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

[vanhees71]

Probabilities predicted in QT are probabilities for finding a certain value when measuring an observable and not that some mathematical auxilliary quantity like $|x \rangle$ occurs. The uncertainty principle clearly tells you that there are no proper eigenvectors of position or momentum operators.

 

[atyy]

I find it strange to insist that there is no physical content to |x>. For we certainly consider it legitimate to write <x|ψ> = ψ(x). If <x| is meaningful in <x|ψ>, then shouldn't we expect that its dual, |x> is also physically meaningful.

This is the very point of the entire discussion! A distribution has physical meaning only when folded with a test function. Note that $|\psi \rangle$ must be in the domain of the position operator which is a dense subspace of the Hilbert space, so that $\langle x|\psi \rangle$ is well-defined!

Well, it could be that position can be exactly measured, so <x|ψ> does make physical sense. The only thing is that after a position measurement, the state cannot collapse to a delta function, since that is not a physical state. So we can have exact position measurement, and we can have collapse after that, but to some other state (consistent with a generalized collapse rule, rather than the textbook projection postulate which cannot deal with continuous variables).

 

[vanhees71]

Position CAN bei exactly measured. It cannot be exactly PREPARED! Since there is no collapse, the no-collapse argument is always true ;-)).

 

[atyy]

Position CAN bei exactly measured. It cannot be exactly PREPARED!

Well said!

Since there is no collapse, the no-collapse argument is always true ;-)).

Here's one way in which one can have collapse after the sharp position measurement:
http://arxiv.org/abs/0706.3526 (edited link)
2.3.2 Ozawa's model of a sharp position measurement"

 

[A. Neumaier]

Position CAN be exactly measured.

How do you measure position exactly (i.e., to an infinite number of decimal places)? I haven't seen any experimental equipment that could do it; not even in a thought experiment. Already the readout of the measurement result would take an infinite time...

 

[friend]

How do you measure position exactly (i.e., to an infinite number of decimal places)? I haven't seen any experimental equipment that could do it; not even in a thought experiment. Already the readout of the measurement result would take an infinite time...

By that logic, we could never confirm theory since we could never measure accurately enough.

Just because QT does not predict |x> doesn't mean that it doesn't happen. Something else is giving rise to particles with which quantum theory works. QFT is giving rise to the particles whose propagation is predicted by QM. And even QFT does not predict where those particles will pop into existence. But it does seem that particles are starting out at some exact location, an atom or nucleus. And the collapse tells us that it is ending exactly somewhere. As I understand it, the collapse is to a point, not to a region.

 

[stevendaryl]

By that logic, we could never confirm theory since we could never measure accurately enough.

Well, and that's true. You never confirm a theory. What you can do is disconfirm (or falsify) a theory by showing that the measured result differs from the predicted result by an amount that is greater than can be plausibly accounted for by measurement imprecision or chance.

 

[friend]

I believe that Quantum Theory predicts that the wave function collapses to a point, not a region. That at least confirms that the <x'| portion of <x'|x> is a physical prediction. So if <x'| is confirmed, then so is its dual |x>

 

[stevendaryl]

I believe that Quantum Theory predicts that the wave function collapses to a point, not a region. That at least confirms that the <x'| portion of <x'|x> is a physical prediction. So if <x'| is confirmed, then so is its dual |x>

No, quantum theory doesn't predict that the wave function collapses to a point. The collapse hypothesis is probably better thought of as a rule of thumb, rather than a necessary postulate of quantum (because there is substantial debate over whether collapse happens at all).

That at least confirms that the <x'| portion of <x'|x> is a physical prediction. So if <x'| is confirmed, then so is its dual |x>

The use of |\phi\rangle and \langle \psi | is just notation. You can't make a deduction about what is sensible based on the fact that you can express it using the notation.

In the mathematics of (single-particle, nonrelativistic, spin-zero, in one spatial dimension for simplicity) quantum mechanics, the possible states of a particle are assumed to be the set of complex-valued functions \psi(x) such that \int dx |\psi(x)|^2 &lt; \infty. In the Dirac notation, |\psi\rangle represents such a function.

Those are the "kets" (sort of a joke based on the word "bracket"). A "bra" is any linear function on "kets" that returns a complex number. One example of a "bra" is the function \langle x| that takes |\psi\rangle and returns \psi(x). That is, it takes the function \psi and returns the value of that function at the location x. Another example is the Fourier transform, the function \langle k| that takes a function \psi and returns \int dx \psi(x) e^{-i k x}. In the Dirac notation, the notation \langle F|B\rangle means the result of applying the bra F to the ket |B\rangle.

