Paradox: The infinite square well vs. the Uncertainty Principle

  • #1
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I've come across an apparent paradox in elementary quantum mechanics, and after a little Googling, haven't found a reference to it. Here goes,

The 1-D infinite square well is a classic problem in introductory QM. We find that the position-space eigenfunctions of the Hamiltonian (the "allowed wave functions") are sine waves with wavelengths such that they vanish at the boundaries of the well. It seems to me that, because these wave functions are pure sine waves, they are eigenstates of momentum, so the momentum uncertainty for these states is 0.

At the same time, the position uncertainty is finite, since the particle is confined inside the well. Then the product of the position and momentum uncertainties is 0, which seems to violate the Uncertainty Principle. If this argument is correct so far, it implies that a perfect infinite square well cannot be constructed in principle. Then one of the assumptions behind the problem is unphysical. There are at least two possibilities:

1. The assumption that the walls are infinitely high.

2. The assumption that the "corners" on the ends are infinitely sharp.

What I am wondering is, which of those two assumptions, or both, is to blame for the paradox? And how can we prove within quantum mechanics that that assumption is unphysical? This isn't homework, I just thought it was interesting. Thanks for any thoughts you have!
 

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  • #2
Dale
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It seems to me that, because these wave functions are pure sine waves, they are eigenstates of momentum, so the momentum uncertainty for these states is 0.
They are not pure sine waves. They are a certain number of half-cycles of a sine wave. I.e. they are sine waves multiplied by a rectangle function.
 
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Okay, but does that affect the momentum resolution? Inside the box, the wavelength is well-defined, so the momentum is also, yes?
 
  • #4
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Mike: This current thread might answer your question:

https://www.physicsforums.com/showthread.php?t=516224

(don't be dissuaded by the several initial posts...we got sidetracked on a tangent that IO think address your question....)

"...the wavelength is well-defined.."


If I understand your question, every individual measurement IS...to any arbitrary level of precision..... and to the momentum as well, but NOT the probability distribution...In the above thread we concluded that there is no uncertainty regarding the arbitrary precision of any single measurement...that the uncertainty actually applies to the distribution of even identically prepared states.
 
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Dale
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Okay, but does that affect the momentum resolution? Inside the box, the wavelength is well-defined, so the momentum is also, yes?
There is no "momentum inside the box" operator. You could certainly make one, but it wouldn't commute with the position operator, so it would not violate the HUP.

The only operator which commutes with the position operator is the standard momentum operator, and the wavefunction for the particle in a box is not an eigenstate of that operator.
 
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@ Naty1,

Hmm... I'm not sure that's a fully satisfactory answer. It seems possible to prepare a particle in a certain energy eigenstate and then measure the momentum to great precision many times, since the energy eigenstates are momentum eigenstates. So the (statistical) momentum error is zero. And just by performing a measurement, I have determined that the particle is inside the box, so we can assign a finite position error.

I believe I understand the definition of the Uncertainty Principle pretty well. I think the resolution to the paradox will have to concern the specific nature of the infinite square well, not the general mathematical definition of the UP.

@ DaleSpam,

Okay, I think I get what you're saying now. Would it be fair to say that the correct wavefunction (meaning, sine wave inside the box and vanishing outside) must be a Fourier series (or integral) composed of many plane waves, which would imply a finite momentum resolution? That seems plausible. I will try to work this out explicitly...
 
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Dale
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Would it be fair to say that the correct wavefunction (meaning, sine wave inside the box and vanishing outside) must be a Fourier series (or integral) composed of many plane waves, which would imply a finite momentum resolution?
Yes. That is correct to my understanding.
 
  • #8
jtbell
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It seems to me that, because these wave functions are pure sine waves, they are eigenstates of momentum, so the momentum uncertainty for these states is 0.

No, they are not eigenstates of momentum and their momentum uncertainty is not zero.

An eigenstate of momentum is a pure sine/cosine wave that extends to infinity in both directions. Square-well wave functions do not extend beyond the walls of the well. Also, they are standing waves, so they contain wave components traveling in both directions.

Try calculating the momentum uncertainty directly from the definition:

[tex]\Delta p = \sqrt {\langle p^2 \rangle - {\langle p \rangle}^2}[/tex]

where

[tex]\langle p \rangle = \int {\Psi^* \left( -i \hbar \frac {\partial}{\partial x} \right) \Psi dx}[/tex]

[tex]\langle p^2 \rangle = \int {\Psi^* {\left( -i \hbar \frac {\partial}{\partial x} \right)}^2 \Psi dx}[/tex]

with the integrals evaluated between the walls of the well. Pick any of the square-well wave functions; the ground state is probably easiest.
 
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  • #9
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jtbell:
.....their momentum uncertainty is not zero

that's my understanding [But I do not interpret the meaning of mathematics. )

except for:

"Also, they are standing waves, so they contain wave components traveling in both directions..." which is clearly accurate....does THAT affect uncertainty??



Is there something mathematicallly unique about the infinite square well.... different from a particle confined in a box or an electron orbital... all haved boundaries and are quantized with vanishing probability of being found at infinity???
 
  • #10
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It seems possible to prepare a particle in a certain energy eigenstate and then measure the momentum to great precision many times, since the energy eigenstates are momentum eigenstates. So the (statistical) momentum error is zero.

The statistical error is NOT zero.

That thread properly concludes that You CAN measure the momentum in any single measurement to as many decimal places as you wish, to any arbitrary level of precision limited only by your measurement apparatus...but NOT accuracy!!!! (This is not HUP).....There WILL be a statistical distribution of measurement results based on the quantum characteristics of the measured particles themselves, not based on measuring device limitations...THAT is limitation of the HUP...there is not a limit on any individual measurement.
 
  • #11
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It seems to me that, because these wave functions are pure sine waves, they are eigenstates of momentum, so the momentum uncertainty for these states is 0.

No, it is not. Consider, Consider say the ground state, you can write the (unnormalized) wave function as: [itex]\phi(x) = \sin \frac{\pi x}{L} = \frac{e^{i \pi x/L} - e^{-i \pi x/L}}{2i}[/itex].

You see that a sine wave consists of two plane waves, one going forward, the other going backward, each with 1/2 probability (if you normalized properly). So you have 1/2 probability that the wave have momentum [itex]p_{\pm} = \pm\frac{\hbar\pi}{L}[/itex]. So you can estimate the uncertainty in momentum as: [itex]\Delta p=\sqrt{\langle p^2 \rangle - \langle p \rangle^2} = \frac{\hbar \pi}{L}[/itex]

The uncertain in position is basically [itex]L[/itex] (you can try to get a more accurate calculation, but I think you are within factor of 2). So [itex]\Delta p \Delta x = \hbar \pi[/itex].
 

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