy is bounded from below it is not possible to have a time operator obeying the commutator relation

$$[x^0,p^0] = -i,$$

where $x$ and $p$ are the above Operators $X$ and $P$ integrated over $\sigma$.

For the case of quantum mechanics this has been proved by Pauli in 1933. The prove goes roughly like this: For an operator $H$ which is bounded from below let there be a Hermitian time operator $T$ with

$$[T,H] = i.$$

It is possible to shift the spectrum of H because of

$$e^{-i\epsilon T} H e^{i\epsilon T} = H - \epsilon,$$

for arbitrary $\epsilon$. That means for an eigenstate $|\psi\rangle$ with

$$H|\psi\rangle = E|\psi\rangle,$$

there is also an eigenstate

$$|\phi\rangle = e^{i\epsilon T}|\psi\rangle,$$

with

$$H|\phi\rangle = (E-\epsilon)|\phi\rangle.$$

For $\epsilon > E$ one has to either conclude that H can not be bounded from below or that T is not a Hermitian operator because it would generate states that lie outside of the Hilbert space.

One possible solution could be to allow negative $p^0$ and to apply some kind of Feynman-Stueckelberg approach by reversing the sign of the world-sheet coordinate $\tau$. In this sense the decay of the vacuum into a positive and negative energy state is simply the propagator of a free state. However, this would not be possible in the supersymmetric case where the states have $p^0>0$.

The other issue related to defining the zeroth component of the commutator is the construction of the Hilbert space of the physical states. In his book String Theory Polchinski starts with a "vacuum" state $|0;k\rangle$ with the norm

$$ \langle 0;k|0;k'\rangle \sim \delta^{26}(k-k').$$

However, when integrating over $k$ one ends up with a superficial delta function because of the additional mass shell constraint. Polchinski solves the issue by introducing a reduced product

$$\langle 0;k|0;k'\rangle \sim \delta^{25}(k-k').$$

This seems to be a logical step because selecting the physical states by

$$L_0|\psi\rangle = \alpha'(p^2+m^2)|\psi\rangle = 0,$$

yields a Hilbert space whose functions depend on 25 momentum components. Still, the full Hilbert space is not normalisable (this differs from what is usually done in QFT during gauge fixing, where the Hilbert space is constructed via an equivalence relation for the gauge and not by selecting a subspace). The question is whether the Hilbert space constructed in such a way does exist at all. Also it seems as if the number of spacetime operators is reduced by one, thus rendering the steps taken for the covariant quantisation needless.

(REMARK: I edited my question to make the point a bit clearer. This is not meant to be a criticism against string theory. It is just that during studying string theory I was confronted by someone with this issue. I just want to understand this detail of the basics because I haven't yet been able to resolve it. I hope somebody can help me.)

Answer 1

Positivity of energy

The probability that a single superstring has a negative value of energy is strictly zero, as implied by supersymmetry. There's no contradiction with the commutation relation. It's easy to see why. Just define the wave function of the superstring in the $P^0$ representation. Then you have $\tilde\psi(P^0)$ and you may assume that $$\tilde\psi(P^0) = 0\mbox{ for } P^0\leq 0.$$ The operator $X^0$ may be defined as $X^0=-i \partial/\partial P^0$ and it's manifest that $$[X^0,P^0]=-i.$$ So your favorite anonymous co-father of quantum mechanics, whether it was Heisenberg, Dirac, or anyone else, had to use another assumption that is not satisfied in string theory to argue that it is not possible. (The assumption has to be violated not only in string theory but any particle-like description of a theory similar quantum field theory.)

It's pretty obvious that you don't quite know what the hypothetical no-go statement is and why it is true, so it may be a sensible idea to forget about this hypothetical statement and discard it as a vague irrational prejudice if you're reading Polchinski's book that is supposed to make everything crystal clear, and I think that it does so. You knowledge of physics can't be built on some vague rumors that you may have heard somewhere. You must actually understand why statements are true.

