Understanding <x' | x> in Quantum Mechanics: Exploring Its Physical Significance

[vanhees71]

It's the same. I don't know, what you are after.

 

[stevendaryl]

It's the same. I don't know, what you are after.

There is a difference between "something can be measured exactly" and "something can be measured arbitrarily accurately", in that if it can be measured exactly, then you know the value after a finite amount of time has passed, while if it can be measured arbitrarily accurately, you only know the value in the limit as the time spent measuring goes to infinity.

I would say that only when there is a discrete set of possibilities is it possible to measure something exactly.

 

[vanhees71]

Ok, then it must always been read as "arbitarily accurately" since there's no continuous quantity that can be measured exactly in this strict sense of the word. It came never to my mind that somebody could think that you can do such a thing.

 

[stevendaryl]

Ok, then it must always been read as "arbitarily accurately" since there's no continuous quantity that can be measured exactly in this strict sense of the word. It came never to my mind that somebody could think that you can do such a thing.

Well, if you naively associate a measurement with a Hermitian operator and use the rule that a measurement always produces an eigenvalue of the corresponding operator, then it might lead to the conclusion that a measurement of position (which is a Hermitian operator) should result in a position eigenvalue (which is a real number, to infinite precision). More realistically, we shouldn't assume that every Hermitian operator corresponds to a measurement.

 

[vanhees71]

The operators should be even self-adjoint ;-)).

Anyway, of course the measurement of a continuous observable always means that you measure it with a certain finite resolution. For any physicist that's self-evident, and it's not due to quantum theory but as valid within classical physics. So this is an empy debate.

 

[stevendaryl]

The operators should be even self-adjoint ;-)).

Anyway, of course the measurement of a continuous observable always means that you measure it with a certain finite resolution. For any physicist that's self-evident, and it's not due to quantum theory but as valid within classical physics. So this is an empy debate.

But one of the participants here, "friend", was misunderstanding this exact point. So it is relevant to remind people of this.

 

[friend]

Do these transition amplitudes have momentum eigenstates? Is there a transform between position basis and momentum basis? Thanks.

 

[A. Neumaier]

Do these transition amplitudes have momentum eigenstates

Please learn first the language of quantum mechanics before dabbling in it. Only linear operators can haven eigenstates, but transition amplitudes are complex numbers, not operators.

 

[secur]

What is <x'|x> in quantum mechanics? I've seen it, but I don't know what it's suppose to mean in physical terms. It this the amplitude for an unspecified particle to go from x to x' ?

To go back to this, maybe I can provide an intuitive explanation that would be semi-correct and interpret this in terms of a "transition amplitude", or probability.

First since position is (generally) continuous there's zero probability of "transitioning" to x' exactly. So replace it with a finite interval around x', say X'. Suppose x is the expected (in the sense of random variable's expected value) position at time 0: viz. center of its position distribution (assuming symmetrical distribution). Remember this would be a special case of the propagator at time 0 (as Nugatory pointed out) - representing where the particle "could be right now". Then the term <X'|x> - or something like it - is simply the probability of finding it, if we measured now, in the interval X'. That's not unlike a "transition probability" for the particle to "go" from its expected position to X'.

I'm trying to mediate between OP's intuition and everybody else's QM math - probably fall between two stools. Undoubtedly not correct, but if you cut me some slack, does it make sense?

 

[friend]

I was thinking more in terms of how position and momentum are Fourier transforms of each other in the calculation of Heisenberg's Uncertainty principle. See this pdf for details. There the wave-function in position space is a gaussian just as is $< x'|U(t)|x >$ when
$< x'|U(t)|x > = {\left( {\frac{m}{{2\pi \hbar it}}} \right)^{{\raise0.5ex\hbox{\scriptstyle 1} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 2}}}}{e^{im{{(x' - x)}^2}/2\hbar t}}$. So I was asking what its conjugate momentum would look like. I think that's a gaussian as well, right?

 

[friend]

I was thinking more in terms of how position and momentum are Fourier transforms of each other in the calculation of Heisenberg's Uncertainty principle. See this pdf for details. There the wave-function in position space is a gaussian just as is $< x'|U(t)|x >$ when
$< x'|U(t)|x > = {\left( {\frac{m}{{2\pi \hbar it}}} \right)^{{\raise0.5ex\hbox{\scriptstyle 1} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 2}}}}{e^{im{{(x' - x)}^2}/2\hbar t}}$. So I was asking what its conjugate momentum would look like. I think that's a gaussian as well, right?

