Given the Hamiltonian $\hat{H}=\hat{x}\hat{p}^2\hat{x}$ with position operator $\hat{x}$ and momentum operator $\hat{p}$, I should give an explicit expression for the measure of the path integral, $$ \int\mathcal{D}[x(t)]\exp\left\{\frac{i}{\hbar}S[x(t)]\right\}, $$ wherein $$ S[x(t)]=\int dt\mathcal{L}(x,\dot{x}), $$ is the action with $\mathcal{L}$ being the Lagrangian.

I thought of two approaches on how to find an explicit expression for the path integral:

1. Reverse the Legendre transformation to find the Lagrangian

We first need to relate the momentum operator with the time derivative of the position operator, e.g. $$ \frac{d\hat{x}}{dt}=[\hat{x},\hat{H}]=2i\hbar\hat{p}, $$ using the canonical commutation relation $[\hat{x},\hat{p}]=i\hbar$.

Finally, the Lagrangian is obtained via, $$ \mathcal{L}(\hat{x},d\hat{x}/dt)=\hat{p}\frac{d\hat{x}}{dt}-H(\hat{x},d\hat{x}/dt)=2i\hbar\left(\frac{1}{2i\hbar}\frac{d\hat{x}}{dt}\right)^2-\hat{x}\left(\frac{1}{2i\hbar}\frac{d\hat{x}}{dt}\right)\hat{x}. $$

I don't know how to proceed from here. Inserting the Lagrangian into the action won't give us any new insights.

2. Derive the path integral formalism from the time evolution operator

This approach is inspired by the derivation used by Shankar in "Principles of Quantum Mechanics 2nd Edition, p. 585".

We start with the time evolution operator in position represenation, $$ U(x_0,x_n,t)=\langle x_0\mid \exp\left[\frac{it}{\hbar}\hat{H}\right]\mid x_n\rangle, $$ and define $\epsilon=t/N$ where we assume $\epsilon\ll1$, thus, $$ U(x_0,x_n,t)=\langle x_0\mid \exp\left[\frac{i\epsilon}{\hbar}\hat{H}\right]^N\mid x_n\rangle=\langle x_0\mid \exp\left[\frac{i\epsilon}{\hbar}\hat{H}\right]\dots\exp\left[\frac{i\epsilon}{\hbar}\hat{H}\right]\mid x_n\rangle+\mathcal{O}(\epsilon^2), $$ where we have used the Baker-Campbell-Hausdorff formula and neglected higher orders of $\epsilon$.

We can now use the completeness identity for the position operator to obtain, $$ U(x_0,x_n,t)=\int dx_1\dots dx_{n-1}\langle x_0\mid \exp\left[\frac{i\epsilon}{\hbar}\hat{H}\right]\mid x_1\rangle\dots\langle x_{n-1}\mid\exp\left[\frac{i\epsilon}{\hbar}\hat{H}\right]\mid x_n\rangle+\mathcal{O}(\epsilon^2). $$

Using the completeness relation in momentum space, we can give an expression for the matrix elements, $$ \langle x_{n-1}\mid\exp\left[\frac{i\epsilon}{\hbar}\hat{H}\right]\mid x_n\rangle=\sum^\infty_{m=0}\frac{(i\epsilon/\hbar)^m}{m!}\langle x_{n-1}\mid\hat{x}\hat{p}^2\hat{x}\mid x_n\rangle=\sum^\infty_{m=0}\frac{(i\epsilon/\hbar)^m}{m!}x_nx_{n-1}\delta^{(2m)}(x_n-x_{n-1}), $$ wherein $\delta^{(2m)}$ denotes the $2m$ derivative with respect to the argument of the $\delta$ distribution.

However, this expression does not appear to be of much help.

Are these approaches the any good? How do I find an explicit expression for the measure of the path integral?

  • 2
    $\begingroup$ @bodokaiser what units are you using? Also $ \int\mathcal{D}[x]\exp\left\{\frac{i}{\hbar}S[x]\right\}, $ can only be used for quadratic Hamiltonian but your Hamiltonian is quartic so you have go for this $ \int\mathcal{D}[x]\mathcal{D}[p]\exp\left\{\frac{i}{\hbar}M[x,p]\right\}, $ where$ M[x,p]=\int dt(p\dot{x}-\mathcal{H}(x,\dot{x})) $ Note that p in above definition of M is not canonical conjugate momentum otherwise M $=$ S. See Mosel Path integrals in Field theory $\endgroup$
    – aitfel
    Dec 3, 2019 at 14:06

1 Answer 1


Just a tangential lark, not the answer to your question. The ordering issues you have and the fact the x-space eigenfunctions of your hamiltonian are unnormalizable monomials in x may entail complications. Nevertheless, you might well get a grip on the problem by considering an obvious point transformation to y-space.

Set $\hbar=1$, i.e. absorb it in variables of your choice, for simplicity. In that case, $$ \hat H= \hat x \hat p^2 \hat x = (\hat x \hat p)^2 -i\hat x \hat p , $$ whose representation in x-space is but $$ -x\partial_x x \partial_x - x\partial_x ~. $$ Now make a point transformation, $$ \hat x = \exp (\hat y), $$ so, in y-space, the Hamiltonian presents as an almost free (complete the square) Hamiltonian, $$ H= -\partial_y^2 -\partial_y . $$ So work in the y-basis, eigenstates of $\ln \hat x$, $$ \hat y | y\rangle = y | y \rangle . $$ The evolution operator then is all but trivial, $$ e^{-it \hat H}= \int \!\! dy ~ | y\rangle e^{it(\partial_y^2 + \partial_y)} \langle y| , $$ so the exact NR propagator is essentially the free one, $$ \langle y''| e^{-it \hat H} | y'\rangle= \int \!\!dy ~ \delta (y'' -y ) e^{it(\partial_y ^2+ \partial_y)} \delta (y-y') \\ = e^{it(\partial_{y''}^2 + \partial_{y''})} \delta (y''-y') =e^{it\partial_{y''}^2 } ~\delta (y''+it-y') \\ \frac{1}{2\pi} \int \! \! dk ~ e^{-ik^2 t} e^{ik(y''+it-y')}= \frac{e^{i(y''+it-y')^2/4t} }{\sqrt{4\pi it}} ~~ . $$

Of course, in y-space the Lagrangian is trivial, as well.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.