What is the physical significance of <x'|x> in quantum mechanics?

[friend]

Well, let's do the calculation

Thank you, vanhees71. But I don't know what question you were trying to answer.

 

[vanhees71]

I thought there were doubts about the correct expression for the free-particle propagator of a Schrödinger particle in the position representation. So I finally gave the derivation, I told you to do yourself for quite some time...

 

[friend]

I've tried to do a Fourier Xform the old fashion way, and I get pretty much the same think you got except for factors of 2π and a minus sign in the exponent. The difference seems to be that you do not use the 2π in your definition of the Fourier Xform. And you do not use the minus sign in the exponent in the second Fourier Xform. Could you please tell me why you seem to be defining your Fourier transforms differently than what I see in wikipedia.org? Thanks.

In the second FT to get &lt; p|U(t)|{p^\prime } &gt;, I need to have a minus sign in the exponent of the definition of the FT as usual. I need to integrate against e^{ - i2\pi k&#039;x&#039;}. But in the process to get &lt; p|U(t)|{x^\prime } &gt; of the first FT, we calculated F(k) = e^{-ik x&#039; -i \frac{k^2}{2m} t}. However, I need the exponent here to have a positive i2\pi kx&#039; (Here I include the 2\pi). Then I can get the difference of - i2\pi (k - k&#039;)x&#039; in the exponent of the integrand that will enable me to get the \delta (k - k&#039;) term in the second FT. So I have to consider ways to turn the F(k) = e^{-ik x&#039; -i \frac{k^2}{2m} t} into F(k) = e^{ik x&#039; +i \frac{k^2}{2m} t}, at least temporarily.

To that end, I wonder if I have to take the complex conjugate of &lt; p|U(t)|{x^\prime } &gt; before applying the second FT, and then afterwards I conjugate the result? Why would I do that? We had &lt; x|U(t)|{x^\prime } &gt;, with the x on the left and the x' on the right before taking the first FT wrt x. This is consistent with &lt; x|\psi &gt; = \psi (x) being a function of x. So to be consistent with this notation, maybe I need to work with &lt; x&#039;|U(t)|p &gt; when taking the second FT wrt x'. I can always conjugate the end result to get &lt; p|U(t)|{p^\prime } &gt;.

 

[stevendaryl]

If you have \langle x | U(t) | x&#039; \rangle, then you can compute \langle p| U(t)|p&#039;\rangle as follows:

\langle p |U(t) |p&#039; \rangle = \int dx \int dx&#039; \langle p|x \rangle \langle x|U(t)|x&#039;\rangle \langle x&#039;|p&#039;\rangle

where \langle x&#039; | p&#039; \rangle = \sqrt{\frac{1}{2\pi}} e^{i p&#039; x&#039;} and \langle p|x \rangle = (\langle x | p \rangle)^* = \sqrt{\frac{1}{2\pi}} e^{-i p x}
(again, letting \hbar = 1 for simplicity).

I wrote the complete derivation, but it's very long, so I deleted it. The result is:

\langle p |U(t) |p&#039; \rangle = e^{-i \frac{p^2}{2mt}} \delta(p&#039;-p)

 

[friend]

If you have \langle x | U(t) | x&#039; \rangle, then you can compute \langle p| U(t)|p&#039;\rangle as follows:

\langle p |U(t) |p&#039; \rangle = \int dx \int dx&#039; \langle p|x \rangle \langle x|U(t)|x&#039;\rangle \langle x&#039;|p&#039;\rangle

where \langle x&#039; | p&#039; \rangle = \sqrt{\frac{1}{2\pi}} e^{i p&#039; x&#039;} and \langle p|x \rangle = (\langle x | p \rangle)^* = \sqrt{\frac{1}{2\pi}} e^{-i p x}
(again, letting \hbar = 1 for simplicity).

I wrote the complete derivation, but it's very long, so I deleted it. The result is:

\langle p |U(t) |p&#039; \rangle = e^{-i \frac{p^2}{2mt}} \delta(p&#039;-p)

That's great. Thank you. Perhaps you could send me a Private Message with the details (I assumed you saved it since it was a lot of work). Or maybe you could give a word description of your procedure with any tricks you had to use? I could probably figure it out for myself since I've been completing the square lately anyway. But just to be clear what were you using for \langle x | U(t) | x&#039; \rangle ? Thanks.PS. Why does it take so long to load these math heavy pages? It's taking my computer 2 minutes to load and the fan start huffing and buffing during that time. Is there anything I can do to speed things up, download something that makes this process faster?