Use the commutation relation to show that the conjugate momentum acts on eigenstates of $\hat{\Phi}$ as $ - i \delta / \delta\phi_a(\mathbf{x})$

Question

This is part (b) of Schwartz's Problem 14.3 in his Quantum Field Theory and the Standard Model textbook.

Suppose that we have a real scalar field operator $\hat{\Phi}(x^0,\mathbf{x})$ with conjugate momentum field $\hat{\Pi}(x^0,\mathbf{x}) := \partial_0 \hat{\Phi}(x^0,\mathbf{x})$. These operators satisfy the equal-time commutation relations $$ [ \hat{\Phi}(x^0,\mathbf{x}), \hat{\Phi}(x^0,\mathbf{y}) ] = 0 \\ [ \hat{\Pi}(x^0,\mathbf{x}), \hat{\Pi}(x^0,\mathbf{y}) ] = 0 \\ [ \hat{\Phi}(x^0,\mathbf{x}), \hat{\Pi}(x^0,\mathbf{y}) ] = i \delta^{(3)}(\mathbf{x} - \mathbf{y}) $$ At some initial time $t=0$ we can define simultaneous (orthonormal) eigenstates of the Schrodinger-picture operator $\hat{\Phi}(0,\mathbf{x})$ and $\hat{\Pi}(0,\mathbf{x})$ which satisfy $$ \hat{\Phi}(0,\mathbf{x}) | \phi_a \rangle = \phi_a(\mathbf{x}) | \phi_a \rangle \ \ \ \ \mathrm{and} \ \ \ \ \hat{\Pi}(0,\mathbf{x}) | \pi_a \rangle = \pi_a(\mathbf{x}) | \pi_a \rangle $$

Question: The task of Schwartz's problem 14.3(b) here is to use the (third) commutation relation to show that $\hat{\Pi}(0,\mathbf{x})$ acts on eigenstates of $\hat{\Phi}(0,\mathbf{x})$ as the variational derivative $- i \delta/ \delta\phi_{a}(\mathbf{x})$.

My Attempt: I think that this means to show that $\langle \phi_a | \hat{\Pi}(0,\mathbf{x}) | \zeta \rangle = - i \dfrac{\delta}{\delta \phi_a(\mathbf{x})} \langle \phi_a | \zeta \rangle$ for any state $|\zeta \rangle$ in the Fock space.

So far what I have shown that the third commutation relation implies that $[ \hat{\Phi}(x^0,\mathbf{x}), \hat{\Pi}(x^0,\mathbf{y})^n ] = i n \delta^{(3)}(\mathbf{x} - \mathbf{y})\hat{\Pi}(x^0,\mathbf{y})^{n-1}$ for any $n \geq 1$. From this it follows that for any number $\epsilon$ we have $[ \hat{\Phi}(x^0,\mathbf{x}), e^{ - i \epsilon \hat{\Pi}(x^0,\mathbf{y}) } ] = \epsilon e^{ - i \epsilon \hat{\Pi}(x^0,\mathbf{y}) } \delta^{(3)}(\mathbf{x} - \mathbf{y})$.

From here, applying this commutator to a field eigenstate $|\phi_a \rangle$ yields $$ \hat{\Pi}(x^0, \mathbf{x}) \big( e^{ - i \epsilon \hat{\Phi}(x^0,\mathbf{y}) } | \phi_a \rangle \big) = \big( \phi_a(\mathbf{x}) + \epsilon \delta^{(3)}(\mathbf{x} - \mathbf{y} ) \big) \big( e^{ - i \epsilon \hat{\Phi}(x^0,\mathbf{y}) } | \phi_a \rangle \big) $$ This is where I get stuck though. I was hoping to use the above to show that $\big( e^{ - i \epsilon \hat{\Phi}(x^0,\mathbf{y}) } | \phi_a \rangle \big) \propto | \phi_a + \epsilon \rangle$, but the extra $\delta$-function doesn't seem to make this work (even without the $\delta$-function problem, this would still just be a proportionality, up to some phase). From there my idea was to consider the inner product $\langle \phi_a | e^{- i \epsilon \hat{\Pi}(0,\mathbf{x}) } | \zeta \rangle$ and take the limit $\epsilon$ to prove the result.

Answer 1

We consider the action of the conjugate momentum operator $\hat{\pi}(t, \vec{x})$ on a field eigenstate $\left|\phi_t\right\rangle$ of a field operator $\hat{\phi}(x)=\hat{\phi}(t, \vec{x})$. We want to answer why this is equal to the the action of the variation of $\phi$ on the same eigenstate: $$ \hat{\pi}(t, \vec{x})\left|\phi_t\right\rangle=-i \frac{\delta}{\delta \phi(x)}\left|\phi_t\right\rangle $$

