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I am following Griffith's Introduction to Quantum Mechanics, and after proving that $$\sigma _A ^2 \sigma _B ^2 \geq \left( \frac{1}{2i} \langle [A,B]\rangle \right)$$ he says

You can certainly measure the position of the particle, but the act of measurement collapses the wave function to a narrow spike, which necessarily carries a broad range of wavelengths in its Fourier decomposition. If you now measure the momentum, the state will collapse to a long sinusoidal wave, with now a well-defined wavelength - but the particle no longer has the position you got in the first measurement.

This is my understanding of the above paragraph: I have a state of interest $\Psi (x,t)$. I make a measurement of, say, position. Just to make things concrete, lets take the simple case of a particle in an infinite square well, $$\Psi (x,t) = \sum _{n=1} ^{\infty} c_n \sqrt{\frac{2}{a}} \sin \left(\frac{n\pi}{a}x\right) e^{-i(n^2\pi^2\hbar/2ma^2)t}$$Mathematically, this measurement is $$\hat{x}\Psi = x\Psi$$ where $\hat{x}$ is the position operator, and $x$ is the eigenvalue. How has the above act, of tacking on an $x$ to the function $\Psi$ collapsed the wavefunction to a narrow spike?

Now, if I find the momentum of my state, $$\hat{p}\hat{x}\Psi = -i\hbar \frac{\partial}{\partial x}(x\Psi)$$ how has this collapsed to a sinusoidal wave?

I would appreciate any advice you have regarding this issue. I have a suspicion that my understanding of what taking a measurement is mistaken...

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It’s just a fundamental postulate of quantum mechanics that after you make a measurement, the system’s quantum state changes to an eigenstate of the observable you measured that corresponds to the measured eigenvalue. There’s no derivation of that fact (although certain interpretations of QM do give somewhat heuristic arguments for why this occurs).

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    $\begingroup$ Thanks for your answer @tparker! However, Griffiths explicitly says that the uncertainty principle is a consequence of the statistical interpretation. Is there a mathematical walkthrough for this? $\endgroup$
    – megamence
    Jun 4, 2021 at 18:15
  • $\begingroup$ What do you mean by "Statistical Inerpretation"? $\endgroup$
    – Young Kindaichi
    Jun 4, 2021 at 18:24
  • $\begingroup$ > If you measure an observable $Q(x,p)$ on a particle in the state $\Psi$, you are certain to get one of the eigenvalues of the hermitian operator $\hat{Q}$. If the spectrum of $\hat{Q}$ is discrete the probability of getting the particular eigenvalue $q_n$ associated with eigenfunction $f_n$. @YoungKindaichi, from the book $\endgroup$
    – megamence
    Jun 4, 2021 at 18:31
  • $\begingroup$ And so How come Uncertainty principle is related to this? $\endgroup$
    – Young Kindaichi
    Jun 4, 2021 at 18:32
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I think your confusion is arising because you may have read (possibly in Zetilli's Quantum Mechanics) that the act of measurement of observable $\hat A$ is mathematically represented by $\hat A \Psi$.

Do not take this literally. You can't think this to mean that applying an operator on the wave function will give you an eigenstate of that operator (i.e. the wavefunction collapse) which will be a result of that measurement. If that were true, you would always get the same answer mathematically , but would your measurement give you the same result everytime?

That would happen only in the case where the wavefunction is already an eigenstate of that operator, in which the above statement is true but trivial.

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  • $\begingroup$ Thanks @IndischerPhysiker for your reply! How else do I think about a measurement being "the wavefunction reduced to a spike"? $\endgroup$
    – megamence
    Jun 5, 2021 at 16:18
  • $\begingroup$ I do not know much in terms of formalism of QM, but I can't think of anything else but this: When you have measured the position once, you have spotted the particle to be there, so it is somewhere, even in accordance with the Copenhagen Interpretation. At this moment, the position is certain, thus the probability density has to be $0$ everywhere else, and non zero at that $x_o$, so $\rho(x)=\delta (x-x_o)$; this gives you normalisation as well. I am sorry if this all sounds too obvious; if you are looking for some mathematical way you can codify wavefunction collapse into an operator, --- $\endgroup$
    – IndischerPhysiker
    Jun 5, 2021 at 16:55
  • $\begingroup$ --- that will not happen, since mathematically, you will $always$ get the same function after operation, and taking that result to even mean something like the measurement result is outright wrong, since that throws away the statistical interpretation down the drain. There is one important thing to mention here: The wavefunction collapse concept is an idealisation. In reality, you will always find a spread in observables. For more refer to Quantum Optomechanics by Warwick $\endgroup$
    – IndischerPhysiker
    Jun 5, 2021 at 17:03
  • $\begingroup$ Thank you for your explanation! @IndischerPhysiker $\endgroup$
    – megamence
    Jun 5, 2021 at 18:09
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If you go through the postulates of Quantum Mechanics, you will find

The result of measuring a physical quantity $\mathcal{A}$ must be one of the eigenvalues of the corresponding observable $\mathcal{A}$.

Now suppose you have taken the position measurement and found the particle to be at $x_0$, How would represent such a wavefunction?

Let's say $$\langle x|\Psi\rangle =\psi(x)=\delta (x-x_0)$$ let's varify that's it's eigenfunction of $X$ with eigenvalue $x_0$.

$$X|\Psi\rangle =X\int |x\rangle \langle x|\Psi\rangle dx=\int x|x\rangle \delta (x-x_0)dx=x_0|x_0\rangle$$ (Note the abuse of notation) How would momentum space wave function look like?

$$\tilde{\psi}(p)\sim \int e^{-ipx/\hbar}\psi(x)dx=\int e^{-ipx/\hbar}\delta(x-x_0)dx=e^{-ipx_0/\hbar}$$ which corresponds to plane wave (A uniform probability distribution).

Similarly, if you take momentum measurement you will get $\tilde{\psi}(p)=\delta(p-p_0)$ and so $\psi(x)\sim e^{ip_0x/\hbar}$.

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  • $\begingroup$ The problem is that $\delta(x)$ is not an element of $L^2(\mathbb{R}^3)$, c.f. the both questions I've linked to the OP. $\endgroup$
    – Tobias Fünke
    Jun 4, 2021 at 19:25
  • $\begingroup$ Put it in another way, something like $|x\rangle$ isn't a physical state, since $\langle x|x\rangle$ is undefined. $\endgroup$
    – Tobias Fünke
    Jun 4, 2021 at 19:32
  • $\begingroup$ @Jakob It's true that $\delta (x)$ or $e^{ipx/\hbar}$ are not physical. Quoting R. Shankar, "But despite all this, we will continue to use the eigenkets $|p\rangle $ because these vectors are so much more convenient to handle mathematically than the proper vectors. It should however be borne in mind that when we say a particle is (coming out of the accelerator) in a state $|p_0\rangle $, it's really in a proper state with a momentum-space wavefunction so sharply peaked at $p=p_0$ that it may be replaced by a delta function $\delta(p-p_0)$." $\endgroup$
    – Young Kindaichi
    Jun 5, 2021 at 4:22

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