I am wondering how to explicitly see the vacuum degeneracy for a free massless scalar field, described by the action $$S = -\frac{1}{2}\int d^4x\,(\partial\phi)^2.$$ This action is invariant under the nonlinearly realized shift symmetry $\phi\rightarrow\phi + c$, and therefore should have a family of degenerate vacua. I would like to understand how to explicitly write an expression for these vacua by solving the functional Schrodinger equation.

The starting point should be the Hamiltonian, which is given by $$H = \frac{1}{2}\int d^3x\,\Pi^2 + (\nabla\phi)^2,$$ where $\Pi = \partial_0\phi$ is the canonical momentum. The functional Schrodinger equation is most easily solved by Fourier transforming the Hamiltonian: $$H = \frac{1}{2}\int d^3k\,\Pi(\vec{k})\Pi(-\vec{k}) + k^2\phi(\vec{k})\phi(-\vec{k}).$$ Then, after identifying $\Pi(\vec{k}) \sim \frac{\delta}{i\delta\phi(\vec{k})}$ in the field basis, one can verify that $$\Psi[\phi] = \exp\Big(-\frac{1}{2}\int d^3k\,\mathcal{E}(k)\phi(\vec{k})\phi(-\vec{k})\Big)$$ solves $H|\Psi\rangle =0$ (ignoring the infinite zero point energy, $E$) for $\mathcal{E}(k) = k$. This solution is intuitive, because the Hamiltonian is a sum of infinitely many decoupled oscillators, and the state above is the product of each of their ground states.

Now, because of the presence of the shift symmetry, I would expect there to be continuous family of vacua, all related by acting with the (exponentiated) shift symmetry generator $$Q = \int d^3x\,\Pi(\vec{x}) = \int d^3k\, \delta^{(3)}(\vec{k})\Pi(\vec{k}),$$ which should be broken by the ground state. However, $Q$ annihilates the previous solution: $$\langle\phi|Q|\Psi\rangle = -\Psi[\phi]\int d^3k\,\delta^{(3)}(\vec{k})\mathcal{E}(k)\phi(\vec{k}) = 0,$$ and therefore, any putative new vacuum is the same as the old: $$|\alpha\rangle = e^{i \alpha Q}|\Psi\rangle = |\Psi\rangle,$$ leading to the question: Is my expression for a vacuum state even correct? If so, why does the "broken" generator annihilate it, and how does one explicitly write down the other vacua?

Because $Q$ only acts at zero momentum, one potential modification I considered is to explicitly add a zero mode to the Hamiltonian. This is achieved by $H\rightarrow H + H_0$, for $H_0 = \frac{1}{2}\Pi(\vec{k} = 0)^2$ and $\Pi(0)\sim \frac{d}{id\phi(0)}$. This mode, which I will call $|0\rangle$, decouples so we can solve its Schrodinger equation $H_0 |0\rangle = E_0|0\rangle$ separately. The solution is $$\langle\phi|0\rangle = \mathcal{N}\exp\Big(iE^{1/2}_0\phi_0\Big),$$ for $\phi_0 = \phi(\vec{k} = 0)$ and some overall normalization $\mathcal{N}$. However for the ground state $E_0 = 0$, so the zero mode is simply an additional multiplicative constant that does not interact with $Q$.

Additional Note: For completeness, I wanted to also point out that the particular form of $|\Psi\rangle$ presented above seems to contradict how we know the charge to act on the field $\phi$, namely $$ e^{-i\alpha Q}\phi e^{i\alpha Q} = \phi + \alpha.$$ If we allow the charges to act on the field, we find $$ \langle \Psi|e^{-i\alpha Q}\phi e^{i\alpha Q}|\Psi\rangle = \langle \Psi|\phi|\Psi\rangle + \alpha.$$ Alternatively, if we allow the charges to act on the states, then we find $$\langle \Psi|e^{-i\alpha Q}\phi e^{i\alpha Q}|\Psi\rangle = \langle \Psi|\phi|\Psi\rangle,$$ leading to a contradiction.


2 Answers 2


There are (at least) three options:

  1. We can retain the field's zero-mode $\phi_0\equiv \int d^3x\ \phi(x)$ as an observable and retain the term $\Pi_0^2$ in the Hamiltonian. Then the model does not have a vacuum state at all. The spectrum of the Hamiltonian does have a finite lower bound, as required by the principles of QFT, but technically the Hilbert space doesn't have any state that exactly achieves that lower bound. In the functional Schrödinger representation, the function with the lowest "eigenvalue" is not normalizable, because it's independent of $\phi_0$, which is one of the integration variables in the functional representation of the inner product. We can still define the expectation value of the field's zero-mode by considering a sequence of (normalizable) states that approaches zero energy. We can choose the sequence so that the expectation value is $\alpha$, for any desired real number $\alpha$. This is is how the "vacuum degeneracy" is manifest, even though the vacuum state(s) itself does not belong to the Hilbert space. This resolves the paradox that was highlighted in the question. The apparent inconsistency between the last two equations in the question is resolved by the fact that $\langle\Psi|\cdots|\Psi\rangle$ is undefined.

