Statement of exercise

On a page 11 of A.Zee's book QFT in a Nutshell, he derives Dirac's formulation of the path integral formulation of QM for a free particle. This starts with the free particle Hamiltonian

$$ H = \frac{\hat{p}^2}{2m} $$

The hat on $\hat{p}$ says that it is meant as an operator. Then he divides the time interval into $N$ parts of length $\delta t$, where each part is

$$ \begin{align} \tag{1}\langle q_{j+1}|e^{-iH \delta t}|q_j\rangle & = \langle q_{j+1}|e^{-i\delta t\,\hat{p}^2 /(2m)}|q_j\rangle \\ \tag{2} & = \int\frac{dp}{2\pi}\langle q_{j+1}|e^{-i\delta t\,\hat{p}^2 /(2m)}|p\rangle\langle p|q_j\rangle \\ \tag{3} & = \int\frac{dp}{2\pi}e^{-i\delta t\,p^2 /(2m)} \langle q_{j+1}|p\rangle\langle p|q_j\rangle \end{align} $$

In $(2)$, he used $1 = \int(dp/2\pi)|p\rangle\langle p|$ with momentum eigenstates $|p\rangle$, and in $(3)$ the fact that $\hat{p}|p\rangle = p|p\rangle$.

Then he goes on to derive the fact that the time integral of the Lagrangian emerges as the exponent in

$$ \langle q_F|e^{-iHt}|q_I\rangle = \int Dq\,e^{i\int dt S} $$

Now Zee poses the exercise to derive the same thing using the Hamiltonian

$$ H = \frac{\hat{p}^2}{2m}+V(\hat{q}) $$

What I tried

I was able to derive the result, but only IF the following is true

$$ \tag{4}\left\langle q_{j+1}\middle|e^{-i\delta t\left\{\hat{p}^2/(2m)+V(\hat{q})\right\}}\middle|q_j\right\rangle = \left\langle q_{j+1}\middle|e^{-i\delta t\left\{\hat{p}^2/(2m)+V(q_j)\right\}}\middle|q_j\right\rangle $$

in other words, if I can replace $V(\hat{q})$ with $V(q_j)$, then I can prove the result, because then, I can repeat the free particle calculation and the potential appears as an extra term in the integral that depends only on $q_j$.

However, I don't know how to prove that $(4)$ is true. $\hat{p}$ and $\hat{q}$ do not commute, so $e^{\hat{p}+V(\hat{q})}$ cannot simply be factorized into $e^{\hat{p}}e^{V(\hat{q})}$.

Why do I think that $(4)$ is true?

The following approximation led me to think that $(4)$ is true:

$$ \begin{align} \tag{5}\left\langle q_{j+1}\middle|e^{-it\left\{\hat{p}^2/(2m)+V(\hat{q})\right\}}\middle|q_j\right\rangle & \approx \langle q_{j+1}|1-it\,(\hat{p}^2/2m+V(\hat{q}))+O(\delta t^2)|q_j\rangle \\ & = \langle q_{j+1}|1-it\,\hat{p}^2/2m+O(\delta t^2)|q_j\rangle - i\delta t\langle q_{j+1}|V(\hat{q})|q_j\rangle \\ \tag{7}& = \langle q_{j+1}|1-it\,\hat{p}^2/2m+O(\delta t^2)|q_j\rangle - i\delta t\langle q_{j+1}|V(q_j)|q_j\rangle \\ & = \langle q_{j+1}|1-it\,(\hat{p}^2/2m+V(q_j))+O(\delta t^2)|q_j\rangle \\ \end{align} \\ $$

$(5)$ should be possible since we're only interested in the limit $\delta t \to 0$. But this seems to be a rather sloppy argument, because in the following steps, before we take the limit $\delta t \to 0$, a Gaussian integral over $p$ is used. So we really need the exponential function and not just some approximation.

Even $(7)$ seems a bit sloppy. Shouldn't I have some limits on the $V(q)$ function to know that $V(\hat{q})|q_j\rangle = V(q_j)|q_j\rangle$?

Any hints?

  • 1
    $\begingroup$ You should use Trotter-Kato theorem. $\endgroup$
    – yuggib
    Commented Oct 20, 2015 at 13:07
  • 1
    $\begingroup$ (and of course you cannot, strictly speaking, do that for any potential, but only for the ones that are sufficiently regular) $\endgroup$
    – yuggib
    Commented Oct 20, 2015 at 13:09

1 Answer 1


Your argument is essentially correct. In the usual physicist's derivation of the path integral, one uses the Zassenhaus formula to expand the exponential as $$ \mathrm{e}^{\delta t(p^2/(2m) + V(q))} = \mathrm{e}^{\delta t p^2/(2m)}\mathrm{e}^{\delta t V(q)}\mathrm{e}^{-\delta t^2 / 2[p^2,V(q)]}\dots$$ where all following terms are at least cubic in $\delta t$. Then, one discards all terms quadratic and higher in $\delta t$ since they vanish in the $\delta t \to 0$ limit.

Another way, where you don't need to "discard" terms by hand, is to use the Lie product formula from the beginning, i.e. to write $$ \mathrm{e}^{p^2/(2m) + V(q)} = \lim_{\delta t \to 0} \left(\mathrm{e}^{\delta t p^2 / (2m)}\mathrm{e}^{\delta t V(q)}\right)^{1/\delta t}$$

  • $\begingroup$ Okay, the Zassenhaus formula seems to be a better way to factorize the exponential. However, in the Gaussian integral there would still be this $\mathrm{e}^{-\delta t^2/2[p^2,V(q)]}$ term. How do I know I can do the Gaussian integral with this term? The Gaussian integral "happens" before we let $\delta t$ go to $0$. $\endgroup$
    – Bass
    Commented Oct 20, 2015 at 13:20
  • 1
    $\begingroup$ @BastianTreichler: That's why I said "usual physicist's derivation". One just leaves out the $\delta t^2$ term before integrating although the limit has not yet been taken. If you want a rigorous derivation of the path integral this whole thing becomes more complicated: You need to continue analytically to Euclidean time and do things with the Wiener measure on the space of continuous paths. $\endgroup$
    – ACuriousMind
    Commented Oct 20, 2015 at 13:24
  • $\begingroup$ OK that's great to know. Just in case you know of a QFT textbook where it's done rigorously, I'd be happy to know, thanks. $\endgroup$
    – Bass
    Commented Oct 20, 2015 at 13:25
  • 1
    $\begingroup$ @BastianTreichler: Off the top of my head, it's in Quantum Physics by Glimm and Jaffe. Beware, that's no physics book ;) $\endgroup$
    – ACuriousMind
    Commented Oct 20, 2015 at 13:28

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