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

[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

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

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.

 

[vanhees71]

${\psi _0}(x')$ 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> ?

I think what you mean is the following. Let's write the time-evolution operator in a more general form, assuming a general initial time $t_0$ rather than setting $t_0=0$. Then we have
$$U(t,x;t_0,x')=\langle x|\exp(-\mathrm{i} \hat{H}(t-t_0)|x' \rangle.$$
Since $\hat{H}$ is time-independent we have for $t<t_1<t_0$
$$\int_{\mathbb{R}} \mathrm{d} x'' U(t,x;t_1,x'') U(t_1,x'';t_0,x') = \int_{\mathbb{R}} \mathrm{d} x'' \langle x|\exp(-\mathrm{i} \hat{H} (t-t_1)|x'' \rangle \langle x''|\exp(-\mathrm{i} \hat{H}(t_1-t_0))|x' \rangle.$$
Now using the completeness relation for the integral over $x''$ and
$$\exp(-\mathrm{i} \hat{H}(t-t_1)) \exp(-\mathrm{i} \hat{H}(t_1-t_0)=\exp(-\mathrm{i} \hat{H}(t-t_0)),$$
where the last step is allowed, because $\hat{H}$ is time-independent and commutes with itself, so that we can use the naive factorization rule for exponential function,
we find
$$\int_{\mathbb{R}} \mathrm{d} x'' U(t,x;t_1,x'') U(t_1,x'';t_0,x') = \langle x|\exp(-\mathrm{i} \hat{H}(t-t_0))|x' \rangle=U(t,x;t_0,x').$$

 

[friend]

$$\int_{\mathbb{R}} \mathrm{d} x'' U(t,x;t_1,x'') U(t_1,x'';t_0,x') = \langle x|\exp(-\mathrm{i} \hat{H}(t-t_0))|x' \rangle=U(t,x;t_0,x').$$

Thank you. But I was trying to get at whether I can equate ${\psi _0}(x')$ to the result of propagating an earlier version of the wavefunction through ${\psi _0}(x')\,\, = \,\,\int_\mathbb{R} {dx''U({t_0},x',x'')} {\psi _{00}}(x'')$. I was wondering if I could ultimately push it back to simply a sort of $< {x_j}|{e^{ - iH\varepsilon /\hbar }}|{x_i} >$.

To that end I suppose the question is whether the wave function, $\psi_0(x')$, in $\psi(t,x)=\int_{\mathbb{R}} \mathrm{d} x' U(t,x,x') \psi_0(x')$ can be a very sharp gaussian, maybe something approaching
$< x|{e^{ - i\varepsilon Ht/\hbar }}|x' > = {\left( {\frac{m}{{2\pi \hbar it}}} \right)^{{\raise0.5ex\hbox{\scriptstyle 1}<br /> \kern-0.1em/\kern-0.15em<br /> \lower0.25ex\hbox{\scriptstyle 2}}}}{e^{ - im{{(x' - x)}^2}/2\hbar t}}$ ? Or maybe the more fundamental question is whether a wave function can be a gaussian?

 

[friend]

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 $p = 0$.

So I take it that p here is the momentum at x (not at x' ). I mean, of course it is, right? If we have a function of x, say f(x), then its Fourier transform is a momentum at that x, right? I guess I get confused because we're dealing with QM and it's values that are spread out.

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:

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

And likewise p' is the momentum at x', right?

 

[stevendaryl]

So I take it that p here is the momentum at x (not at x' ). I mean, of course it is, right?

No, there is no "x" in the expression $\langle p|U(t)|x'\rangle$. Roughly speaking, this is the probability amplitude that a particle initially located at $x'$ will be found later to have a momentum $p$.

If we have a function of x, say f(x), then its Fourier transform is a momentum at that x, right?

No, that's not correct at all. The point of a Fourier transform is to represent a function $f(x)$ as a linear combination of functions of the form $e^{ikx}$. You already know, I assume, that sines and cosines can be written as combinations of complex exponentials:

$sin(\frac{2 \pi x}{L}) = \frac{1}{2i} e^{i\frac{2 \pi x}{L}} + \frac{-1}{2i} e^{-i\frac{2\pi x}{L}}$
$cos(\frac{2 \pi x}{L}) = \frac{1}{2} e^{i\frac{2 \pi x}{L}} + \frac{1}{2} e^{-i\frac{2\pi x}{L}}$

Well, any function that is periodic with period $L$ (that is, $f(x + L) = f(x)$) can be written as a combination of exponentials:

$f(x) = A_0 + A_{-1} e^{-i \frac{2\pi x}{L}} + A_1 e^{+i \frac{2\pi x}{L}} + A_{-2} e^{-2 i \frac{2\pi x}{L}} + A_2 e^{+2i \frac{2\pi x}{L}} + ... = \sum_n A_n e^{n i \frac{2 \pi x}{L}}$

The Fourier transform of $f(x)$ is just the coefficients of the expansion of $f$: $A_n$. With a Fourier integral, we replace the sum by an integral, and instead of

$f(x) = \sum_n A_n e^{i n \frac{2 \pi x}{L}}$

we write:

$f(x) = \int dk A(k) e^{i k x}$

(The latter can sort of be thought of as the limit of the first, as $L \rightarrow \infty$). The Fourier transform of $f(x)$ is $A(k)$. That is just a number, (complex, in general). It is not a momentum. How Fourier transforms relate to momenta is this:

If a particle is initially in a state described by the wave function $f(x)$, and you measure its momentum, then the probability that the result will be $p \pm \delta p$ is proportional to $|A(p/\hbar)|^2 \delta p$. The Fourier transform of a wave function doesn't give a momentum, it gives a probability distribution on momenta.

 

[friend]

If a particle is initially in a state described by the wave function $f(x)$, and you measure its momentum, then the probability that the result will be $p \pm \delta p$ is proportional to $|A(p/\hbar)|^2 \delta p$. The Fourier transform of a wave function doesn't give a momentum, it gives a probability distribution on momenta.

Thank you, stevendaryl, for your insights. So let me see if I got it.

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

doesn't say that the particle will necessarily be at x or x', only that there is the probability of finding x at x' . Or does it say (is this equal to saying) that there is a probability of finding x' at x ? It gets a little more difficult from there.

We have

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

seems to be a gaussian distribution in the momentum p centered at p = 0 with what looks like an plane wave of moment p. Or is it a plane wave whose amplitude is modulated by a gaussian?

And then there is

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

which seems to be a gaussian distribution of the moment centered at p=p'. It's harder to tell what this is saying. Is it saying that there is only one momentum being considered, p=p' ? Or do we have two momentums to consider and possibly correlate to the distribution of x and/or x' ?Thanks for your help.

 

[vanhees71]

To repeat it once more! The propagator is a distribution, and it's modulus squared is NOT a probability. Only folding it with a square-integrable wave function, the initial state of the particle, leads to a wave function whose modulus squared gives a probability.

Mathematically the propagator is the unitary time evolution of the quantum system, i.e., it's a unitary map from the initial wave function to the wave function at time $t$.

 

[friend]

Only folding it with a square-integrable wave function, the initial state of the particle, leads to a wave function whose modulus squared gives a probability.

Right. So there seems to be some controversy about whether a gaussian can be a wave function. In my quantum mechanics text, after introducing the algebra of Hilbert spaces, etc. the first wave function they use as an example is a gaussian. But it is not a complex gaussian. So they show that it is square integrable as they normalize it. In this case the exponent of the gaussian is just doubled, then they apply the standard techniques to evaluate it. However, if the exponent of the gaussian were complex, then multiplying it by its complex conjugate will result in an exponent of 0, which means that the probability density is 1 for all x. This is not square integrable. Do you have any insight that might clear this up? Thanks.

 

[stevendaryl]

Right. So there seems to be some controversy about whether a gaussian can be a wave function.

It's definitely true that a Gaussian can be a wave function: For example: $\sqrt{\frac{2\lambda}{\pi}} e^{-\lambda x^2}$ is a perfectly good wave function. The distinction that vanhees71 is making is this: A wave function $f(x)$ is square-integrable. That means $\int |f(x)|^2 dx < \infty$. In the case of the function $f(x) = \sqrt{\frac{m}{2\pi \hbar i t}} e^{\frac{-i m (x-x')^2}{2 \hbar t}}$, if you take the absolute square, you get $\frac{m}{2\pi \hbar t}$, and the integral gives $\infty$. So it can't be a wave function.

 

[friend]

It's definitely true that a Gaussian can be a wave function: For example: $\sqrt{\frac{2\lambda}{\pi}} e^{-\lambda x^2}$ is a perfectly good wave function. The distinction that vanhees71 is making is this: A wave function $f(x)$ is square-integrable. That means $\int |f(x)|^2 dx < \infty$. In the case of the function $f(x) = \sqrt{\frac{m}{2\pi \hbar i t}} e^{\frac{-i m (x-x')^2}{2 \hbar t}}$, if you take the absolute square, you get $\frac{m}{2\pi \hbar t}$, and the integral gives $\infty$. So it can't be a wave function.

