I am studying Srednicki book of Quantum Field theory. In chapter 6 regarding the path integral there was derived equation of transition probability for hamiltonian of type:

$$H(\hat{P},\hat{Q})= \frac{1}{2m}\hat{P}^2+V(\hat{Q})\tag{6.1}$$

between two adjacent space points (eq. 6.5):

$$<q_2|e^{-iH\delta t}|q_1> = \int \frac{dp_1}{2 \pi} e^{-i H(p_1,q_1)\delta t} e^{ip_1(q_2 - q_1)}.\tag{6.5}$$

Now here comes the part which I do not understand. When we have a more general Hamiltonian (which have term that contains both operators $\hat{P}$ and $\hat{Q}$ ) Then we have to adopt Weyl-ordering. Srednicki says that this implies that we simply need to replace $H(p_1,q_1)$ with $H(p_1,\bar{q}_1)$ where $\bar{q}_1 = \frac{1}{2}(q_1+q_2)$ so the solution should look like:

$$<q_2|e^{-iH\delta t}|q_1> = \int \frac{dp_1}{2 \pi} e^{-i H(p_1,\bar{q}_1)\delta t} e^{ip_1(q_2 - q_1)}.$$

My questions are:

  1. How to derive the need of use midpoint rule in our solution, when we apply Weyl transformation?

  2. Can you please show me an example of hamiltonian, where not conducting replacement $q_1 \rightarrow \bar{q}_1$ will result in incorrect transition amplitude in the limit of $\delta t \rightarrow 0$ ?


1 Answer 1


Concerning question 1: If we define

$$\hat{U}~:=~ \exp\left[-\frac{i\epsilon}{\hbar}\hat{H}\right]$$

then we know from the Weyl map that

$$\langle q_{+} \mid \hat{U}\mid q_{-}\rangle ~=~ \int \frac{d^np}{(2\pi\hbar)^n}~\exp\left[\frac{i}{\hbar}p(q_{+}-q_{-})\right] U_{W}\left(\frac{q_{+}+q_{-}}{2},p\right).$$

Next use that

$$U_{W}~=~\exp\left[-\frac{i\epsilon}{\hbar}H_W\right] +{\cal O}(\epsilon^2) $$

to derive the sought-for formula [1]. See also this related Phys.SE post.


  1. F. Bastianelli and P. van Nieuwenhuizen, Path Integrals and Anomalies in Curved Space, 2006; eq. (2.1.8).
  • $\begingroup$ Thanks for explanation. So the midpoint apprently comes from the Weyl map definition itself. About the second question I guess it is hard to find suitable Hamiltonian (maybe it does not exist at all) and the only difference is in how fast the solution converges to actual one as $\delta t \rightarrow 0 $ $\endgroup$
    – rodrykbyk
    Jun 21, 2020 at 20:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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