  2. We can exclude $\phi_0$ from the set of observables and delete the term $\Pi_0^2$ from the Hamiltonian. Then we don't need to integrate over $\phi_0$ in the definition of the inner product, so the vacuum state exists — but it's unique, not degenerate. This is analogous to the situation for the free quantum electromagnetic field.

  3. A hybrid approach: We can keep $\phi_0$ in the set of observables but delete the term $\Pi_0^2$ from the Hamiltonian. Then we still need to integrate over $\phi_0$ in the definition of the inner product, but the $\phi_0$-dependence of the state is arbitrary (as long as it's normalizable) because the Hamiltonian is oblivious to $\phi_0$. This option seems contrived, at least for the free-scalar model being considered here, which is why I didn't even mention it in the earlier version of this answer, but I decided to add it for completeness.

Details for option 1

The math becomes more clear if we discretize space. Consider the model described in the question, but in a discrete space with a finite number $N$ of points and periodic boundary conditions. Then, in units where the distance between neighboring points is $1$, the action has the form $$ S\propto \sum_j (\dot \phi_j)^2 -\sum_{(j,k)}(\phi_j-\phi_k)^2 \tag{1} $$ where the index $j$ runs over the $N$ spatial points and $(j,k)$ runs over all nearest-neighbor pairs of points. The Hamiltonian is $$ H\propto \sum_j \Pi_j^2 +\sum_{(j,k)}(\phi_j-\phi_k)^2. \tag{2} $$ The operator $$ Q\propto \sum_j\Pi_j \tag{3} $$ commutes with the Hamiltonian because it commutes with all of the differences $\phi_j-\phi_k$. Now let's simplify things even further by taking $N=2$. Then the Hamiltonian is \begin{align} H &\propto \Pi_1^2+\Pi_2^2+(\phi_1-\phi_2)^2 \\ &= \frac{1}{2}(\Pi_1+\Pi_2)^2+ \frac{1}{2}(\Pi_1-\Pi_2)^2+(\phi_1-\phi_2)^2 \tag{4} \end{align} and $Q$ is $$ Q\propto \Pi_1+\Pi_2. \tag{5} $$ The Hilbert space should consist of normalizable functions of the real variables $\phi_1$ and $\phi_2$, but then the vacuum state does not belong to the Hilbert space. Equation (4) shows that the vacuum state must be independent of the combination $\phi_1+\phi_2$, which means it can't be normalizable, so it can't belong to the Hilbert space. The Hilbert space includes states for which the expectation value of $H$ is arbitrarily close to zero, but not equal to zero. For example, it includes states of the form $$ \Psi(\phi_1,\phi_2)=\exp\big(-\beta(\phi_1+\phi_2-\alpha)^2\big)\Psi_0(\phi_1-\phi_2) \tag{6} $$ where $\beta>0$, $\alpha$ is arbitrary, and $\Psi_0$ is the vacuum state of $$ H_0\propto \frac{1}{2}(\Pi_1-\Pi_2)^2+(\phi_1-\phi_2)^2. \tag{7} $$ Now we see the resolution of the paradox: in a state of the form (6), the expectation value of $\phi_1+\phi_2$ is $\alpha$, which is arbitrary. The question missed this by assuming that the vacuum state belongs to the Hilbert space. That would correspond to taking $\beta\to 0$, in which case the $\alpha$-dependence disappears, but that also makes the state (6) non-normalizable. We can still define the expectation value of $\phi_1+\phi_2$ by computing it when $\beta$ is finite and then taking the limit of the expectation value as $\beta\to 0$. The result is $\alpha$, which is arbitrary.

Details for option 2

If we exclude $\phi_1+\phi_2$ from the set of observables, then we can replace the Hamiltonian (4) with (7), and we can take the Hilbert space to consist of normalizable functions of just the single variable $\phi_1-\phi_2$. Then the theory has a vacuum state, and it's unique, because the shift symmetry now acts trivially on the model's observables: it's a gauge symmetry.


Technicalities like this aren't just something that pure mathematicians worry about. In Lectures on the Infrared Structure of Gravity and Gauge Theory arXiv:1703.05448, Strominger reviews some of the interesting physics that we would have missed if we had glossed over the "technicalities."

