Suppose you have the transition amplitude in the presence of a source [itex] <q''t''|q't'>_{f} [/itex](adsbygoogle = window.adsbygoogle || []).push({});

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:

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

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# Extracting ground state in path integral

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