# More Path Integral questions

1. Aug 14, 2009

### maverick280857

Hi everyone,

In chapter 5 of Lewis Ryder's book on QFT, the expression for the propagator as a path integral is derived. Equation 5.7, which is the expression for the propagator over a small path $(q_{j+1} t_{j+1};q_{j}t_{j})$, reads

$$\langle q_{j+1} t_{j+1} |q_{j}t_{j}\rangle = \frac{1}{2\pi\hbar}\int dp \exp{\left[\frac{i}{\hbar}p(q_{j+1}-q_j)\right]} - \frac{i\tau}{\hbar}\langle q_{j+1}|H|q_{j}\rangle$$

where $\tau = t_{j+1}-t_{j}$. This expression holds quite generally, but equation 5.13, which reads

$$\langle q_{f} t_{f} |q_{i}t_{i}\rangle = \int \frac{\mathcal{D}q\mathcal{D}p}{h}\exp{\frac{i}{\hbar}\left[\int dt p\dot{q}-H(p,q)\right]}$$

is derived under the assumption that H is of the form

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

This allows us to express the propagator as a function of the action S[q(t)] in the above expression.

But what if H is not of this form? What does the propagator look like there? I suppose it depends on the specific case (the author points out one example of a Lagrangian $L = f(q)\dot{q}^2/2$ which requires the introduction of an effective action different from $\int L dt$), but are there any general rules or classes of systems where one can write the above expression, but which do not have the canonical form of H given above?

The author also states that Feynman began with the above expression for the propagator, which is not a very rigorous thing to do, given the counterexample in the previous paragraph.

Thanks.

2. Aug 14, 2009

### maverick280857

I got a partial answer on page 281 of Peskin and Schroeder.

3. Aug 14, 2009

### Avodyne

In general the formula is correct if the classical H(p,q) is the Weyl symbol of the quantum hamiltonian, defined as

$$H(p,q)\equiv\int ds\,e^{ips/\hbar}\langle q{-}{\textstyle{1\over2}}s|\hat H|q{+}{\textstyle{1\over2}}s\rangle.$$

4. Aug 15, 2009

### RedX

I think Avodyne is correct, as p^2/(2m)+V(q) is not the most general Hamiltonian you can have. You should be able to have H(p,q)=ap^2+bp+pf(q)+V(q) for constants a, b and arbitrary function f(q). This ought to be the most general Hamiltonian that allows one to safely pass into the Lagrangian scheme. The only example of such a cross term pf(q) I can think of are in 3-gluon vertices in non-Abelian gauge theories (or just boson 3-vertices in general).

Last edited: Aug 15, 2009
5. Aug 15, 2009

### maverick280857

Thanks RedX and Avodyne.