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

[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 $\epsilon \rightarrow 0^+$.