The key is to consider the action of $$ \hat{U}=e^{i \lambda \int \mathrm{d}^3 \vec{y} \hat{\pi}(t, \vec{y}) f(\vec{y})}, $$ on an eigenstate of $\hat{\phi}$, for a test function $f$. Since $\hat{\pi}$ and $\hat{\phi}$ are canonical conjugates, we can use their canonical commutation relations, to see that \begin{equation}\label{expr} e^{-i \lambda \int \mathrm{d}^3 \vec{y} \hat{\pi}(t, \vec{y}) f(\vec{y})} \hat{\phi}(x) e^{i \lambda \int \mathrm{d}^3 \vec{y} \hat{\pi}(t, \vec{y}) f(\vec{y})}=\hat{\phi}(t, \vec{x})+\lambda f(\vec{x})+o(\lambda) \end{equation} where $f$ is a test function. From this follows $$ \hat{\phi}(x)\left(e^{i \lambda \int \mathrm{d}^3 \vec{y} \hat{\pi}(t, \vec{y}) f(\vec{y})}\left|\phi_t\right\rangle\right)=(\phi(x)+\lambda f(\vec{x}))\left(e^{i \lambda \int \mathrm{d}^3 \vec{y} \hat{\pi}(t, \vec{y}) f(\vec{y})}\left|\phi_t\right\rangle\right), $$ which in turn tells us that that applying $e^{i \lambda \int \mathrm{d}^3 \vec{y} \hat{\pi}(t, \vec{y}) f(\vec{y})}$ to an eigenstate of $\hat{\phi}$ yields $$ e^{i \lambda \int \mathrm{d}^3 \vec{y} \hat{\pi}(t, \vec{y}) f(\vec{y})}\left|\phi_t\right\rangle=\left|\phi_t+\lambda f\right\rangle $$ Finally, let us consider $$ \left\langle\phi_t\left|e^{i \lambda \int \mathrm{d}^3 \vec{y} \hat{\pi}(t, \vec{y}) f(\vec{y})}\right| \Psi\right\rangle=\Psi\left[\phi_t-\lambda f\right] . $$

Evaluating the right-hand side and left-hand side for small $\lambda$, we find $$ \hat{\pi}(t, \vec{x})\left|\phi_t\right\rangle=-i \frac{\delta}{\delta \phi(x)}\left|\phi_t\right\rangle $$

Answer 2

First, a correction to the main result in your original post. It should be

$$ \hat{\Phi}(\mathbf{x}) \big( e^{ - i \epsilon \hat{\Pi}(\mathbf{y}) } | \phi_a \rangle \big) = \big( \phi_a(\mathbf{x}) + \epsilon \delta^{(3)}(\mathbf{x} - \mathbf{y} ) \big) \big( e^{ - i \epsilon \hat{\Pi}(\mathbf{y}) } | \phi_a \rangle \big). $$

(For simplicity, we focus on one point of time and thus omit the time argument.)

There are two separate questions.

The first one is about the mysterious $\delta^{(3)}(\mathbf{x} - \mathbf{y} )$. I will explain why it must be there in two ways.

First, this essentially says that the field $e^{ - i \epsilon \hat{\Pi}(\mathbf{y}) } | \phi_a \rangle$ has the same value as $| \phi_a \rangle$ everywhere, except for the spatial point $\mathbf{y}$, where its value is shifted by $\epsilon \delta^{(3)}(0)$. The $\delta^{(3)}(0)$ must be there, because otherwise it would be exactly the same field. We are dealing with the Fock space here. Any two states that differ only in a set of zero measure (in this particular case a single point) are equivalent.

Alternatively, let's write $|\phi_a\rangle$ formally as

$$|\phi_a\rangle = \int d^3 \mathbf{z} \phi(\mathbf{z}) |\mathbf{z}\rangle,$$

where $|\mathbf{z}\rangle$ is some kind of state such that $|\phi_a\rangle$ is a field that has value $\phi(\mathbf{z})$ at $\mathbf{z}$.

Let's see what $\frac{\delta}{\delta \mathbf{\phi(x)}}$ does to $|\phi_a\rangle$. We have

$$\big( 1 + \epsilon \frac{\delta}{\delta \phi(\mathbf{x})}\big) |\phi_a\rangle = \int d^3 \mathbf{z} \big(\phi(\mathbf{z}) + \epsilon \delta^3(\mathbf{x}-\mathbf{z}) \big) |\mathbf{z}\rangle, $$

exactly the same $\delta^3$ appeares. We can thus draw the conclusion that

$$ \hat{\Pi}(\mathbf{x})|\Phi\rangle = i\frac{\delta}{\delta \phi(\mathbf{x})}|\Phi\rangle, $$

up to a phase.

(If you wonder what the eigenstate actually is, see this post.)

What does it mean physically? It means that $e^{ - i \epsilon \hat{\Pi}(\mathbf{y}) } | \phi_a \rangle$ is not really an eigenstate of $\hat{\Phi}(\mathbf{y})$. We will need some kind of $\int d^3\mathbf{z} p(\mathbf{z}) e^{ - i \epsilon \hat{\Pi}(\mathbf{z}) } | \phi_a \rangle$, with an appropriate distribution $p(\mathbf{z})$, to get an eigenstate of $\hat{\Phi}(\mathbf{y})$.

The second question is about the phase. We can simply redefine $|\Phi\rangle$ to make the phase constant.