Say one has an unperturbed system that can be described with a wavefunction that is a superposition of many eigenstates. How does one make the wavefunction collapse into an eigenstate of say, the energy operator rather than say, an eigenstate of the momentum operator? In other words, what is the mechanism that decides into which eigenstate's operator the wavefunction collapses when a measurement is made?

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    $\begingroup$ by making a measurement of the energy, one gets the eigenvalue for the energy operator $\endgroup$ – anna v Mar 15 '18 at 19:38
  • $\begingroup$ If you are measuring energy (momentum), you will get an energy (momentum) eigenstate. $\endgroup$ – jobe Mar 15 '18 at 19:39
  • $\begingroup$ I'd appreciate more details. How can one measure the energy instead of say the momentum? $\endgroup$ – AccidentalBismuthTransform Mar 15 '18 at 19:39
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    $\begingroup$ I wrote an answer with mathematical detail. If you want more intuition, please leave a comment under the answer. $\endgroup$ – DanielSank Mar 15 '18 at 20:58

It's commonly (and correctly) said that quantum states collapse in a particular operator's basis when we "measure" the system in that basis. We can understand what's going on much better if we understand what a measurement really is.

Call the main system $S$ and call the operator into whose basis you want to collapse $\mathcal{O}_S$. Now add in another system $E$. Suppose we let $S$ interact with $E$ through an interaction operator $\mathcal{O}_S \otimes \mathcal{O}_E$ where $\mathcal{O}_E$ is some operator on $E$.

If $E$ and $\mathcal{O}_E$ are chosen such that after the interaction, the state of $E$ depends strongly on the initial state of $S$, then we have measured $S$. From the perspective of observations made on just $S$, $S$ will have collapsed in basis $\mathcal{O}_S$.


Suppose $S$ and $E$ are both two-level quantum systems. Let them interact through operator $$H_\text{interaction} / \hbar = g \left( \sigma_z \otimes \sigma_x \right) \, .$$

Now suppose $S$ starts off in state $$ \left \lvert \Psi \right \rangle_S = \frac{1}{\sqrt{2}} \left( \left \lvert 0 \right \rangle + \left \lvert 1 \right \rangle \right)$$ and $E$ starts off in the state $$ \left \lvert \Psi \right \rangle_E = \left \lvert 0 \right \rangle \, .$$ Using the basis $\{00, 01, 10, 11 \}$, we can write the combined initial state as $$\left \lvert \Psi(t=0) \right \rangle = \frac{1}{\sqrt{2}} \left( \begin{array}{c} 1 \\ 0 \\ 1 \\ 0 \end{array} \right) \, .$$ Under $H_\text{interaction}$, the system evolves in time to \begin{align} \left \lvert \Psi(t) \right \rangle &= \frac{1}{\sqrt{2}} \exp \left[ -i gt \left( \sigma_x \otimes \sigma_x \right) \right] \left( \begin{array}{c} 1 \\ 0 \\ 1 \\ 0 \end{array} \right) \\ &= \frac{1}{\sqrt{2}} \left( \begin{array}{cccc} \cos(gt) & -i\sin(gt) & 0 & 0 \\ -i \sin(gt) & \cos(gt) & 0 & 0 \\ 0 & 0 & \cos(gt) & i\sin(gt) \\ 0 & 0 & \sin(gt) & \cos(gt) \end{array}\right) \left( \begin{array}{c} 1 \\ 0 \\ 1 \\ 0 \end{array} \right) \\ &= \frac{1}{\sqrt{2}} \left( \begin{array}{c} \cos(gt) \\ -i \sin(gt) \\ \cos(gt) \\ i\sin(gt) \end{array} \right) \, . \end{align} When $gt = 0$, we get the same state we started with, as we should. When $gt = \pi/2$, we get $$\left \lvert \Psi(t=\pi/2g)\right \rangle = \left( \begin{array}{c} 0 \\ -i \\ 0 \\ i \end{array} \right) = \frac{-i}{\sqrt{2}} \left( \left \lvert 00 \right \rangle - \left \lvert 11 \right \rangle \right) \, .$$ This is an entangled state. In particular, it is a Bell state. Note that if we were to measure the state of $E$, we'd know for sure what is the state of $S$. In other words, $E$ has measured $S$.

Formally, you can write down the information available about $S$ in the absence of knowledge about $E$ by forming the density matrix and then tracing over $E$. The density matrix is $$\rho \equiv \left \lvert \Psi \right \rangle \left \langle \Psi \right \rvert = \frac{1}{2} \left( \left \lvert 00 \right \rangle \left \langle 00 \right \rvert +\left \lvert 11 \right \rangle \left \langle 11 \right \rvert -\left \lvert 01 \right \rangle \left \langle 10 \right \rvert -\left \lvert 10 \right \rangle \left \langle 01 \right \rvert \right) \, .$$ Tracing over $E$ means keeping only terms where the state of $E$ is the same in the ket and the bra, which leaves $$\rho_S = \frac{1}{2} \left( \left \lvert 0 \right \rangle \left \langle 0 \right \vert +\left \lvert 1 \right \rangle \left \langle 1 \right \vert \right)$$ which represents a classical probability distribution where $S$ has 1/2 probability of being in state $\left \lvert 0 \right \rangle$ and 1/2 probability of being in state $\left \lvert 1 \right \rangle$.

In other words, $S$ has collapsed in the $z$ basis!

  • $\begingroup$ "Tracing over $E$" doesn't mean "keeping only terms where the state of $E$ is the same in the ket and the bra". It means forming $$\rho_{ii'} := \sum_j \rho_{iji'j}.$$ The resulting density matrix only acts on the Hilbert space $\mathcal{H}_S$, not on the composite Hilbert space. So you should have $$\rho_S = \frac{1}{2} (|0 \rangle \langle 0| + |1 \rangle \langle 1|).$$ It's also rather misleading to say that $S$ has collapsed in the $z$ basis, because this maximally mixed qubit has the same coordinate decomposition in any orthonormal basis. $\endgroup$ – tparker Mar 16 '18 at 1:08
  • $\begingroup$ @tparker and you're right about the collapse axis. I should pick a better example. I was going to do $\sigma_z n$ where $n$ is the number operator on a harmonic oscillator, but I didn't want to require too much mathematical background from the reader. Any suggestions? $\endgroup$ – DanielSank Mar 16 '18 at 1:26
  • $\begingroup$ Nope. Except you should fix the "Tracing over E means keeping only terms where the state of E is the same in the ket and the bra" part $\endgroup$ – tparker Mar 16 '18 at 2:10
  • $\begingroup$ @tparker Yes yes I know! :-) I'm thinking of how to go about this in a way that's accessible and accurate. Index notation is best but non-standard. Books and professors in lecture tend to write stuff like this $$\left \langle n \cdot \right \rvert \left( \left \lvert i j \right \rangle \left \langle k l \right \rvert \right) \left \lvert n \cdot \right \rangle $$... or not. I don't know what the standard notation really is. $\endgroup$ – DanielSank Mar 16 '18 at 2:12
  • $\begingroup$ @tparker I wonder if we need to start in a different state to show the decay toward the z-axis. Interesting... I haven't thought about this much before. $\endgroup$ – DanielSank Mar 16 '18 at 18:39

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