# Handling the $\nabla \phi$ term in the Hamiltonian in a path integral

Let the scalar Hamiltonian be of the form $$H = \int d^3x \left [\hat{\pi}^2 + (\nabla \hat{\phi})^2 + m^2\hat{\phi}^2 \right ]$$.

We wish to evaluate the quantity $$\langle \phi_f | \exp(-iHt) | \phi_i \rangle$$ via path-integral. The standard procedure is to divide the time into $$N$$ slices, of width $$\epsilon$$, and insert a complete set of eigenstates $$I = \int |\pi\rangle \langle \pi| \frac{d\pi}{2\pi} \times \int |\phi \rangle \langle \phi | d\phi$$ between each time slice.

One then makes use of the relation $$\hat{\phi}(x)|\phi \rangle = \phi(x)|\phi\rangle$$, to convert the Schroedinger picture operator $$\hat{\phi}(x)$$ to the scalar function $$\phi(x)$$ in the exponent of $$\langle \phi_f | \exp(-iHt) | \phi_i \rangle$$. One can do a similar thing for $$\hat{\pi}(x)$$.

So far so good. Now, my question is that we can use the relation $$\hat{\phi}(x)|\phi \rangle = \phi(x)|\phi(x)\rangle$$ to convert $$\hat{\phi}(x)$$ to $$\phi(x)$$. But what about $$\nabla \hat{\phi}$$ ? The eigenket, $$|\phi\rangle$$, is not an eigenket of $$\nabla \hat{\phi}$$. How does one reduce the operator $$\nabla \hat{\phi}(x)$$ to a scalar $$\nabla\phi(x)$$?

Are we using the following trick?

$$\hat{\phi}(x)|\phi \rangle = \phi(x)|\phi\rangle$$ ....................(1)

$$\hat{\phi}(x + \delta)|\phi \rangle = \phi(x + \delta)|\phi\rangle$$ ........ (2)

Subtracting (1) from (2) and dividing by $$\delta$$, and taking the limit $$\delta \rightarrow 0$$, we get:

$$\nabla \hat{\phi}(x)|\phi \rangle = \nabla\phi(x)|\phi\rangle$$.

Is the above reasoning correct?

• What does the path integral approach (title) have to do with the body of the question? Commented Sep 23, 2022 at 14:59
• The quantity $\langle \phi_f| exp(-iHt)|\phi_i \rangle$ is evaluated using path integral Commented Sep 24, 2022 at 23:38

## 1 Answer

Yes, your reasoning is correct. To spell it out a bit more, the gradient $$\nabla$$ is an operator on position space, but doesn't act on Hilbert space. The field $$\hat{\phi}(x)$$ is an operator-valued function of position space (technically an operator valued distribution).

So we can write as an operator equation $$$$\nabla \hat{\phi}(x) = \lim_{\epsilon\rightarrow 0} \frac{\hat\phi(x+\epsilon) - \hat\phi(x)}{\epsilon}$$$$ Then we can act on a field eigenstate $$|\lambda\rangle$$ (I've changed the name from $$\phi(x)$$ to $$\lambda(x)$$ just to avoid notational confusion, but the label of the eigenstate doesn't matter). This state satisfies $$$$\hat\phi(x) | \lambda \rangle = \lambda(x) | \lambda \rangle$$$$ Note that I've chosen notation where the state $$|\lambda\rangle$$ does not have an explicit $$x$$ dependence. This is because the state is not a function of position, directly -- it is just a ray in Hilbert space. In other words, we don't have a different state at every position, there is just one state describing the whole system.

Putting this together, we can evaluate $$\begin{eqnarray} \nabla \hat{\phi}(x) | \lambda\rangle &=& \left[\lim_{\epsilon\rightarrow 0} \frac{\hat\phi(x+\epsilon) - \hat\phi(x)}{\epsilon} \right] | \lambda\rangle \\ &=& \lim_{\epsilon\rightarrow 0} \frac{\hat\phi(x+\epsilon) |\lambda\rangle - \hat{\phi}(x)|\lambda\rangle}{\epsilon} \\ &=& \left[\lim_{\epsilon\rightarrow 0} \frac{\lambda(x+\epsilon) - {\lambda}(x)}{\epsilon} \right] | \lambda \rangle \\ &=& \nabla \lambda(x) | \lambda\rangle \end{eqnarray}$$

In other words, an eigenstate $$|\lambda\rangle$$ of the field operator $$\hat\phi(x)$$ is also an eigenstate of the gradient-of-field operator $$\nabla\hat\phi(x)$$. If you think about it, it's actually hard to imagine that this would not be the case. In a state where you know the value of the field everywhere, surely you also know the value of the gradient of the field everywhere.