Now, there is a special type of "bra" which can be obtained from a ket: If |\phi\rangle is any ket, then you can define the corresponding bra \langle \phi | to be that function that takes a ket |\psi\rangle and returns \int dx \phi^*(x) \psi(x).

Note that not every bra comes from a ket. So the existence of a bra \langle F| does not imply that there is a corresponding ket |F\rangle. The two most common ones are \langle x| and \langle k|. They do not correspond to any state in the Hilbert space.

Now, often people do use the Dirac notation to represent what seems like the duals to \langle x| and \langle k|. They define |k\rangle to be the ket corresponding to the function e^{i k x}, and they define |x&#039;\rangle to be the ket corresponding to the "function" \delta(x-x&#039;). But that's playing loose with the notation. Those are not elements of the hilbert space, and are not possible states of a particle.

 

[friend]

No, quantum theory doesn't predict that the wave function collapses to a point. The collapse hypothesis is probably better thought of as a rule of thumb, rather than a necessary postulate of quantum (because there is substantial debate over whether collapse happens at all).

Whatever you call it, collapse, many worlds, etc., QT predicts a probability distribution, but the result of measurement is only one of those possibilities, even when the spectrum is continuous. Also, the way we prepare a particle in a state is to do a measurement. So measuring the position prepares the particle in that state. We may not be able to prepare a state in an exact position because the probability density to measure in that state is infinitesimal. But that doesn't preclude the fact that a measurement of position will put it in an exact position.

 

[stevendaryl]

Whatever you call it, collapse, many worlds, etc., QT predicts a probability distribution, but the result of measurement is only one of those possibilities, even when the spectrum is continuous.

I disagree with that. Here's a way of thinking of wave function collapse that sort of explains how a measurement works:

Suppose you have a system described by a wave function |\psi\rangle. Then you make an observation of the system. Then what you can do is to rewrite the wave function as the sum of two parts: |\psi\rangle = \alpha |\psi_{yes}\rangle + \beta |\psi_{no}\rangle, where |\psi_{yes}\rangle and |\psi_{no}\rangle are orthogonal states such that |\psi_{yes}\rangle is consistent with your observation, and |\psi_{no}\rangle is not. Then wave function collapse amounts to the replacement of the full |\psi\rangle by just |\psi_{yes}\rangle. You're not going to get a state with a precise value for an observable O unless your observation is only consistent with that one, precise value.

If the observable has a discrete spectrum (such as spin, or the energy of a bound system), then it's possible to have an observation that uniquely determines the observable's value. But if the observable is continuous, then a single observation can't possibly determine the precise value of the observable, and so your observation is consistent with a range of values, and so the "collapse" will result in a state with a range of values.

 

[friend]

But if the observable is continuous, then a single observation can't possibly determine the precise value of the observable, and so your observation is consistent with a range of values, and so the "collapse" will result in a state with a range of values.

Now we're talking about meter accuracy as opposed to what theory predicts. Does theory predict a collapse to only one particular value in the continuous spectrum case? Perhaps this question gets to whether distributions have some inherent existence in and of themselves? Or are distributions always made up of individual samples that are the thing that inherently exist?

 

[A. Neumaier]

Does theory predict a collapse to only one particular value in the continuous spectrum case?

No. Definitely not.

 

[stevendaryl]

Now we're talking about meter accuracy as opposed to what theory predicts. Does theory predict a collapse to only one particular value in the continuous spectrum case?

As I said, collapse is a contentious part of the quantum formalism. Von Neumann's collapse hypothesis was that if you measure \hat{O} and get the value \lambda, then the system will collapse into an eigenstate of \hat{O} with eigenvalue \lambda. But my claim is that there is no way to measure position. The best you can do is to measure a range of positions. So instead of an operator \hat{x}, you might use some operator \hat{x}_{\delta x} which returns a discrete set of possibilities:

• 0, meaning that the particle is somewhere between x=-\delta x/2 and x=+\delta x/2
• 1, meaning that the particle is somewhere between x=\delta x/2 and x = 3/2 \delta x
• etc.

Measuring the system to have \hat{x}_{\delta x} = n would then cause the wave function to collapse from \psi(x) to \psi_n(x), where

1. \psi_n(x) = C \psi(x), when (n-1/2) \delta x &lt; x &lt; (n + 1/2) \delta x
2. \psi_n(x) = 0, otherwise.

where C would be chosen so that \int_{(n-1/2)\delta x}^{(n+1/2)\delta x} dx C^2 |\psi(x)|^2 = 1

 

[friend]

Now we're talking about meter accuracy as opposed to what theory predicts. Does theory predict a collapse to only one particular value in the continuous spectrum case? Perhaps this question gets to whether distributions have some inherent existence in and of themselves? Or are distributions always made up of individual samples that are the thing that inherently exist?