The article I link to here uses the symbols σ_x and σ_y in their gaussian distributions for the position and momentum. But I seem to be using √(-ħt/2mi) in place of σ_x. Can I still take the Fourier transform of $< x'|U(t)|x >$ and get a momentum?

 

[secur]

The article by Andre Kessler is odd: he derives an "Uncertainty Principle" with only a cosine function, i.e. only the real part of psi. You shouldn't follow that for a model. Uncertainty principle is not that important. either. Follow some standard textbook instead. Note he's also doing a time-independent solution, which is normal.

Whereas you're using the time evolution function, with the regular QM wave function, viz. an exponential with imaginary exponent.

You can take Fourier Transform of anything but instead of <x′|U(t)|x>, normally you FT the time-independent wave function in position coordinates to get momentum representation, and so on. In that case yes, a gaussian distribution transforms to another one. That's called a wave packet.

The expression you're using instead of σx is not directly comparable because of the differences between Kessler's and your approaches.

Start with a standard textbook. You'll get to the proper treatment of Uncertainty Principle, and a lot of other things, within a few chapters.

 

[friend]

The article I link to here uses the symbols σ_x and σ_y in their gaussian distributions for the position and momentum. But I seem to be using √(-ħt/2mi) in place of σ_x. Can I still take the Fourier transform of $< x'|U(t)|x >$ and get a momentum?

So we can at least say that
$< x|U(t)|x' > = {\left( {\frac{m}{{2\pi \hbar it}}} \right)^{{\raise0.5ex\hbox{\scriptstyle 1} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 2}}}}{e^{im{{(x - x')}^2}/2\hbar t}}$
is a function of the variable x. And then according to this page, the Fourier transform of a function of position gives a function of momentum. Then we have from this page that the Fourier transform of a gaussian is another gaussian. I think the only thing left is to show that this new gaussian can be expressed as $< p|U(t)|p' > = {\left( {\frac{m}{{2\pi \hbar it}}} \right)^{{\raise0.5ex\hbox{\scriptstyle 1} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 2}}}}{e^{im{{(p - p')}^2}/2\hbar t}}$ or something like it.

 

[vanhees71]

Somehow everything is messed up now. For a free particle you have
$$\hat{U}(t)=\exp \left (-\frac{\mathrm{i} \hat{p}^2 t}{2m \hbar} \right),$$
and thus
$$\langle p|\hat{U}(t)|p' \rangle=\exp \left (-\frac{\mathrm{i} p^2 t}{2m \hbar} \right) \delta(p-p').$$
Now you can do a Fourier transform. You can regularize the integral by introducing a small imaginary part for $t$ ("infinitesimal Wick rotation"), i.e., you set $t \rightarrow t-\mathrm{i} \epsilon$ with $\epsilon>0$. After the Fourier transform you can make $\epsilon \rightarrow 0^+$.

 

[friend]

Somehow everything is messed up now. For a free particle you have
$$\hat{U}(t)=\exp \left (-\frac{\mathrm{i} \hat{p}^2 t}{2m \hbar} \right),$$
and thus
$$\langle p|\hat{U}(t)|p' \rangle=\exp \left (-\frac{\mathrm{i} p^2 t}{2m \hbar} \right) \delta(p-p').$$
Now you can do a Fourier transform. You can regularize the integral by introducing a small imaginary part for $t$ ("infinitesimal Wick rotation"), i.e., you set $t \rightarrow t-\mathrm{i} \epsilon$ with $\epsilon>0$. After the Fourier transform you can make $\epsilon \rightarrow 0^+$.

It seems the Dirac delta in your post ruins the gaussian nature of $\langle p|\hat{U}(t)|p' \rangle$. So I don't see how your equation is a gaussian which is the Fourier transform of a gaussian.

 

[vanhees71]

Do the calculation, and you'll find the Gaussian in position representation. I can't help it, the math is as it is!

 

[friend]

Do the calculation, and you'll find the Gaussian in position representation. I can't help it, the math is as it is!