My guess is that she or more likely he (the founder of QM) assumed that the Fourier transform of $\tilde \psi(P^0)$, namely the function $\psi(X^0)$, also has to vanish for negative $X^0$. Those two conditions - vanishing of both the functions and its Fourier transform for negative values of the argument - are not compatible conditions. However, there is no condition that $\psi(X^0)$ vanishes for a negative value of $X^0$ for a single superstring (or any single particle, for that matter), so the contradiction disappears and the very construction above proves that there can't be any inconsistency.

String theory is usually treated in the "first-quantized" way, so we obtain one-particle and multi-particle states "directly" and "directly" calculate their scattering amplitudes. That was the treatment I was assuming above; the negative-energy one-particle states are strictly absent.

However, string theory may also be treated in the "string field theory" formalism that makes it look like a quantum field theory in the spacetime, with infinitely many fields corresponding to the excitations of the string. Then the string fields may be Fourier expanded into components with various values of $P^0$, and the operators in front of the waves with a positive/negative value of $P^0$ are labeled as annihilation/creation operators, just like in any quantum field theory. Only the creation operators may produce nonzero particle states out of the vacuum - which is why particles always carry non-negative or positive values of $P^0$.

Normalization of eigenstates of momentum

The second question is purely about the normalization of some vectors and their existence. There is no real subtlety here. The off-shell states $|0;k'\rangle$ where $k$ doesn't have to satisfy that $k_\mu k^\mu=m^2$ may be assumed to exist for any $k$. However, the physical - Virasoro - conditions force the $k$ to be on-shell.

So, in mathematical language, if you assume a particular type of a particle and its wave function is written as $$|\psi \rangle = \int d^{26} k\,\tilde\psi(k) |0;k'\rangle,$$ then the Virasoro condition guarantees that $\tilde\psi(k)=0$ for all $k_\mu$ such that $k_\mu k^\mu\neq m^2$. To obtain states of a finite norm that satisfy the on-shell condition, you must pick $$\tilde\psi(k) = \sqrt{\delta(k_\mu k^\mu - m^2) } \tilde\psi_{reduced} (\vec k)$$ where the reduced wave function only depends on the $25$ spatial components of the momentum. I had to add the square root of the delta function because the total norm of the states above will contain the $\delta$-function that, when integrated over $k^0$, will set $k^0$ to the right on-shell value given by the usual square root. In my notation, the vector that only has the 25-dimensional $\delta$-function as the inner product may be obtained from the previous one multiplied by the square root of the $\delta$-function as follows: $$||0;k\rangle = \int_0^\infty dk^{\prime 0} \sqrt{ \delta\left( k^{\prime 0} -\sqrt{m^2+\vec k^2} \right) } |0; (k^{\prime 0},\vec k)\rangle $$ The $\delta$-function in the formula above is the same one as previous one, up to powers of $k^{\prime 0}$ and numerical constants that you should be able to calculate (and the second $\delta$-function also contains $\theta(k^0)$ relatively to the first one - the first one also incorrectly allows negative values of $k^0$). I chose this explicit form here because using the last displayed formula above, you may easily derive the second inner product - with the 25-dimensional $\delta$-function - from the first one. The square root of the $\delta$-function becomes an ordinary one and disappears after the integration over $k^0$, leaving the usual second inner product for the on-shell states.

Those square roots of the $\delta$-function look awkward and they're not really necessary in any way. It's clear that the right physical space is a simple combination of the vectors $||0;\vec k\rangle$ that only depend on 25 components of the momentum. It's just a physical fact that there is a single on-shell condition, so the number of independent components of the energy-momentum vector is 25 rather than 26. If you pick 25 coordinates out of 26, you inevitably violate the "totally manifest" Lorentz symmetry. Kinematics always does so, even in ordinary quantum field theory where cross sections have the $1/2E$ factors etc. that have to multiply the squared Lorentz-invariant amplitudes.

