Extracting ground state in path integral

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geoduck
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Suppose you have the transition amplitude in the presence of a source [itex]<q''t''|q't'>_{f}[/itex]

To extract the ground state, we change the Hamiltonian to [itex]H-i\epsilon[/itex] , 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 [itex]t' \rightarrow -\infty[/itex] due to the imaginary term we added. So we have what we want: [itex]|0 t'>[/itex]

But shouldn't the above really be [itex]|q't'>_{f=0}=<0|q> |0 t'>_{f=0}[/itex]?

It seems what we really need is:

$$|q't'>_f=<0|q>e^{iHt'-if(t')xt'} |0>=<0|q>|0t'>_f $$
 
Absolutely. I also made some mistakes. I hope this is more clear:

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

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

To extract the ground state, we change the Hamiltonian to [itex]H-i\epsilon[/itex] , 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 [itex]t' \rightarrow -\infty[/itex] due to the imaginary term we added. So we have what we want: [itex]|0 t' \rangle[/itex] instead of [itex]|q't' \rangle[/itex], up to a coefficient of [itex]\langle 0|q' \rangle[/itex].

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

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:

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

Does the [itex]H \rightarrow H-i\epsilon\[/itex] work if your Hamiltonian includes a source?