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I'm having some trouble understanding one part of the derivation of path integral formulation of QM, say I have the propagator, then we can break it up in N parts s.t.

$$[x',t_1;x_0,t_0]=[x';e^{i\textbf{H}\Delta t/\hbar}...._{N\ times}....e^{i\textbf{H}\Delta t/\hbar};x_0], \tag{1}$$

defining $[a;b]$ to be bra-kets, respectively and $\textbf{H}$ our hamiltonian operator, where we're taking particle from point $x_0$ at $t_0$ to point $x'$ at $t_1$ and the intervals are broken giving us $\Delta t=(t_1-t_0)/N$.

Thus, we can rewrite as

$$[x',t_1;x_0,t_0]=[x_N;e^{i\textbf{H}\Delta t/\hbar};x_{N-1}]...[x_1;e^{i\textbf{H}\Delta t/\hbar};x_{0}] \tag{2}$$ For me, we'd have $$[x_N;e^{i\textbf{H}\Delta t/\hbar};x_{N-1}]=\int dx_{N-1}[x_N;e^{i\textbf{H}\Delta t/\hbar};x_{N-1}][x_{N-1};x_{N-1}]; \tag{3}$$ Where I used the identity $$1 = \int dx\ ;x_N][x_N; \tag{4}$$

I think we would have something like this equation after solving (2) in integral form:

$$[x',t_1;x_0,t_0]= $$ $$ \int dx_{N-1}[x_N;e^{i\textbf{H}\Delta t/\hbar};x_{N-1}][x_{N-1};x_{N-1}] \int dx_{N-2}[x_{N-1};e^{i\textbf{H}\Delta t/\hbar};x_{N-2}][x_{N-2};x_{N-2}]... \\ \times\int dx_{0}[x_1;e^{i\textbf{H}\Delta t/\hbar};x_{0}][x_{0};x_{0}] \qquad (5) $$

but books are saying the following:

$$[x',t_1;x_0,t_0]=$$ $$ \int dx_1 ... \int dx_{N-1}[x_N;e^{i\textbf{H}\Delta t/\hbar};x_{N-1}][x_{N-1};e^{i\textbf{H}\Delta t/\hbar};x_{N-2}]...[x_2;e^{i\textbf{H}\Delta t/\hbar};x_{1}] [x_1;e^{i\textbf{H}\Delta t/\hbar};x_{0}] $$

And I quite can't see why.

So here are my questions:

  1. Did I do any mistake in my derivation?

  2. Why last equation holds? Instead of having the propagators inside of each integral.

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    $\begingroup$ Just a note on convention - the standard bra-ket notation typically utilizes the angle brackets $\langle$ (\langle) and $\rangle$ (\rangle) rather than square brackets for state vectors. $\endgroup$
    – J. Murray
    Commented Sep 25, 2017 at 2:14

2 Answers 2

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When you insert the identity operator in between each of your infinitesimal propagators, you need to integrate over all intermediate states. In other words,

$$\langle x_N|e^{-iH\Delta t} e^{-iH \Delta t} \ldots e^{-iH\Delta t}|x_0\rangle =$$ $$ \langle x_N|e^{-iH\Delta t}\left(\int dx_{N-1}|x_{N-1}\rangle\langle x_{N-1}|\right)e^{-iH\Delta t}\left(\int dx_{N-2}|x_{N-2}\rangle\langle x_{N-2}|\right)e^{-iH\Delta t}\ldots |x_{0}\rangle$$ When you performed this step, you did not integrate over all of the intermediate states. I'm not sure exactly what you meant to do - you recycled dummy variables and inserted new sets of states afterward or something.

From there, you can pull all of the integral signs to the left (this doesn't do anything, it just de-clutters the notation) and you find that this equals

$$\int dx_{N-1}\int dx_{N-2}...\int dx_1 \langle x_N|e^{-iH\Delta t}|x_{N-1}\rangle\langle x_{N-1}|e^{-iH\Delta t}|x_{N-2}\rangle\langle x_{N-2}|\ldots|x_1\rangle\langle x_1|e^{-iH\Delta t}|x_0\rangle$$

just as the book claims.

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  • $\begingroup$ Thank you so much, now it makes sense to me why we flush the integrals to the left and the rest can be thrown to the right side of the equation, I was misunderstanding what happens after the first equality, now it's all clear. $\endgroup$
    – Orbitz
    Commented Sep 25, 2017 at 17:15
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I believe that you have a problem in step 2. The $|x_{N-1} \rangle$ that comes out of nowhere in your derivation, actually appears because we use completeness relation (as in your step 4). We insert this integral between two $e^{i\mathbf{H}\Delta t/\hbar}$.

The states are normalized: $\langle i| i \rangle = 1$. It doesn't matter how many of them are present in the final expression.

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