- #1
ianhoolihan
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Dear all,
I have decided to begin working my way through Quantum Field Theory in a Nutshell (Zee), and hopefully thoroughly. Here is one simple question from page 11, relating to the path integral formulation of QM.
Equation (4) is
[tex]<q_F|e^{-iHT}|q_i>=\int{}Dq(t)e^{i\int_0^T dt \frac{1}{2}m{\dot{q}}^2} [/tex]
where
[tex]\int{}Dq(t)=\lim_{N\to\infty}\left(\frac{-im}{2\pi \delta t}\right)^{\frac{N}{2}}\left(\prod_{k=1}^{N-1}\int dq_k\right)[/tex]
and then Zee states that
I was just wondering how one can tell that this is an integral over all possible paths? As far as I can tell, it could just be a multi-dimensional volume integral (i.e. the [itex]\prod_{k=1}^{N-1}\int dq_k[/itex] term).
Could someone please dot my "i"s?
Ianhoolihan
I have decided to begin working my way through Quantum Field Theory in a Nutshell (Zee), and hopefully thoroughly. Here is one simple question from page 11, relating to the path integral formulation of QM.
Equation (4) is
[tex]<q_F|e^{-iHT}|q_i>=\int{}Dq(t)e^{i\int_0^T dt \frac{1}{2}m{\dot{q}}^2} [/tex]
where
[tex]\int{}Dq(t)=\lim_{N\to\infty}\left(\frac{-im}{2\pi \delta t}\right)^{\frac{N}{2}}\left(\prod_{k=1}^{N-1}\int dq_k\right)[/tex]
and then Zee states that
this tells us that to obtain [itex]<q_F|e^{-iHT}|q_i>[/itex] we simply integrate over all possible paths [itex]q(t)[/itex] such that [itex]q(0)=q_i[/itex] and [itex]q(T)=q_F[/itex].
I was just wondering how one can tell that this is an integral over all possible paths? As far as I can tell, it could just be a multi-dimensional volume integral (i.e. the [itex]\prod_{k=1}^{N-1}\int dq_k[/itex] term).
Could someone please dot my "i"s?
Ianhoolihan