I think this is the particle/wave debate. I'm understanding that particles are point particles with 0 radius. So if there are such things as particles, then they exist at a particular point.

 

[stevendaryl]

I think this is the particle/wave debate. I'm understanding that particles are point particles with 0 radius. So if there are such things as particles, then they exist at a particular point.

Saying that something is a point-particle is to say that it has no internal structure (unlike a proton, which has quarks inside it). Whether that means that it exists at a point or not is dependent on the interpretation of QM you are using. But it's never the case that a particle is in a definite state of position. There's no way to put it into such a state, and there is no way to detect it as being in such a state.

 

[A. Neumaier]

Measuring the system to have $\hat{x}_{\delta x} = n$ would then cause

But then the state after the measurement would depend on which accuracy you ascribe to your position measurement...

In reality, the analysis of concrete measurements, the best result to be assigned, and the determination of final state are complicated calibration issues, and the talk about measurement in the foundations is a heavy idealization of the real situation. People in quantum optics who have ot estimate the efficiency and accuracy of what they do always work with density matrices rather than pure states, and employ Lindblad equations or rather than crude collapse arguments to model the dynamics of the state.

 

[friend]

Saying that something is a point-particle is to say that it has no internal structure (unlike a proton, which has quarks inside it). Whether that means that it exists at a point or not is dependent on the interpretation of QM you are using. But it's never the case that a particle is in a definite state of position. There's no way to put it into such a state, and there is no way to detect it as being in such a state.

Which brings up the other question about QM. Does QM tell us what actually is or only what we can measure. Perhaps the "actually is" are the particles. And what we can measure is the distribution of the waves.

What I'm beginning to think is that there are such things as point particles, but we can never measure exactly where they are because their position is continually being passed around between the virtual particles that surround it so that the uncertainty of its position accumulates over time. And in this view particles are true points, and things like charge, mass, and spin are caused by how it interacts with the surrounding virtual particles.

 

[vanhees71]

There's nothing that prevents measuring the position of a quantum as accurately as you like, at least not in principle. It may be difficult to measure a position very accurately, but it's not impossible. To verify the standard deviation $\Delta x$ for a given state you have to measure position on an ensemble of equally prepared particles much more accurate than $\Delta x$. The same holds true for $\Delta p$, and you can never measure momentum and position accurately on the same particle, but you can prepare an ensemble to measure position very accurately and then another ensemble to measure momentum very accurately to check the uncertainty relation $\Delta x \Delta p \geq 1/2$ (this is for non-relativistic QT, for the relativistic case, see the first chapter of Landau Lifshitz vol. 4).

 

[A. Neumaier]

There's nothing that prevents measuring the position of a quantum as accurately as you like

But that's quite different from your earlier claim (in post #53) that one can measure it exactly.

 

[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 &lt; x&#039;|U(t)|x &gt; when
&lt; x&#039;|U(t)|x &gt; = {\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&#039; - 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 &lt; x&#039;|U(t)|x &gt; when
&lt; x&#039;|U(t)|x &gt; = {\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&#039; - 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 &lt; x&#039;|U(t)|x &gt; 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 &lt; x&#039;|U(t)|x &gt; and get a momentum?

So we can at least say that
&lt; x|U(t)|x&#039; &gt; = {\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&#039;)}^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 &lt; p|U(t)|p&#039; &gt; = {\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{{(p - p&#039;)}^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

&lt; x|U(t)|x&#039; &gt; = {\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&#039;)}^2}/2\hbar t}}.

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

&lt; x|U(t)|x&#039; &gt; \,\,\,\,\, \to \,\,\,\,\, &lt; x|x&#039; &gt; \,\, = \,\,\,\delta (x - x&#039;).

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

&lt; p|U(t)|p&#039; &gt; \,\,\,\, \to \,\,\,\, &lt; p|p&#039; &gt; \,\, = \,\,\delta (p - p&#039;).

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{$\scriptstyle 1$}<br /> \kern-0.1em/\kern-0.15em<br /> \lower0.25ex\hbox{$\scriptstyle 2$}}}}{e^{im{{(x - x&#039;)}^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

&lt; x|U(t)|x&#039; &gt; = f(x) = {\left( {\frac{m}{{2\pi \hbar it}}} \right)^{\frac{1}{2}}}{e^{im{{(x - x&#039;)}^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&#039;\rangle = \exp \left( { - \frac{{{\rm{i}}{p^2}t}}{{2m\hbar }}} \right)\delta (p - p&#039;).