It seems that both $\sqrt{\frac{2\lambda}{\pi}} e^{-\lambda x^2}$ and $\sqrt{\frac{m}{2\pi \hbar i t}} e^{\frac{-i m (x-x')^2}{2 \hbar t}}$ are telling us that the amplitude of finding a particle at x has a gaussian distribution. And since they are both gaussian, it would be easy to make a Dirac delta out of either of them. One with a real exponent, the other with a complex exponent. So are these functions equal in the limit where they both become a Dirac delta function? After all, $< x|x' > = \delta (x - x')$. Does it matter how we express that $\delta (x - x')$ ?

 

[stevendaryl]

It seems that both $\sqrt{\frac{2\lambda}{\pi}} e^{-\lambda x^2}$ and $\sqrt{\frac{m}{2\pi \hbar i t}} e^{\frac{-i m (x-x')^2}{2 \hbar t}}$ are telling us that the amplitude of finding a particle at x has a gaussian distribution. And since they are both gaussian, it would be easy to make a Dirac delta out of either of them. One with a real exponent, the other with a complex exponent. So are these functions equal in the limit where they both become a Dirac delta function? After all, $< x|x' > = \delta (x - x')$. Does it matter how we express that $\delta (x - x')$ ?

The normalization is different for the two:
For $\psi(x) = \sqrt{\frac{m}{2\pi \hbar i t}} e^{\frac{-i m (x-x')^2}{2 \hbar t}}$, it is normalized so that $\int \psi(x,t) dx = 1$.

For $\psi(x) = \sqrt{\frac{2\lambda}{\pi}} e^{-\lambda x^2}$, it is normalized so that $\int \psi^*(x) \psi(x) dx = 1$

 

[friend]

So let me summarize my concerns from all this.

$< x|U(t)|x' > \,\, = \,\, < x|{e^{ - iHt/\hbar }}|x' >$

and

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

But we can insert the resolution of the identity many times to get propagator

$< x|{e^{ - iHt/\hbar }}|x' > \,\, = \,\,\int_{ - \infty }^{ + \infty } {\int_{ - \infty }^{ + \infty } {\int_{ - \infty }^{ + \infty } { \cdot \cdot \cdot \int_{ - \infty }^{ + \infty } { < x|{e^{ - i\varepsilon Ht/\hbar }}|{x_1} > } } } } < {x_1}|{e^{ - i\varepsilon Ht/\hbar }}|{x_2} > < {x_2}|{e^{ - i\varepsilon Ht/\hbar }}|{x_3} > \cdot \cdot \cdot < {x_n}|{e^{ - i\varepsilon Ht/\hbar }}|x' > d{x_1}d{x_2}d{x_3} \cdot \cdot \cdot d{x_n}$

vanhess71 in post #111 tells us that the propagator is a distribution and its modulus squared is not a probability. It needs to be folded (integrated) with a square-integrable functions. But I have the intuition that the propagator, all by itself, has some physical interpretation. It's called the transition amplitude for a particle to go from x' to x. But if it is not square-integrable, then it is not a wave function.

So I tried to suggest that maybe the wave function that is folded with propagator might be the result of an even earlier wave function folded with a prior propagator. Is it possible to then find even earlier propagators until the wave function it is folded with is actually of the form of $< {x_n}|{e^{ - i\varepsilon Ht/\hbar }}|x' >$ ?

But according to stevendaryl in post #97,
$< x|{e^{ - iHt/\hbar }}|x' > \,\,\, = \,\, < x|U(t)|x' > \,\, = \,\,{\left( {\frac{m}{{2\pi \hbar it}}} \right)^{{\raise0.5ex\hbox{\scriptstyle 1}<br /> \kern-0.1em/\kern-0.15em<br /> \lower0.25ex\hbox{\scriptstyle 2}}}}{e^{ - im{{(x - x')}^2}/2\hbar t}}$.

This means that
$< x|{e^{ - i\varepsilon Ht/\hbar }}|x' > \,\,\,\, = \,\,{\left( {\frac{m}{{2\pi \hbar it}}} \right)^{{\raise0.5ex\hbox{\scriptstyle 1}<br /> \kern-0.1em/\kern-0.15em<br /> \lower0.25ex\hbox{\scriptstyle 2}}}}{e^{ - i\varepsilon m{{(x - x')}^2}/2\hbar t}}$

which is in the form of a gaussian.