In position space we have

$< x|U(t)|x' > = {\left( {\frac{m}{{2\pi \hbar it}}} \right)^{{\raise0.5ex\hbox{\scriptstyle 1} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 2}}}}{e^{im{{(x - x')}^2}/2\hbar t}}$.

which is a gaussian. So as t→0, we have that

$< x|U(t)|x' > \,\,\,\,\, \to \,\,\,\,\, < x|x' > \,\, = \,\,\,\delta (x - x')$.

I'm expecting that we should also have that as some parameter (1/t ?) approaches zero, we would have a gaussian as the Fourier transform of the position gaussian such that

$< p|U(t)|p' > \,\,\,\, \to \,\,\,\, < p|p' > \,\, = \,\,\delta (p - p')$.

Maybe $U(t)$ is not the right operator to insert in momentum space, and I should look for another. What is simply the Fourier transform of

${\left( {\frac{m}{{2\pi \hbar it}}} \right)^{{\raise0.5ex\hbox{\scriptstyle 1} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 2}}}}{e^{im{{(x - x')}^2}/2\hbar t}}$ ?

 

[vanhees71]

I've given the answer in #84. In momentum space it's very simple to evaluate the propagator without much calculation, because the momentum eigenvectors are energy eigenvectors for a free particle. You can simply set $t=0$ in the equation and get $\delta(p-p')$ as it must be. I've also given you the hint, how to do the Fourier transformation from momentum to position space. Of course you can do the same in opposite direction. In any case you have to regularize the propagator before doing so. The reason is that it is not a function but a distribution (generalized function).

 

[friend]

I've given the answer in #84. In momentum space it's very simple to evaluate the propagator without much calculation, because the momentum eigenvectors are energy eigenvectors for a free particle. You can simply set $t=0$ in the equation and get $\delta(p-p')$ as it must be. I've also given you the hint, how to do the Fourier transformation from momentum to position space. Of course you can do the same in opposite direction. In any case you have to regularize the propagator before doing so. The reason is that it is not a function but a distribution (generalized function).

What you seem to have shown is that you can Fourier transform the momentum to get the position. What I'm looking for is how to transform the position to get the momentum. I've tried to do that and got something that's starting to look like your $\langle p|\hat{U}(t)|p' \rangle$ equation in post #84. Give me a little while to write up the math and I'll show my progress, and maybe I can get some help finishing it. Thanks.

 

[vanhees71]

Sure! That's a good exercise. As I said, you have to regularize the integral before you can make sense of the Fourier transform. In this case you can set $m \rightarrow m+\mathrm{i} \epsilon$ with $\epsilon>0$ for $t>0$. At the very end of the Fourier transform (still of a Gaussian with this regularization) you take the limit $\epsilon \rightarrow 0^+$.

 

[friend]

Sure! That's a good exercise. As I said, you have to regularize the integral before you can make sense of the Fourier transform. In this case you can set $m \rightarrow m+\mathrm{i} \epsilon$ with $\epsilon>0$ for $t>0$. At the very end of the Fourier transform (still of a Gaussian with this regularization) you take the limit $\epsilon \rightarrow 0^+$.

Wait! Are you saying that you've already seen the Fourier transform from position to momentum of <x|U(t)|x'>? Is this online somewhere?

 

[friend]

I start with

$< x|U(t)|x' > = f(x) = {\left( {\frac{m}{{2\pi \hbar it}}} \right)^{\frac{1}{2}}}{e^{im{{(x - x')}^2}/2\hbar t}}$.

I want to take the Fourier transform of this function and see if it is the momentum

$\langle p|\hat U(t)|p'\rangle = \exp \left( { - \frac{{{\rm{i}}{p^2}t}}{{2m\hbar }}} \right)\delta (p - p')$.

So the Fourier transform of $f(x)$ is

$\Im [f(x)] = \int_{ - \infty }^{ + \infty } {f(x){e^{ - 2\pi ikx}}dx = } \int_{ - \infty }^{ + \infty } {{{\left( {\frac{m}{{2\pi \hbar it}}} \right)}^{\frac{1}{2}}}{e^{im{{(x - x')}^2}/2\hbar t}} \cdot {e^{ - 2\pi ikx}}dx}$

but,

${e^{ - 2\pi ikx}} = \cos (2\pi kx) - i\sin (2\pi kx)$.

