How does one make the wavefunction collapse into an eigenstate of a particular operator?

Question

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?

Answer 1

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 $\mathcal{O}_S$.

Example

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!