And stevendaryl assures us in post #113 that a gaussian is a legitimate wave function that can be folded against the propagator. But
${\left( {\frac{m}{{2\pi \hbar it}}} \right)^{{\raise0.5ex\hbox{\scriptstyle 1}<br /> \kern-0.1em/\kern-0.15em<br /> \lower0.25ex\hbox{\scriptstyle 2}}}}{e^{ - i\varepsilon m{{(x - x')}^2}/2\hbar t}}$ has a complex exponent. So it is not square-integrable to form a probability. How do we get a gaussian with a real exponent so it can be a proper wave function (in a gaussian from) that can be folded with a propagator to yeild a wave function that can give a probability?

Let me suggest this possibility for your consideration:

Let's try to regulate the first $< {x_n}|{e^{ - i\varepsilon Ht/\hbar }}|x' >$ by letting $\varepsilon \,\, = \,\,{\varepsilon _r} + i\delta$ . Then

$< {x_n}|{e^{ - i({\varepsilon _r} + i\delta )Ht/\hbar }}|x' > \,\,\,\, = \,\,{\left( {\frac{m}{{2\pi \hbar it}}} \right)^{{\raise0.5ex\hbox{\scriptstyle 1}<br /> \kern-0.1em/\kern-0.15em<br /> \lower0.25ex\hbox{\scriptstyle 2}}}}{e^{ - i({\varepsilon _r} + i\delta )m{{({x_n} - x')}^2}/2\hbar t}}$ .

As you can see there is now a gaussian with a real exponent. Can this serve the function of a proper wave function to fold against the propagator in order to yield a time advanced wave function that will give a probability? Can we work through all the math with $\delta$ not equal to zero until we get a formula for a probability and then take the limit as $\delta \to 0$ ?

 

[friend]

Let's try to regulate the first $< {x_n}|{e^{ - i\varepsilon Ht/\hbar }}|x' >$ by letting $\varepsilon \,\, = \,\,{\varepsilon _r} + i\delta$ . Then...

Maybe this is equivalent to proving that we necessarily must start with a distributed wave function (not a Dirac delta) in order that any propagated wave function would result in non-zero probabilities. Perhaps if we try to start our propagation from a single point it always results in a zero probability density of finding it at a later point.

 

[friend]

$\langle x | U(t) | x' \rangle = \frac{1}{2\pi} \int (e^{-ik x' - i \frac{k^2}{2m} t}) e^{i k x} dk$You can Fourier-transform again over $x'$:

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

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 of the FT. 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.

 

[stevendaryl]

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 of the FT. 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.

I may have made a mistake. There are different conventions for the Fourier transform and its inverse.

 

[vanhees71]

Well, let's do the calculation
$$U(t,x,x')=\langle x|\exp \left (-\frac{\mathrm{i} \hat{p}^2 t}{2m} \right)|x' \rangle = \int_{\mathbb{R}} \mathrm{d} p \langle x|\exp \left (-\frac{\mathrm{i} \hat{p}^2 t}{2m} \right)|p \rangle \langle p|x' \rangle = \int_{\mathbb{R}} \mathrm{d} p \frac{1}{2 \pi}\exp \left [-\frac{\mathrm{i} p^2 t}{2m}+\mathrm{i} p (x-x') \right].$$
Es explained countless times in this thread, the integral doesn't make sense as it stands but has to be regularized this we do by introducing a small imaginary part into $t$: $t \rightarrow t-\mathrm{i} \epsilon$ with $\epsilon>0$. Then you have a simple Gaussian integral, leading to
$$U_{\epsilon}(t,x,x')=\frac{1}{\sqrt{2 \pi}} \exp \left (-\frac{m(x-x')^2}{2(\epsilon+\mathrm{i} t)} \right ) \frac{1}{\sqrt{(\epsilon+\mathrm{i} t)/m}}.$$
For $\epsilon \rightarrow 0^+$ you have $\sqrt{\epsilon+\mathrm{i} t} \rightarrow \exp(\mathrm{i} \pi/4) \sqrt{t}$ and thus
$$U(t,x,x')=\sqrt{\frac{m}{2 \pi t}} \exp(-\mathrm{i} \pi/4) \exp \left (\frac{\mathrm{i} m(x-x')^2}{2 \mathrm{i} t} \right ).$$