 |
|
| Science Forum Index » Physics - Research Forum » Path integral question... |
|
Page 1 of 1 |
|
| Author |
Message |
| metaweta... |
 Posted: Mon Oct 26, 2009 11:22 am |
|
|
|
Guest
|
In A. Zee's excellent QFT in a nutshell, he starts off by describing
the path integral formulation of quantum mechanics. When he gets to
the part where it's time to calculate
<q_{j+1}| exp(-i H \delta t) |q_j>
he uses the free Hamiltonian H=p^2/2m and then sticks in an integral
over momentum so the time evolution operator is acting on
eigenvectors. This lets him pull out a scalar to the left. He does
the integral and with some appropriate definitions, gets for the
probability amplitude at time T
\int Dq exp( i \int_0^T dt 1/2 m (dq(t)/dt)^2 ).
He then leaves it as an exercise to show that had we used an arbitrary
Hamiltonian H=p^2/2m + V(x), that we would have had the Lagrangian in
the exponent there:
\int Dq exp( i \int_0^T dt [1/2 m (dq(t)/dt)^2 - V(q(t))] ).
Now momentum doesn't commute with energy, so I can't just follow the
same process. What do I do next?
========== Moderator's note =======================================
Have a look at
http://theorie.physik.uni-giessen.de/~hees/publ/lect.pdf |
|
|
| Back to top |
|
|
|
| Igor Khavkine... |
| Posted: Tue Oct 27, 2009 3:03 am |
|
|
|
Guest
|
On Oct 26, 10:22 pm, metaweta <metaw... at (no spam) gmail.com> wrote:
[quote]In A. Zee's excellent QFT in a nutshell, he starts off by describing
the path integral formulation of quantum mechanics. When he gets to
the part where it's time to calculate
<q_{j+1}| exp(-i H \delta t) |q_j
he uses the free Hamiltonian H=p^2/2m and then sticks in an integral
over momentum so the time evolution operator is acting on
eigenvectors. This lets him pull out a scalar to the left. He does
the integral and with some appropriate definitions, gets for the
probability amplitude at time T
\int Dq exp( i \int_0^T dt 1/2 m (dq(t)/dt)^2 ).
He then leaves it as an exercise to show that had we used an arbitrary
Hamiltonian H=p^2/2m + V(x), that we would have had the Lagrangian in
the exponent there:
\int Dq exp( i \int_0^T dt [1/2 m (dq(t)/dt)^2 - V(q(t))] ).
Now momentum doesn't commute with energy, so I can't just follow the
same process. What do I do next?
========== Moderator's note =======================================
Have a look at
http://theorie.physik.uni-giessen.de/~hees/publ/lect.pdf
[/quote]
Hendrik's notes contain a great introduction to path integrals
including the answer to your question. However, for simplicity, I'll
summarize the answer below.
Replace the first expectation value you wrote above by
<q_{j+1}| exp(-i H \delta t) |p_j><p_j|q_j>,
where I've inserted a resolution of identity in the momentum basis |
p_j> (omitting the integration over p_j). As you wrote, you are
already familiar with this step. If H were a function of p alone, then
acting on |p_j> on the right its dependence on p would be replaced by
p_j. For more general Hamiltonians, you have to notice that the same
trick works whenever H can be written as a sum of terms that look like
A(q)B(p), where A and B are arbitrary functions of the operators q and
p. Then, the presence of <q_{j+1}| on the left can be taken advantage
of as well,
<q_{j+1}| A(q)B(p) |p_j> = A(q_{j+1})B(p_j) <q_{j+1}|p_j> .
Both H=p^2/2m and H=p^2/2m-V(q) are of this form, since p^2 = p^2 * 1
and V(q) = 1 * V(q). From here, you can follow the same steps as for
the free particle Hamiltonian to get to the Lagrangian form of the
path integral.
Hope this helps.
Igor |
|
|
| Back to top |
|
|
|
|
|
All times are GMT - 5 Hours
The time now is Mon Dec 07, 2009 1:36 am
|
|