So,

$\Im [f(x)] = \int_{ - \infty }^{ + \infty } {{{\left( {\frac{m}{{2\pi \hbar it}}} \right)}^{\frac{1}{2}}}{e^{im{{(x - x')}^2}/2\hbar t}} \cdot (\cos (2\pi kx) - i\sin (2\pi kx))dx}$

But ${\sin (2\pi kx)}$ is an odd function and the exponential is even. So the integral over symmetrical limits will be zero and all we have left is,

${\left( {\frac{m}{{2\pi \hbar it}}} \right)^{\frac{1}{2}}}\int_{ - \infty }^{ + \infty } {{e^{im{{(x - x')}^2}/2\hbar t}} \cdot \cos (2\pi kx)dx}$

So let's make a change of variable, $u = x - x'$, which means, $x = u + x'$, and $dx = du$. Then the Fourier transform becomes,

${\left( {\frac{m}{{2\pi \hbar it}}} \right)^{\frac{1}{2}}}\int_{ - \infty }^{ + \infty } {{e^{im{{(u)}^2}/2\hbar t}} \cdot \cos (2\pi k(u + x'))du}$

But we have the trigonometric identity that,

$\cos (\alpha \pm \beta ) = \cos \alpha \cdot \cos \beta \mp \sin \alpha \cdot \sin \beta$.

And here $\alpha = 2\pi ku$ and $\beta = 2\pi kx'$ so that the transform becomes,

${\left( {\frac{m}{{2\pi \hbar it}}} \right)^{\frac{1}{2}}}\int_{ - \infty }^{ + \infty } {{e^{im{u^2}/2\hbar t}} \cdot (\cos 2\pi ku \cdot \cos 2\pi kx' - \sin 2\pi ku \cdot \sin 2\pi kx')du}$.

But here again ${\sin 2\pi ku}$ is an odd function so that the integral over symmetric limits would give zero, leaving,

${\left( {\frac{m}{{2\pi \hbar it}}} \right)^{\frac{1}{2}}}\cos (2\pi kx')\int_{ - \infty }^{ + \infty } {{e^{im{u^2}/2\hbar t}}\cos (2\pi ku)du}$

And since the cosine and the exponential are both even, the integral from -∞ to +∞ is twice the integral from 0 to ∞, so the transform is,

$2{\left( {\frac{m}{{2\pi \hbar it}}} \right)^{\frac{1}{2}}}\cos (2\pi kx')\int_0^\infty {{e^{im{u^2}/2\hbar t}}\cos (2\pi ku)du}$.

But we have from the integration tables that

$\int_0^\infty {{e^{ - {a^2}{u^2}}}\cos (bu)du = \frac{{\sqrt \pi }}{{2a}}{e^{ - {b^2}/4{a^2}}}}$,

where here

$a = {\left( { - \frac{{im}}{{2\hbar t}}} \right)^{\frac{1}{2}}}$,

and

$b = 2\pi k$

so that the transform becomes,

$2{\left( {\frac{m}{{2\pi \hbar it}}} \right)^{\frac{1}{2}}}\cos (2\pi kx')\frac{{\sqrt \pi }}{{2{{\left( { - \frac{{im}}{{2\hbar t}}} \right)}^{\frac{1}{2}}}}}{e^{ - {{(2\pi k)}^2}/4\left( { - \frac{{im}}{{2\hbar t}}} \right)}}$

and cancelling terms gives,

$\cos (2\pi kx') \cdot {e^{ - i2\hbar t{\pi ^2}{k^2}/m}}$.

So the question is how do I get this to look like, or function like, what you have

$\langle p|\hat U(t)|p'\rangle = \exp \left( { - \frac{{{\rm{i}}{p^2}t}}{{2m\hbar }}} \right)\delta (p - p')$ ?

I suppose I can get the exponent to look more like it if I use $k = p/\hbar$. But I don't know how I can get the Dirac delta out of the cosine function. Or maybe I don't have to if I do a reverse Fourier transform and discover that I get the original function back again. Any help is appreciated.

 

[vanhees71]

That's way too complicated. As I said, first regularize the integral as said in my previous posting. Then you see that it's a Gaussian integral. Complete the square in the exponential and use the formula for the Gaussian integral!