  • $\begingroup$ Thanks for the answer. My example was motivated with SSB in mind, and this model is the decoupling limit of the standard $U(1)$ linear sigma model, which has degenerate vacua. I would have also expected the linear sigma model to possess a zero mode, i.e. a very soft goldstone. I may be misunderstanding, but should I take your answer to imply that none of the vacua are normalizable? Moreover, if I do not insist that the goldstone zero mode is observable, does the vacuum become unique? $\endgroup$ Oct 3, 2021 at 4:59
  • $\begingroup$ Perhaps it is not sensible to ignore the goldstone zero mode though, if this would require us to view the broken $U(1)$ as gauge symmetry. $\endgroup$ Oct 3, 2021 at 5:09
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    $\begingroup$ @bittermania Regarding the linear sigma model: I have not thought carefully enough about how to take the limit in which only the goldstone modes remain. (I assume that's the decoupling limit you're talking about.) However, generally speaking, theories with massless particles don't have vacuum states in the strict sense of a (normalizable) state-vector in the Hilbert space unless we exclude the zero-momentum operator(s) from the set of observables, like we normally do for the quantum electromagnetic field. $\endgroup$ Oct 3, 2021 at 16:40
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    $\begingroup$ @bittermania The "vacuum degeneracy" phenomenon is still physically meaningful, because we can still take the low-energy limit of expectation values like I explained in the answer. $\endgroup$ Oct 3, 2021 at 16:40
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    $\begingroup$ @bittermania It's because of option (2) in your comment. Taking the zero-mode factor of the state to be an infinite plane wave (that is, independent of the zero-mode variable) makes the state non-normalizable, so the $\alpha$ term in your second-to-last equation might as well be zero because the other terms are "infinite". Loosely speaking, the equations $\delta(0)=\delta(0)+\alpha$ and $\delta(0)=\delta(0)$ are perfectly consistent with each other, because $\delta(0)=\infty$. $\endgroup$ Oct 4, 2021 at 0:48

@Chiral Anomaly has already posted a very nice answer which I have marked correct. I just wanted to add a few more words which connect the discrete toy model provided to the model in the original question more explicitly. Essentially, it boils down to treating an IR cutoff carefully.

The solution to the functional Schrodinger equation above is given by $$\Psi[\phi] = \exp\Big(-\frac{1}{2}\int d^3k\,\mathcal{E}(k)\phi(\vec{k})\phi(-\vec{k})\Big).$$ In the product above, every oscillator mode is normalizable except for the mode at $\vec{k} =0$. Thus, overall this solution is not normalizable. The state does not exist in the Hilbert space, and it is not a well-defined vacuum state, leading to the paradoxes outlined in the question.

We can cure this by handling the zero mode more carefully. Introduce a small vector $\vec{\beta}$ whose magnitude $\beta$ will serve as an IR cutoff. Write the solution to the Schrodinger equation as $$\Psi[\phi] = \exp\Big(-\frac{1}{2}\int_{0}^{\beta}d^3k\,\mathcal{E}(k)\phi(\vec{k})\phi(-\vec{k})-\frac{1}{2}\int_{\beta}^{\infty}d^3k\,\mathcal{E}(k)\phi(\vec{k})\phi(-\vec{k})\Big)$$ Using the fact that $\beta$ is small, we may approximate this expression as follows: $$\Psi[\phi]\approx \exp\Big(-\frac{\beta^2}{2}\phi_0^2-\frac{1}{2}\int_{\beta}^{\infty}d^3k\,\mathcal{E}(k)\phi(\vec{k})\phi(-\vec{k})\Big)\equiv\Psi_{\beta}[\phi_0,\phi],$$ where for clarity I have made the dependence on the zero mode explicit. $\Psi_{\beta}[\phi_0,\phi]$ does not minimize the Hamiltonian, but it is normalizable and thus exists in the Hilbert space. Defining the conjugate momenta of the zero mode as in the question, one can check that the zero mode piece transforms exactly as we want it to: $$\langle \phi|e^{i\alpha C}|\Psi_{\beta}\rangle = e^{\alpha\frac{d}{d\phi_0}}\Psi_{\beta}[\phi_0, \phi] = \Psi_{\beta}[\phi_0 - \alpha, \phi].$$ Note however, that the $\alpha$ dependence will vanish when the IR cutoff is taken to zero.

So long as we take the IR cutoff to zero at the end of the computation, we may compute sensible expectation values, even if we let the broken generator act on the state instead of the field: $$\frac{1}{\langle\Psi_{\beta}|\Psi_{\beta}\rangle} \langle \Psi_{\beta}|e^{-i\alpha C}\phi_0 e^{i\alpha C}|\Psi_{\beta}\rangle = \frac{1}{\langle\Psi_{\beta}|\Psi_{\beta}\rangle}\int d\phi_0 \int \mathcal{D}\phi_{k>\beta}\exp\Big(-\beta^2(\phi_0 - \alpha)^2-\int_{\beta}^{\infty}d^3k\,\mathcal{E}(k)\phi(\vec{k})\phi(-\vec{k})\Big)\phi_0 = \alpha.$$ As required, the IR cutoff has dropped out and can safely be taken to zero.

  • $\begingroup$ I like your answer best (+1)! It's more concise and more aesthetically pleasing. $\endgroup$ Oct 5, 2021 at 13:15

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