# Superposition in the density matrix formalism

1. Nov 9, 2014

### VantagePoint72

Suppose I have a two level system with the states labeled $|0\rangle$ and $|1\rangle$. In this basis, these correspond to density matrices:
$\rho_0 = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \quad \rho_1 = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$

I can create a coherent superposition of my basis states like $|\psi\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$ or any other pure state on the Bloch sphere. This particular one has density matrix:
$\rho_{\psi} = |\psi\rangle\langle\psi| = \frac{1}{2}\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$

If all I had were $\rho_0$ and $\rho_1$ and I decided I wanted to produce to a coherent superposition of those two states, is there any operation I can do to $\rho_0$ and $\rho_1$ directly to produce $\rho_{\psi}$ (or the density matrix of some other state in the Bloch sphere)? Or is it necessary to always first factor the density matrices into the outer products of vectors, produce the superposition with the vectors, and then convert the new state vector back into a density matrix? Clearly it's not just a matter of taking some kind of linear combination of $\rho_0$ and $\rho_1$ because at best that gives you a mixed state and at a worst doesn't even give you something positive, Hermitian, and of trace 1.

Equivalently (more or less): if I handed you $\rho_0$, $\rho_1$, and $\rho_{\psi}$ you can easily check that all three are pure states by confirming that they're idempotent without needing to compute their eigenvectors. Can you similarly confirm that $\rho_{\psi}$ is a superposition of the others from the matrices alone?

2. Nov 9, 2014

### Orodruin

Staff Emeritus
If you know that $\rho_0$ and $\rho_1$ are orthogonal (check by $\rho_0 \rho_1 = 0$), then you can construct the projection operator $P = \rho_0 + \rho_1$ which will project $\rho_\psi$ onto the subspace spanned by the 0 and 1 states. As $P \rho_\psi P = \rho_\psi$, it is a superposition of the two (in this case this is trivial since $P = \mathbb 1$, but I imagine you want something more general). This only tells you that it is a combination of the two, not necessarily that it is a pure state, which you will have to check separately.

3. Nov 9, 2014

### naima

$\rho_\psi$ is equal to $U(\theta) \rho_0 U^+(\theta)$ where U is a rotation in the hilbert space.
What do you call the superposition of density matrices?

4. Nov 9, 2014

### VantagePoint72

Thanks, Orodruin. naima, that doesn't really have anything to do with my question. I'm not asking if there's an operation that turns $\rho_0$ or $\rho_1$ into $\rho_{\psi}$. If asking if there's one that takes both of them and spits out a pure state superposition like $\rho_{\psi}$.

5. Nov 10, 2014

### naima

I think that i understang what you are looking for.
To get the probability that a state $\psi$ "hits" the screen at x to take $\psi (x)$ and square its absolute value.
If the state is given by its density matrix $\rho$ the probability is $Tr (\rho |x> <x|)$ so it seems nearer the concept of probabilities.
Many people think that density matrices states are more fundamental than amplitudes of probabilities. There is a group of physicists who try to avoid them using only things like your $\rho$
Take a beam splitter. the input signal has the choice between two channels and you can describe what evolves along the different paths with amplitudes or with density matrices.
It is harder when those paths meet again.
If you like amplitudes you simply add them at the meeting point.
If you dislike them you will try to hide them under the carpet.
I do not know something like Feynmann path integral for density matrix.
If $\rho_0$ and $\rho_1$ meet and that you hate amplitudes, you can say that the only way to describe the nature of their interaction in a real setup is to give their resulting density matrix!