2
$\begingroup$

Say I have a state $|\Psi_0\rangle$.

I measure observable $\hat{A}$, the wavefunction collapses to one its eigenstates. I can write $|\Psi_0\rangle = \sum_j \alpha_j|\psi^A_j\rangle$, where $|\psi^A_j\rangle$ are the eigenstates of $\hat{A}$.

Suppose I do not have access to the results of the experiment though.

I now apply a second operator $\hat{B}$.

I can see two ways of proceeding to know how the system will evolve in time:

1) I can write $|\Psi^B_t\rangle = \hat{B}|\Psi_t\rangle = \sum_j \alpha_j\hat{B}|\psi^A_j(t)\rangle$, where I have to work out how each of the $|\psi^A_j\rangle$ evolves with time;

2) Defining the density matrix $\rho = \sum_i p_i|\psi^A\rangle\langle\psi^A|$, to then use $\frac{\partial \rho}{\partial t} = \frac{1}{i\hbar}[\rho, H]$.

I agree option #2 is easier, but option #1 is still feasible.

Is there any situation where I cannot use method 1 (pure state time evolution) but have to use use the density matrix approach?

$\endgroup$
  • $\begingroup$ If your initial state is not pure... Like for instance a 50/50 mixture of $\vert +\rangle_z$ and $\vert +\rangle_y$. $\endgroup$ – ZeroTheHero Jan 15 '18 at 0:42
  • $\begingroup$ Can I not write $|\Psi\rangle = 1/\sqrt{2} |+\rangle_z + 1/\sqrt{2} |+\rangle_y$? Or a similar combination with an orthogonal basis. $\endgroup$ – SuperCiocia Jan 15 '18 at 0:47
  • $\begingroup$ $|\psi\rangle = \frac{1}{\sqrt{2}} |+\rangle_z + \frac{1}{\sqrt{2}} | + \rangle_y$ is a pure state, a superposition of two states. A 50/50 mixture cannot be describet by a pure state: 50% of the system is composed of $|+\rangle_z$ and 50% of $|+\rangle_z$. This is different from a superposition. $\endgroup$ – Patrick Jan 15 '18 at 0:59
  • $\begingroup$ So if I did not want to use the density matrix method, I'd have to carry out 2 distinct calculations? $\endgroup$ – SuperCiocia Jan 15 '18 at 1:15
  • $\begingroup$ I see your point though. This means that for my example, option #2 is wrong then? $\endgroup$ – SuperCiocia Jan 15 '18 at 1:21
1
$\begingroup$

Sorry, I have misunderstood your question in the comments. To be concrete, consider the Stern-Gerlach experiment. Assume you have an initial state $|\psi\rangle = c_0|z_+\rangle + c_1|z_-\rangle$ and pass it through a SG apparatus in the z direction. Then you know for sure that the end states are either $|z_+\rangle$ or $|z_-\rangle$. If you put your SG inside a black box so that you cannot know which of the end states it is, I believe you can treat this as a mixed state with $|c_0|^2$ probability of being $z_+$ and $|c_1|^2$ probability of being $z_-$: $$\rho = |c_0|^2|z_+\rangle\langle z_+| + |c_1|^2 |z_-\rangle\langle z_-| $$ So you can use this density matrix to evolve the system in time as you described.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.