- #1
maverick280857
- 1,789
- 5
Hi again everyone,
I have some doubts about the path integral expressions given in Section 9.1 of Peskin and Schroeder (pg 281 and 282).
For a Weyl ordered Hamiltonian H, the propagator has the form given by equation 9.11, which reads
[tex]U(q_{0},q_{N};T) = \left(\prod_{i,k}\int dq_{k}^{i}\int \frac{dp_{k}^{i}}{2\pi}\right)\exp{\left[i\sum_{k}\left(\sum_{i}p_{k}^{i}(q_{k+1}^{i}-q_{k}^{i})-\epsilon H\left(\frac{q_{k+1}+q_{k}}{2},p_{k}\right)\right)\right]}[/tex]
Now, as the authors point out, for the case when [itex]H = \frac{p^2}{2m} + V(q)[/tex], we can do the momentum integral, which is (taking the potential term out)
[tex]\int\frac{dp_{k}}{2\pi}\exp{\left(i\left[p_k(q_{k+1}-q_{k})-\epsilon\frac{p_{k}^2}{2m}\right]\right) = \frac{1}{C(\epsilon)}\exp{\left[\frac{im}{2\epsilon}(q_{k+1}-q_{k})^2\right]}[/tex]
where
[tex]C(\epsilon) = \sqrt{\frac{2\pi\epsilon}{-im}}[/tex]
Now, I do not understand how the distribution of factors [itex]C(\epsilon)[/itex] equation 9.13 (given below) comes up. To quote the authors:
So, my question is: how did we get this term:
[tex]\left(\frac{1}{C(\epsilon)}\prod_{k}\int\frac{dq_{k}}{C(\epsilon)}\right)[/tex]
PS -- Is it because the momentum index goes from 0 to N-1 and the coordinate index goes from 1 to N-1? The momentum integral product produces N terms, so to write the coordinate integral with the same indexing as before (in the final expression), i.e. k = 1 to N-1, we factor a [itex]C(\epsilon)[/itex] out?
Also, in equation 9.12,
[tex]U(q_{a},q_{b};T) = \left(\prod_{i}\int \mathcal{D}q(t)\mathcal{D}p(t)\right)\exp{\left[i\int_{0}^{T}dt\left(\sum_{i}p^{i}\dot{q}^{i}-H(q,p)\right)\right]}[/tex]
shouldn't we just write
[tex]\left(\int \mathcal{D}q(t)\mathcal{D}p(t)\right)[/tex]
instead of
[tex]\left(\prod_{i}\int \mathcal{D}q(t)\mathcal{D}p(t)\right)[/tex]
since the [itex]\mathcal{D}[/itex] itself stands for [itex]\prod[/itex]?
I have some doubts about the path integral expressions given in Section 9.1 of Peskin and Schroeder (pg 281 and 282).
For a Weyl ordered Hamiltonian H, the propagator has the form given by equation 9.11, which reads
[tex]U(q_{0},q_{N};T) = \left(\prod_{i,k}\int dq_{k}^{i}\int \frac{dp_{k}^{i}}{2\pi}\right)\exp{\left[i\sum_{k}\left(\sum_{i}p_{k}^{i}(q_{k+1}^{i}-q_{k}^{i})-\epsilon H\left(\frac{q_{k+1}+q_{k}}{2},p_{k}\right)\right)\right]}[/tex]
Now, as the authors point out, for the case when [itex]H = \frac{p^2}{2m} + V(q)[/tex], we can do the momentum integral, which is (taking the potential term out)
[tex]\int\frac{dp_{k}}{2\pi}\exp{\left(i\left[p_k(q_{k+1}-q_{k})-\epsilon\frac{p_{k}^2}{2m}\right]\right) = \frac{1}{C(\epsilon)}\exp{\left[\frac{im}{2\epsilon}(q_{k+1}-q_{k})^2\right]}[/tex]
where
[tex]C(\epsilon) = \sqrt{\frac{2\pi\epsilon}{-im}}[/tex]
Now, I do not understand how the distribution of factors [itex]C(\epsilon)[/itex] equation 9.13 (given below) comes up. To quote the authors:
Notice that we have one such factor for each time slice. Thus we recover expression (9.3), in discretized form, including the proper factors of [itex]C[/itex]:
[tex]
U(q_{a},q_{b};T) = \left(\frac{1}{C(\epsilon)}\prod_{k}\int\frac{dq_{k}}{C(\epsilon)}\right)\exp\left[i\sum_{k}\left(\frac{m}{2}\frac{(q_{k+1}-q_{k})^2}{\epsilon}-\epsilon V\left(\frac{q_{k+1}+q_{k}}{2}\right)\right)\right]
[/tex]
So, my question is: how did we get this term:
[tex]\left(\frac{1}{C(\epsilon)}\prod_{k}\int\frac{dq_{k}}{C(\epsilon)}\right)[/tex]
PS -- Is it because the momentum index goes from 0 to N-1 and the coordinate index goes from 1 to N-1? The momentum integral product produces N terms, so to write the coordinate integral with the same indexing as before (in the final expression), i.e. k = 1 to N-1, we factor a [itex]C(\epsilon)[/itex] out?
Also, in equation 9.12,
[tex]U(q_{a},q_{b};T) = \left(\prod_{i}\int \mathcal{D}q(t)\mathcal{D}p(t)\right)\exp{\left[i\int_{0}^{T}dt\left(\sum_{i}p^{i}\dot{q}^{i}-H(q,p)\right)\right]}[/tex]
shouldn't we just write
[tex]\left(\int \mathcal{D}q(t)\mathcal{D}p(t)\right)[/tex]
instead of
[tex]\left(\prod_{i}\int \mathcal{D}q(t)\mathcal{D}p(t)\right)[/tex]
since the [itex]\mathcal{D}[/itex] itself stands for [itex]\prod[/itex]?
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