However, aside from these kinemetical factors that are universal (independent on the particle type and its interactions) and easy to learn, there's a lot of actual dynamics - the scattering amplitudes that depend on the momenta and on the particle types; and the infinitely many internal "Hagedorn tower" excitations of the string by the stringy oscillators. It's the "difficult" objects from the previous sentence that remain manifestly Lorentz-covariant in the covariant quantization which is why the quantization is called covariant.

In an alternative quantization, such as the light cone gauge, the Lorentz symmetry becomes really hard to prove. In fact, you need to prove that $[J^{i-},J^{j-}]=0$ in the light-cone coordinates and this condition itself, because of various "double commutator terms", will actually force you to prove that $D=26$ as well. So the calculation of this commutator is not equivalent to the classical Poisson bracket calculations; it knows something about the one-loop behavior of the world sheet theory, too.

So you must learn how the kinematical factors - such as the inner products of the physical states and the extra coefficients that you have to insert to the squared Lorentz-invariant amplitudes to obtain the actual cross sections etc. - work because these are basic and essentially trivial things. The nontrivial physics, i.e. the dynamics, hides in the excitations of the string and the interactions of the strings (or interactions of quantum fields, in similar problems involving quantum field theory), and for those more difficult dynamical questions, it matters whether you want to make unitarity manifest (like in light-cone gauge); or Lorentz symmetry manifest (like in the covariant quantization).

Answer 2

This is an idea I had a while back about time operators, which might work. I have tried to reconstruct it. Define the time operator $$ {\hat T}~=~i\hbar\sum_{j\ne k}{{|E_j\rangle\langle E_k|}\over{E_j~-~ E_k }}. $$ that acts upon a ket $|t\rangle~=~N^{1/2}\sum_nexp(iE_nt/\hbar)$ as $$ {\hat T}|t\rangle~=~i\hbar\sum_{j\ne k}{{|E_j\rangle\langle E_k|}\over{E_j~-~E_k }}|t\rangle~=~iN^{-1/2}\hbar\sum_{j\ne k}(E_j - E_k)^{-1}|E_j\rangle e^{-iE_kt/ħ}. $$ This is a Fourier summation, which in the continuum limit gives $t|t\rangle$ by the Cauchy integral formula. Now compute matrix elements of $[T,~H]$ for $|\psi\rangle~=~\sum_ja_j|E_j\rangle$ $$ \sum_ja_j\langle E_i|[T,~H]|E_j\rangle~=~i\hbar\sum_{j,k,l}a_j\langle E_i|(E_k~-~E_l)^{-1}|E_k\rangle\langle E_l |E_j\rangle $$ $$ =~i\hbar\sum_{j,k}a_j\langle E_i| (E_k~-~E_j)^{-1}|E_k\rangle, $$ where the matrix element $\langle E_i|(E_k~-~E_j)^{-1}|E_k\rangle~=~\delta_{ik}(E_k~-~E_j)^{-1}$. $|E_i\rangle$ is not in the projective summation of the time operator for $[{\hat T},~H]~=~i\hbar$, and generally $[{\hat T},~ H]~=~0$. \vskip.12in A Cauchy sequence of states will converge to a bounded state $|\psi\rangle~=$ $\sum_{j=0}^Na_j(N)|E_j\rangle$, for $N$ the bound on the complete set. For the coefficient $\sim~(1/j)$ as $N~\rightarrow~\infty$ the accumulation point contains a dense set of points with $E~+~\delta E$ energy eigenvalues which satisfy $\sum_j\langle E_j|\psi\rangle~=~0$. This means that the commutator $[T,~H]~=~i\hbar$ holds on a measure zero set, and for the function $\psi(t)$ an almost periodic function.

Answer 3

Since there has been no useful information up to now I figured out some answers myself.

First of all, the argument could be made even stronger. It is not only the time operator but also the position operators that appear to be problematic because an arbitrary shifted state $|0;k+\epsilon\rangle$ might lie outside the mass shell since generally $k^2 \neq (k+\epsilon)^2$. In usual quantum field theories (e.g. QED) it is possible to construct position operators (if one really has a need for them) which fulfill the Heisenberg algebra with the three-momentum. Unlike the case of string theory they do not commute with the energy operator, i.e. the mass shell condition is preserved. Unfortunately, these position operators do not obey Lorentz transformations and for massless particles they cannot be constructed for a particle spin greater than $1/2$.