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&#039;)}^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&#039;)}^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&#039;)}^2}/2\hbar t}} \cdot \cos (2\pi kx)dx}

So let's make a change of variable, u = x - x&#039;, which means, x = u + x&#039;, 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&#039;))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&#039; 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&#039; - \sin 2\pi ku \cdot \sin 2\pi kx&#039;)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&#039;)\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&#039;)\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&#039;)\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&#039;) \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&#039;\rangle = \exp \left( { - \frac{{{\rm{i}}{p^2}t}}{{2m\hbar }}} \right)\delta (p - p&#039;) ?

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&#039;\rangle: x and x&#039;. If you perform a Fourier transform over x, you don't get \langle p|U(t)|p&#039; \rangle, you get \langle p | U(t) | x&#039; \rangle. You have to perform a second Fourier transform, over x&#039;, to get \langle p |U(t) | p&#039; \rangle.

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

\langle x | x&#039; \rangle = \delta(x-x&#039;) = \frac{1}{2\pi} \int dk e^{ik (x-x&#039;)}

So it's a linear combination of plane waves. If you let a plane wave e^{i k (x-x&#039;)} evolve with time, it turns into:

e^{i k (x-x&#039;) - i \frac{k^2}{2m} t} (letting \hbar = 1). So we have:

\langle x | U(t) | x&#039; \rangle = \frac{1}{2\pi} \int dk e^{i k (x-x&#039;) - i \frac{k^2}{2m} t}

We can pull out a factor of e^{-ik x&#039;} to get:

\langle x | U(t) | x&#039; \rangle = \frac{1}{2\pi} \int (e^{-ik x&#039; - 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&#039;\rangle, and F(k) = e^{-ik x&#039; -i \frac{k^2}{2m} t}

So (switching back to p from k):

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

This is a Gaussian, but a Gaussian centered on p = -\frac{m x&#039;}{t}, not p = 0. You can Fourier-transform again over x&#039;:

\langle p |U(t)|p&#039; \rangle = \int dx&#039; e^{i p&#039; x&#039;} e^{-ip x&#039; - i \frac{p^2}{2m} t}

Factoring out the part that doesn't depend on x&#039; gives:

\langle p |U(t)|p&#039; \rangle = e^{-i \frac{p^2}{2m}t} \int dx&#039; e^{i (p&#039; - p) x&#039;}

That integral is a representation for \delta(p&#039;-p). So we get:

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

If you're wondering why doesn't have the same form (with x \Rightarrow p and x&#039; \Rightarrow p&#039;) as \langle x |U(t)|x&#039; \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&#039;)} evolve with time, it turns into:

e^{i k (x-x&#039;) - 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&#039;\rangle can be obtained from the form \frac{1}{2\pi} \int dk\ e^{-i \frac{k^2}{2m} t + i k (x-x&#039;)}

Note that - \frac{i t k^2}{2m} + i k (x-x&#039;) = - (\sqrt{\frac{it}{2m}} k - \sqrt{\frac{m}{2 i t}}(x-x&#039;))^2 + \frac{m}{2it} (x-x&#039;)^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&#039;), then that integral becomes:
\frac{1}{2\pi} e^{\frac{m}{2it}(x-x&#039;)^2} \sqrt{\frac{2m}{it}} \int du\ e^{-u^2} = \frac{1}{2\pi} e^{\frac{m}{2it}(x-x&#039;)^2} \sqrt{\frac{2m}{it}}\sqrt{\pi}= \sqrt{\frac{m}{2\pi it}}e^{\frac{m}{2it}(x-x&#039;)^2}

 

[friend]

Now I'm not so sure which of \langle p | U(t) | x&#039; \rangle or \langle p|U(t)|p&#039; \rangle I'm interested in. If \langle x|U(t)|x&#039;\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&#039; \rangle because that is its momentum after the transition? Or do I want \langle p|U(t)|p&#039; \rangle because \langle x|U(t)|x&#039;\rangle has an undetermined momentum at both x' and x? Does \langle x|U(t)|x&#039;\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&#039;)\,\,\, = \,\,\, &lt; x|{e^{ - iHt/\hbar }}|x&#039; &gt; \,\,\, = \\<br /> \int_{ - \infty }^{ + \infty } {\int_{ - \infty }^{ + \infty } {\int_{ - \infty }^{ + \infty } { \cdot \cdot \cdot \int_{ - \infty }^{ + \infty } { &lt; x|{e^{ - iH\varepsilon /\hbar }}|{x_1} &gt; } } } } &lt; {x_1}|{e^{ - iH\varepsilon /\hbar }}|{x_2} &gt; &lt; {x_2}|{e^{ - iH\varepsilon /\hbar }}|{x_3} &gt; \cdot \cdot \cdot &lt; {x_n}|{e^{ - iH\varepsilon /\hbar }}|x&#039; &gt; 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.