Quantum Mechanics – Pure State vs Mixed State Explained

density-operatorquantum mechanicsquantum-statesstatistical mechanicswavefunction

Consider, I have a quantum state $|\Psi\rangle$, such that :
$$|\Psi\rangle=c_1|\psi_1\rangle+c_2|\psi_2\rangle$$

This is defined as a pure state, since I have complete information about the system. Before measurement ( collapse ), the system is in state $|\Psi\rangle. $ After measurement, there would be a $|c_1|^2$ probability that the state is now $|\psi_1\rangle,$ and a $|c_2|^2$ probability that the system is now in state $|\psi_2\rangle$. Let us have some operator $\hat{H}$ such that $\hat{H}|\psi_n\rangle=\lambda_n|\psi_n\rangle$.

Suppose I have a friend, who performs this measurement. If he gets $\lambda_1$, it means the state has now become $\psi_1$ and vice-versa.

However, imagine now, that my friend was coming over to tell me the results. However, due to some weird phenomena, he disappeared off the face of the earth, before he could tell me the result.

So, all I know now is that before the measurement, the state was $\Psi$, and an experiment was conducted, and the wavefunction collapsed to either $\psi_1$ or $\psi_2$. Since my friend couldn't tell me the results, I don't know which state is the system actually in. From the initial state, I can vaguely say that there is a $|c_1|^2$ probability of the final state being $\psi_1$ and a $|c_2|^2$ probability of it being $\psi_2$.

I cannot say that the system is in superposition since I know that a measurement was carried out, and one of the two values was obtained.

Since I no longer have complete information about my system, can I consider this an example of a mixed state? So, can I use a density matrix to describe this final state, since this seems like an example of mixed state?

Best Answer

Yes your reasoning is entirely correct. You would now describe the state at hand with the density operator

$$ \rho = |c_1|^2 |\psi_1 \rangle \langle \psi_1| + |c_2|^2 |\psi_2 \rangle \langle \psi_2|$$

Note that when dealing with mixed states, the description of the particle depends on the knowledge of the observer. Your friend would have described the state as $ |\psi_1 \rangle \langle \psi_1| $ or $|\psi_2 \rangle \langle \psi_2|$. This might seem really weird, but the point is that if you would repeat the whole experiment ( including your friend and his disappearence) then your description would yield the correct statistical outcomes.

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