Extracting ground state in path integral

[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&#039; \rightarrow -\infty due to the imaginary term we added. So we have what we want: |0 t&#039;&gt;

But shouldn't the above really be |q&#039;t&#039;&gt;_{f=0}=&lt;0|q&gt; |0 t&#039;&gt;_{f=0}?

It seems what we really need is:

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

 

[Avodyne]

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

 

[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&#039;&#039;t&#039;&#039;|q&#039;t&#039; \rangle_{f}.

For example in terms of path integrals \langle q&#039;&#039;t&#039;&#039;|q&#039;t&#039; \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&#039; \rightarrow -\infty due to the imaginary term we added. So we have what we want: |0 t&#039; \rangle instead of |q&#039;t&#039; \rangle, up to a coefficient of \langle 0|q&#039; \rangle.

But shouldn't the relationship we just derived really be |q&#039;t&#039; \rangle_{f=0}= \langle 0|q&#039; \rangle |0 t&#039; \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&#039;&#039;t&#039;&#039;|q&#039;t&#039; \rangle_{f} =[\langle q&#039;&#039;|0 \rangle \langle 0|q&#039; \rangle ] *\, \,_f \langle 0 t&#039;&#039; |0t&#039; \rangle_f

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