# Extracting ground state in path integral

1. Nov 15, 2014

### geoduck

Suppose you have the transition amplitude in the presence of a source $<q''t''|q't'>_{f}$

To extract the ground state, we change the Hamiltonian to $H-i\epsilon$ , because we can write:
$$|q't'>=e^{iHt'} |n><n|q> \rightarrow e^{iE_0t'} |0><0|q>=<0|q>e^{iHt'} |0>=<0|q> |0 t'>$$

where only the ground state survives when $t' \rightarrow -\infty$ due to the imaginary term we added. So we have what we want: $|0 t'>$

But shouldn't the above really be $|q't'>_{f=0}=<0|q> |0 t'>_{f=0}$?

It seems what we really need is:

$$|q't'>_f=<0|q>e^{iHt'-if(t')xt'} |0>=<0|q>|0t'>_f$$

2. Nov 21, 2014

### Greg Bernhardt

Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?

3. Nov 21, 2014

### Avodyne

I can't follow the equations. Please learn the tex symbols \langle and \rangle.

4. Nov 22, 2014

### geoduck

Absolutely. I also made some mistakes. I hope this is more clear:

Suppose you have the transition amplitude in the presence of a source $\langle q''t''|q't' \rangle_{f}$.

For example in terms of path integrals $\langle q''t''|q't' \rangle_{f} =\int [dx(t)]\, e^{i\{ L+f(t)x(t)\}dt}$

To extract the ground state, we change the Hamiltonian to $H-i\epsilon$ , because we can write:
$$|q't' \rangle=e^{iHt'} \sum_n |n \rangle \langle n|q' \rangle \rightarrow e^{iE_0t'} |0 \rangle \langle 0|q' \rangle= \langle 0|q' \rangle e^{iHt'} |0 \rangle= \langle 0|q' \rangle |0 t' \rangle$$

where only the ground state survives when $t' \rightarrow -\infty$ due to the imaginary term we added. So we have what we want: $|0 t' \rangle$ instead of $|q't' \rangle$, up to a coefficient of $\langle 0|q' \rangle$.

But shouldn't the relationship we just derived really be $|q't' \rangle_{f=0}= \langle 0|q' \rangle |0 t' \rangle_{f=0}$?

It seems what we really want is this relationship:

$$|q't' \rangle_f= \langle 0|q' \rangle e^{i \int \{H-f(t)x\} dt} |0 \rangle= \langle 0|q' \rangle |0t' \rangle_f$$

as it'll allow us to write:

$\langle q''t''|q't' \rangle_{f} =[\langle q''|0 \rangle \langle 0|q' \rangle ] *\, \,_f \langle 0 t'' |0t' \rangle_f$

Does the $H \rightarrow H-i\epsilon\$ work if your Hamiltonian includes a source?