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Suppose I have a particle whose momentum I measure to be $p$ with uncertainty $\delta p$. Right after the measurement we know that its wave function is given by $\psi(x)=\int g(p)e^{ipx/\hbar}dp$ (restricting to 1D for simplicity) with $g$ a certain function which is $0$ outside $p-\delta p$, $p+\delta p$ and with $\int |g(p)|^2dp=1$.

What more can we say about that function $g$?

I was ok with not being able to say anything more about it but I have to admit that, for instance, we know that $g$ is never much more concentrated than it has to be. For else if we'd take for instance the canon from the Stern-Gerlach experiment (which throws silver atoms at circa 500 m/s) and direct it at the middle of a door, we should see atoms bouncing back from time to time. Indeed if $g$ is extremely concentrated, the resulting wave can get quite spread out and one could see diffraction effects. But we never do.

Qmechanic
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student_du_03
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2 Answers2

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It depends on the state before the measurement. If the pre-measurement wavefunction is given by some $\psi_i(x)$, then the post-measurement wavefunction $\psi_f(x)$ is the projection of $\psi_i(x)$ onto the subspace of the Hilbert space which is consistent with your measurement results.

More concretely, let $$\psi_i(x) = \frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^\infty A(p) e^{ipx/\hbar}dp$$

If you measure the momentum to be in the interval $p\in[p_0-\delta p, p_0 + \delta p]$ then the post-measurement state is simply $$\psi_f(x) = \frac{1}{\sqrt{2\pi\hbar}}\int_{p_0-\delta p}^{p_0 + \delta p} A(p) e^{ipx/\hbar}dp$$

As a side note, the normalization is not automatically preserved by measurement, but that's okay because quantum states are elements of the projective Hilbert space and are only well-defined up to an overall multiplicative constant anyway. It's usually convenient to normalize the wavefunction, but note that you'll have to re-normalize it after a projective measurement.

"If the interval in which you measure is small enough, then you can approximate as a constant function on that interval" Why?

Because as long as $A$ is at least continuous, then for a small interval this $g$ enter image description here

looks rather a lot like this one enter image description here

That's if you assume, as my textbook very implicitly does, that one can in fact approximate with a constant, but I don't see this assumption anywhere in the postulates.

No, this is a straightforward application of the uncertainty principle to a wave packet. In order to observe diffraction effects, you need the spread of your wavepacket to be at least on the same order of magnitude as the size of the hole. For e.g. a Gaussian wave packet which saturates the uncertainty relation, the spread in $p$ is inversely proportional to the spread in $x$, implying that a large spatial delocalization corresponds to a fantastically localized momentum.

J. Murray
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  • How is this relevant to my question? – student_du_03 Aug 28 '20 at 16:25
  • @student_du_05 How is it not? The $g$ you refer to is precisely determined by the pre-measurement state and the results of the measurement you perform. What else are you looking for? – J. Murray Aug 28 '20 at 16:29
  • Your statement doesn't give any new restriction on $g$ and in particular doesn't explain why there is no diffraction with a Stern Gerlach canon shooting at a door. Or you'd have to tell me something about your $A$ that I don't know. – student_du_03 Aug 28 '20 at 17:38
  • @student_du_05 Clearly there is some miscommunication happening here. If you want to know precisely what $g$ will be after a measurement, you need to know what $A$ was before it. If the interval in which you measure $p$ is small enough, then you can approximate $g$ as a constant function on that interval, but without knowing $A$ then you can't be exact. – J. Murray Aug 28 '20 at 17:59
  • As far as the diffraction goes, in order for a wave packet to be delocalized on the spatial scale of a doorway ($\sim 1$ m), then the momentum would have to be localized to within approximately $10^{-35}$ kg m/s, corresponding to a velocity uncertainty of 0.3 nm/s for a silver atom. – J. Murray Aug 28 '20 at 18:03
  • "If the interval in which you measure is small enough, then you can approximate as a constant function on that interval" Why? That's my question. – student_du_03 Aug 28 '20 at 18:13
  • "as far as the diffraction goes, in order for a wave packet to be delocalized on the spatial scale of a doorway (∼1 m), then the momentum would have to be localized to within approximately 10−35 kg m/s, corresponding to a velocity uncertainty of 0.3 nm/s for a silver atom" That's if you assume, as my textbook very implicitly does, that one can in fact approximate $g$ with a constant, but I don't see this assumption anywhere in the postulates. – student_du_03 Aug 28 '20 at 18:15
  • Maybe it is that after a relatively long time in the atom oven all the $A$'s look alike and so give similar $g$'s? I know that wave packets spread with time (so after a long time we can assume that $\psi$ is such and such) but I don't know that their Fourier spectrum (ie. $g$) smoothens to a nearly constant function. – student_du_03 Aug 28 '20 at 18:16
  • By the way thank you for your time – student_du_03 Aug 28 '20 at 18:21
  • @student_du_05 Please see my update to address your additional questions. Note that sampling individual particles from a thermal distribution (which is what happens in Stern-Gerlach) is not a trivial matter to treat theoretically, and my answer addresses only the effects of a projective measurement on a state when the observable has a continuous spectrum, as per your original question. – J. Murray Aug 28 '20 at 19:27
  • Why do you assume $A$ is a gaussian? Also yes I know you applied the uncertainty principle, I forgot a part of what I meant to say: "that's problematic if you assume that $g$ is a constant", sorry. – student_du_03 Aug 28 '20 at 19:51
  • @student_du_05 Let's discuss this in chat. – J. Murray Aug 28 '20 at 19:54
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I think I finally found an answer to my question. The atoms in a Stern-Gerlach-type oven can't have too wide wave functions (space wise) because else they would interact a lot with each other (they would be 'very close everywhere') and this configuration is too energetic.

student_du_03
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  • That's a complete misunderstanding of quantum mechanics. The "spreading" of a wave function has nothing to do with inter-atomic forces. – FlatterMann Jun 17 '23 at 19:23