 

[stevendaryl]

Wait! Are you saying that you've already seen the Fourier transform from position to momentum of <x|U(t)|x'>? Is this online somewhere?

First, note that there are two parameters in $\langle x|U(t)|x'\rangle$: $x$ and $x'$. If you perform a Fourier transform over $x$, you don't get $\langle p|U(t)|p' \rangle$, you get $\langle p | U(t) | x' \rangle$. You have to perform a second Fourier transform, over $x'$, to get $\langle p |U(t) | p' \rangle$.

The quick way to Fourier-transform over $x$ is just to note:

$\langle x | x' \rangle = \delta(x-x') = \frac{1}{2\pi} \int dk e^{ik (x-x')}$

So it's a linear combination of plane waves. If you let a plane wave $e^{i k (x-x')}$ evolve with time, it turns into:

$e^{i k (x-x') - i \frac{k^2}{2m} t}$ (letting $\hbar = 1$). So we have:

$\langle x | U(t) | x' \rangle = \frac{1}{2\pi} \int dk e^{i k (x-x') - i \frac{k^2}{2m} t}$

We can pull out a factor of $e^{-ik x'}$ to get:

$\langle x | U(t) | x' \rangle = \frac{1}{2\pi} \int (e^{-ik x' - i \frac{k^2}{2m} t}) e^{i k x} dk$

This is ALREADY in the form of a Fourier transform. If you have $f(x) = \frac{1}{2\pi} \int F(k) e^{i k x} dk$, then the Fourier transform of $f$ is just $F$. So in our case, $f(x) = \langle x |U(t)|x'\rangle$, and $F(k) = e^{-ik x' -i \frac{k^2}{2m} t}$

So (switching back to $p$ from $k$):

$\langle p|U(t)|x'\rangle = e^{-ip x' - i \frac{p^2}{2m} t}$

This is a Gaussian, but a Gaussian centered on $p = -\frac{m x'}{t}$, not $p = 0$. You can Fourier-transform again over $x'$:

$\langle p |U(t)|p' \rangle = \int dx' e^{i p' x'} e^{-ip x' - i \frac{p^2}{2m} t}$

Factoring out the part that doesn't depend on $x'$ gives:

$\langle p |U(t)|p' \rangle = e^{-i \frac{p^2}{2m}t} \int dx' e^{i (p' - p) x'}$

That integral is a representation for $\delta(p'-p)$. So we get:

$\langle p |U(t)|p' \rangle = 2 \pi e^{-i \frac{p^2}{2m}t} \delta(p'-p)$

If you're wondering why doesn't have the same form (with $x \Rightarrow p$ and $x' \Rightarrow p'$) as $\langle x |U(t)|x' \rangle$, the answer is: $U(t)$ has the form $e^{-i (\hat{p}^2/2m) t}$. So it doesn't treat $\hat{x}$ and $\hat{p}$ symmetrically.

On the other hand, if instead of a free particle, you consider a harmonic oscillator, then $\hat{H} = \frac{1}{2m} \hat{p}^2 + \frac{K}{2} \hat{x}^2$. That is more symmetric between $\hat{x}$ and $\hat{p}$.

 

[friend]

If you let a plane wave $e^{i k (x-x')}$ evolve with time, it turns into:

$e^{i k (x-x') - i \frac{k^2}{2m} t}$ (letting $\hbar = 1$).

Very interesting. Thank you for your response. I didn't quite get where you got the $- i \frac{k^2}{2m} t$ term to get time evolution.

 

[stevendaryl]

Very interesting. Thank you for your response. I didn't quite get where you got the $- i \frac{k^2}{2m} t$ term to get time evolution.

In coordinate representation, if you have a wave function $\psi(x)$ at time $t=0$, then its value at a later time is given by: (again, letting $\hbar= 1$)

$\psi(x,t) = U(t) \psi(x) = e^{-i \hat{H} t} \psi(x) = e^{-i \frac{\hat{p}^2}{2m} t} \psi(x)$

In the case $\psi(x) = e^{ikx}$,

$\hat{p} \psi = k \psi$.