Well, my original question contains a false assumption which is that the Hilbert space would only contain states being on the mass shell. Actually, the Hilbert space of string theory contains all possible four-momenta, but the quantisation procedure ensures that only states with a momentum fulfilling the mass shell condition contribute to physical amplitudes. I am however not sure about the negative energy states but I guess they also do not contribute to physical amplitudes. In quantum theory and standard quantum field theory the Hilbert space (of free fields) is explicitly made up of positive energy states being on the mass shell. There, Pauli's argument applies that it is impossible to construct a time operator. This is not true for string theory as described above.

However, as a consequence the usual norm of its Hilbert space is not well defined. It seems that the only possible solution is a redefinition of the norm. The approach seems to be common also in other approaches to quantum gravity theories with a covariant quantisation mechanism (see e.g. arxiv/1102.1148 and references therein). Not being able to work it out in more detail at the moment, I just take it for granted.

Answer 4

The difference between string theory and quantum field theory is that QFT is the theory of point excitation whereas string theory is the theory of 1d strings. When we talk about $[X^0 , H]=-i$ and thereafter the construction of operators like $e^{-iX^0ϵ}He^{iX^0ϵ} = H - ϵ$ to construct states with arbitrary energy. We tend to forget a very critical result that the time operator $X^0$ is assumed to have had the representation of $i\partial_{P^0}$ but that is not the case in String Theory. When never really explicitly assume the representation of operators! We start out with Lagrangian and then use the variational principle to extract the equation of motion from it. We solve that equation of motion to arrive at a solution in terms of the Fourier coefficient. In quantum mechanics, if we had $[X^0 , H] = -i$, then the action of $X^0$ would result in the eigenvalue at a point on the particle's worldline. However, in string theory action of operators like $X^0$ is to fundamentally change the excitation mode of the string! Remember in string theory $X(σ, τ)$ is not just a spacetime coordinate but the quantum field on the worldsheet. Whenever $\hat{X}$ acts on the string it excites the relevant mode in the spectrum at that one point with coordinate $(\sigma,\tau)$. Pauli Theorem is indeed true but from this perspective but only when we assume the form of an operator like we do in quantum mechanics. Let's just take a look at the Theorem

\begin{align} He^{iT\epsilon}|x\rangle &= e^{iT\epsilon}H|x\rangle+[H,e^{iT\epsilon}]|x\rangle\\ &=e^{iT\epsilon}E|x\rangle -\epsilon e^{iT\epsilon}|x\rangle\\ &= (E-\epsilon)e^{iT\epsilon}|x\rangle \end{align}

Now if I wonder how we got $[H,e^{iT\epsilon}]=\epsilon e^{iT\epsilon}$ then the direct and obvious answer would be if I assumed the representation of the Hamiltonian operator as $i\partial_{T}$. But why do that when we can just leave it as it is? The constraints put on $X^\mu$ and $P^\mu$ can be interpreted as quantum field theory on the worldsheet. String Hamiltonian and everything else comes out beautifully from quantum field theory on the worldsheet without us ever having to worry about the representation. Just calculate $X^\mu$ and $P^\mu$ from the equation of motion, impose the equal time commutator constraint on it, and pass down the constraint from $X^\mu$ and $P^\mu$ to the creation and annihilation operator on the worldsheet then express everything in terms of worldsheet variables. One interesting remark here is that doing all of this never leads to $[\tau,i\partial_{\tau}]=-i$ where $i\partial_{\tau}$ would have been worldsheet hamiltonian. However, since we were considering the $X^0$, the excitation modes are timelike and hence have a negative norm. These states are eliminated as a result of old covariant quantization or BRST cohomology. Thus they do not appear in the string spectrum. Thereafter the spectrum of Hamiltonian for strings has discrete eigenvalues and energy stays quantized.