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I have a few short questions about an interpretation of what happens with position and momentum wave functions described in literature I am using. Given momentum space wave function and position space wave function:

$$\Phi(p,t) = \frac{1}{\sqrt{2 \pi \hbar}}\int^{\infty}_{-\infty}e^{-(ipx/\hbar)}\Psi(x,t)dx$$

$$\Psi(p,t) = \frac{1}{\sqrt{2 \pi \hbar}}\int^{\infty}_{-\infty}e^{(ipx/\hbar)}\Phi(p,t)dp$$

In the literature the following is stated: "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 (hence momenta) 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."

  1. As I understand, $\Phi$ is the Fourier transform of $\Psi$ but, referring to the equations above, why does this imply that if $\Psi$ collapses to spike that the Fourier transform $\Phi$ is broad?

  2. What I understand is that the measurement of momentum gives you some eigenvalue in the continuous spectra of the momentum operator as a measurement. So what happens is that the position wave function collapses to a narrow range about the measured value, say $p_1$, so the wave collapses to something like:

$$\Psi(x,t) = \frac{1}{\sqrt{2 \pi \hbar}}\int^{\alpha}_{\beta}e^{(ipx/\hbar)}\Phi(p,t)dp~~~~\text{for }p_1 \in [\alpha, \beta]$$

Is this a correct interpretation?

  1. How does this correspond to a collapse to a long sinusoidal wave, with a well-defined wavelength as is quoted from the literature?

Thanks a lot for any assistance.

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I think you are getting something wrong in the 2nd and 3rd point, but i am going to try and give you an explanation. If i got something wrong in your questions, please do point it out.

1) If you measure the position, then Ψ(x)=δ(x-z) where z is the position that you measured(not a variable). If you plug this into the Fourier transform of Ψ(the first relation for Φ(p)), then the integral gives you exp[ipz/hbar], which is a monochromatic wave(with well-defined wavelength-momentum).

2) If you measure the momentum, you do not a wave function in the position space that collapses to a narrow range but you again get a complex exponential exp[iAx/hbar] with A being the momentum that you measured(again, not a variable). It is the same this as the previous point but in the coordinate space rather than momentum space.

3) It follows from (1) and (2) again with plugging Φ(p)=δ(p-A) into your second relation for finding Ψ(x)

For the second point, in order to get a function that has a narrow range which is a Gaussian function, its Fourier transform must also be a Gaussian wavepacket. This is also consistent with the uncertainty principle. The narrower the wavefunction is, the wider its Fourier transform is. This goes for the other way around.

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  • $\begingroup$ Please see my answer, which gives detail to my revised understanding based on your response, thanks. $\endgroup$ – Alex Apr 11 '16 at 11:28
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The act of measuring cannot be represented by an operation in Linear Algebra, because postulate of QM says that upon measurement of observable $A$ with eigenvectors and eigenvalues given by $|a_i>$ and $a_i$ respectively on $|\psi>$ . $$|\psi>=\sum_i c_i |a_i>$$ Gives $|\psi_{after}>$.

where $|\psi_{after}>$ can be any of $|a_i>$ with probability $|c_i|^2$. Now considering the momentum operator case $\phi(p,t)$ will collapse to $\delta(p-\alpha)$. where $\alpha$ is randomly selected from the distribution $|\phi(p,t)|^2$.

So for your point 2, position space wave function $\psi(x,t)$ will be Fourier transform of $\delta(p-\alpha)$ will be a wave with wavelength $\alpha/\hbar$.

Hope it clears other points too.

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After revising the (very informative) responses to my previous queries, my revised understanding (which addresses my queries) is:

If we measure the position then $\Psi(x,t)$ collapses to $\Psi(x,t) = \delta(x-z)$ (where $z$ is the measured position), using the Fourier transform we get: $$\Phi(p,t) = \frac{1}{\sqrt{2 \pi \hbar}} \int_{-\infty}^{\infty}e^{-\frac{ipx}{\hbar}}\delta(x-z)dx = \frac{1}{\sqrt{2 \pi \hbar}} e ^{-\frac{ipz}{\hbar}}$$ which is a monochromatic wave. At this point we have $$\Psi(x,t) = \delta(x-z)~~\text{and}~~\Phi(p,t) = \frac{1}{\sqrt{2 \pi \hbar}}e^{-\frac{ipz}{\hbar}}.$$ If a subsequent measurement of of momentum gives the value $A$, the position wave function collapses to the corresponding eigenvector, hence now $$\Psi(x,t) = \frac{1}{\sqrt{2 \pi \hbar}}e^{\frac{iAx}{\hbar}}.$$ Plugging this into the Fourier transform yields $$\Phi(p,t) = \frac{1}{2 \pi \hbar}\int_{-\infty}^{\infty}e^{-\frac{ipx}{\hbar}}e^{\frac{iAx}{\hbar}}dx = \frac{1}{2 \pi \hbar}\int^{\infty}_{-\infty}e^{\frac{i(A-p)}{\hbar}}dx = \delta(A-p).$$

As noted also in one of the responses "The narrower the wavefunction is, the wider its Fourier transform is".

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    $\begingroup$ Yes, this is exactly right. But, you also have to understand it based on intuition of the basic quantum mechanical principles. That means, via the uncertainty principle. Then these mathematical results become something which is expected. $\endgroup$ – TheQuantumMan Apr 11 '16 at 23:13
  • $\begingroup$ Also, go a little but further. If the wavefunction in coordinate space(Ψ(x)) is two delta functions at different positions, then what is its fourier transform(the wavefunction in momentum space Φ(p))? Try to guess it first. Then what do you expect when the two positions of each delta function are close together and what if they are further apart? Draw the resulting Φ(p) in a program(say wolfram alpha--google search it, it's free) in each case and check if the results satisfy your intuition about the uncertainty principle! $\endgroup$ – TheQuantumMan Apr 11 '16 at 23:17

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