So $e^{-i \frac{\hat{p}^2}{2m} t} \psi = e^{-i \frac{k^2}{2m}} \psi$

An integral is just like a superposition, so:

$U(t) \frac{1}{2\pi} \int dk e^{ikx} = \frac{1}{2\pi} \int dk U(t) e^{ikx} = \frac{1}{2\pi} \int dk e^{-i \frac{k^2}{2m} t} e^{ikx}$

 

[stevendaryl]

The form of $\langle x|U(t)|x'\rangle$ can be obtained from the form $\frac{1}{2\pi} \int dk\ e^{-i \frac{k^2}{2m} t + i k (x-x')}$

Note that $- \frac{i t k^2}{2m} + i k (x-x') = - (\sqrt{\frac{it}{2m}} k - \sqrt{\frac{m}{2 i t}}(x-x'))^2 + \frac{m}{2it} (x-x')^2$

This is what vanhees71 meant by completing the square. So if we make the substitution $u = \sqrt{\frac{it}{2m}} k - \sqrt{\frac{m}{2 i t}}(x-x')$, then that integral becomes:
$\frac{1}{2\pi} e^{\frac{m}{2it}(x-x')^2} \sqrt{\frac{2m}{it}} \int du\ e^{-u^2} = \frac{1}{2\pi} e^{\frac{m}{2it}(x-x')^2} \sqrt{\frac{2m}{it}}\sqrt{\pi}= \sqrt{\frac{m}{2\pi it}}e^{\frac{m}{2it}(x-x')^2}$

 

[friend]

Now I'm not so sure which of $\langle p | U(t) | x' \rangle$ or $\langle p|U(t)|p' \rangle$ I'm interested in. If $\langle x|U(t)|x'\rangle$ represents the transition amplitude of a particle from x' to x, then what is its conjugate momentum? Do I want $\langle p | U(t) | x' \rangle$ because that is its momentum after the transition? Or do I want $\langle p|U(t)|p' \rangle$ because $\langle x|U(t)|x'\rangle$ has an undetermined momentum at both x' and x? Does $\langle x|U(t)|x'\rangle$ have the momentum p' at x' and the momentum p at x ? Or maybe I'm looking for p-p'. Thanks.

 

[vanhees71]

Note that this is the propagator, i.e., a generalized function and not representing a state. It's used to solve the initial value problem for the free Schrödinger equation, i.e., given the wave function at $t=0$, $\psi_0(x)$, the wave function at a later time $t>0$ is given by
$$\psi(t,x)=\int_{\mathbb{R}} \mathrm{d} x' U(t,x,x') \psi_0(x').$$
Of course, you can use any other representation of the time-evolution operator depending on your problem. E.g., if you've given the wave function at $t=0$ in momentum representation, $$\tilde{\psi}_0(p)=\langle p|\psi_0 \rangle,$$ but you want to have the wave function in position representation you just use the appropriate completeness relations
$$\psi(t,x)=\langle x|\psi(t) \rangle=\langle x|\hat{U}(t)|\psi_0 \rangle=\int_{\mathbb{R}} \mathrm{d} p \langle x|\hat{U}(t)|p \rangle \langle p|\psi_0 \rangle.$$
Now
$$U(t,x,p)=\langle x|\exp\left (-\frac{\mathrm{i} \hat{p}^2 t}{2m} \right )|p \rangle=\exp\left (-\frac{\mathrm{i} p^2 t}{2m} \right ) \langle x|p \rangle =\frac{1}{\sqrt{2 \pi}} \exp\left (-\frac{\mathrm{i} p^2 t}{2m} \right ) \exp(\mathrm{i} p x).$$
Thus you get
$$\psi(t,x)=\int_{\mathbb{R}} \mathrm{d} p \frac{1}{\sqrt{2 \pi}} \exp\left (-\frac{\mathrm{i} p^2 t}{2m} \right ) \exp(\mathrm{i} p x) \tilde{\psi}_0(p).$$

 

[friend]

Just as an aside, we can start with

$\psi (t,x) = \int_\mathbb{R} {\text{d}} x'U(t,x,x'){\psi _0}(x')$,

where ${\psi _0}(x')$ is the wave function at the starting time t₀. But it seems arbitrary where t₀ starts. And so we might just as easily assume that

${\psi _0}(x')\,\, = \,\,{\psi _0}({t_0},x')\,\, = \,\,\int_\mathbb{R} {dx''U({t_0},x',x'')} {\psi _{00}}(x'')$,

where ${\psi _{00}}(x'')$ starts at an even earlier time than t₀ and is propagated to t₀ . Since a general propagator is

