Position and momentum measurement effects on wave functions

Question

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?

3. 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.

Answer 1

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.

Answer 2

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.

Answer 3

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".