Pedagogy and HUP

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Main Question or Discussion Point

HUP was taught at least to me, as a brute fact that came into existence when my lecturer wrote it on the board...with chalk.

I was fortunate enough to have already had some background in Fourier Transforms.

When doing a basic course on SWE and the link between it and wave solutions, the HUP then seemed obvious and mundane.

Many students however that had not already studied the FT went into philosophical crisis.

The solution seems obvious;

Do not mention the HUP until students have seen a lot of concrete examples of FT's in mundane applications.

My slogan would be:

FT's first.....then wave mechanics.....then the HUP.

Thoughts?​
 
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Answers and Replies

  • #2
Demystifier
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FT's first.....then wave mechanics.....then the HUP.
I agree. :smile:
 
  • #3
vanhees71
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HUP was taught at least to me, as a brute fact that came into existence when my lecturer wrote it on the board...with chalk.

I was fortunate enough to have already had some background in Fourier Transforms.

When doing a basic course on SWE and the link between it and wave solutions, the HUP then seemed obvious and mundane.

Many students however that had not already studied the FT went into philosophical crisis.

The solution seems obvious;

Do not mention the HUP until students have seen a lot of concrete examples of FT's in mundane applications.

My slogan would be:

FT's first.....then wave mechanics.....then the HUP.

Thoughts?​
In my opinion you should not start teaching QM with wave mechanics but use the representation free approach with the advantage that you can discuss the most simple case of spin-1/2 observables (2D unitary vector space instead of the full separable infinite-dimensional Hilbert space). The HUP just follows from positive definiteness of the scalar product. It's not limited to conjugate pairs of observables and thus not limited to (generalized) Fourier transformations.
 
  • #4
A. Neumaier
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The HUP just follows from positive definiteness of the scalar product. It's not limited to conjugate pairs of observables and thus not limited to (generalized) Fourier transformations.
But it needs noncommuting oservables.
 
  • #5
jtbell
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When doing a basic course on SWE
Just to clarify: SWE = ?
 
  • #6
vanhees71
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But it needs noncommuting oservables.
I referred to the usual Heisenberg-Robertson uncertainty relation which holds for any pair of observables
$$\Delta A \Delta B \geq \frac{1}{2} |\langle [\hat{A},\hat{B}] \rangle|.$$
If the observables ##A## and ##B## are compatible, i.e., if the corresponding representing self-adjoint operators ##\hat{A}## and ##\hat{B}## commute, of course, there's no restriction by the uncertainty relation, i.e., you can prepare the system in states, where both observables have determined values (represented, e.g., by a common eigenvector of the operators).
 
  • #7
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Just to clarify: SWE = ?
Schrödinger wave equation.
 

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