$\begin{array}{l} U(t,x,x')\,\,\, = \,\,\, < x|{e^{ - iHt/\hbar }}|x' > \,\,\, = \\ \int_{ - \infty }^{ + \infty } {\int_{ - \infty }^{ + \infty } {\int_{ - \infty }^{ + \infty } { \cdot \cdot \cdot \int_{ - \infty }^{ + \infty } { < x|{e^{ - iH\varepsilon /\hbar }}|{x_1} > } } } } < {x_1}|{e^{ - iH\varepsilon /\hbar }}|{x_2} > < {x_2}|{e^{ - iH\varepsilon /\hbar }}|{x_3} > \cdot \cdot \cdot < {x_n}|{e^{ - iH\varepsilon /\hbar }}|x' > d{x_1}d{x_2}d{x_3} \cdot \cdot \cdot d{x_n} \end{array}$,

It seems that integrating this against ${\psi _0}(x')$ would just be a process of inserting even more resolutions of identity to cover the propagation from t₀₀ to t₀ (from x" to x' ).

The point being that ultimately it seems a wave function of a particle will start its propagation from a single point |x> so that <x|U(t)|x'> does represent a wave function in the traditional sense.

 

[vanhees71]

No, $U(t,x,x')$ cannot represent a wave function, because it is not square integrable. It doesn't live in Hilbert space but in the dual of a smaller dense subspace that is the domain of the position operator. This dual space is larger than the Hilbert space and thus contains proper generalized functions like this propagator!

 

[friend]

No, $U(t,x,x')$ cannot represent a wave function, because it is not square integrable. It doesn't live in Hilbert space but in the dual of a smaller dense subspace that is the domain of the position operator. This dual space is larger than the Hilbert space and thus contains proper generalized functions like this propagator!

I think my point was that ${\psi _0}(x')$ can always be put in the form $\int_\mathbb{R} {dx''U({t_0},x',x'')} {\psi _{00}}(x'')$ so that ultimately ${\psi _{00}}(x'')$ approaches (but never equal to) $< x''|{e^{ - iH\varepsilon /\hbar }}|{x_{00}} >$ , if you get my drift.

 

[vanhees71]

I don't understand what you mean. What's $\psi_{00}$ supposed to be?

 

[friend]

I don't understand what you mean. What's $\psi_{00}$ supposed to be?

${\psi _0}(x')$ is supposed to be the initial wave-function at $t=0$, or $t_0$. But it occurs to me that in some circumstances, other events from a previous time could have lead to ${\psi _0}(x')$. In that case we could just as easily write ${\psi _0}(x')\,\, = \,\,\int_\mathbb{R} {dx''U({t_0},x',x'')} {\psi _{00}}(x'')$, where ${\psi _{00}}(x'')$ is the wave function from a previous time and $U({t_0},x',x'')$ propagates it from that previous time to $t_0$. I have to wonder if perhaps this idea can be iterated back further in time yet again. And where can we say those iterations must stop. Must they stop where the iterated initial wave function begins to look like another one of those $< {x_j}|{e^{ - iH\varepsilon /\hbar }}|{x_i} >$ that are introduced by inserting the resolution of identity in yet another propagator as we iterate this process?

I suppose in some circumstances where ${\psi}(t,x)$ oscillates one possibility could be that the ${\psi _{00}}(x'')$ at some time earlier than $t_0$ might be the same as some time after $t_0$. But as I understand it, a particle will start from an infinitesimal point (at least conceptually), in which case doesn't that mean it would start from some |x> ?

 

[friend]

Now I'm not so sure which of $\langle p | U(t) | x' \rangle$ or $\langle p|U(t)|p' \rangle$ I'm interested in. If $\langle x|U(t)|x'\rangle$ represents the transition amplitude of a particle from x' to x, then what is its conjugate momentum? Do I want $\langle p | U(t) | x' \rangle$ because that is its momentum after the transition? Or do I want $\langle p|U(t)|p' \rangle$ because $\langle x|U(t)|x'\rangle$ has an undetermined momentum at both x' and x? Does $\langle x|U(t)|x'\rangle$ have the momentum p' at x' and the momentum p at x ? Or maybe I'm looking for p-p'. Thanks.

This is what I'm really interested in.