Can density matrices be coherently added without having to go back to the schrodinger picture?

Question

If I have two coherent superpositions:

$$ |ψ\rangle = |0\rangle + |1\rangle \\ |φ\rangle = |0\rangle - |1\rangle $$

Each of these two wavefunctions has density matrices, which can be written (in the 0,1 basis) as:

$$ \rho_{\psi} = \frac{1}{2}\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$$

$$ \rho_{\phi} = \frac{1}{2}\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}$$

If I look at a coherent superposition of my two superpositions in ket notation: $$\frac{1}{\sqrt{2}}(|ψ\rangle+|φ\rangle) \implies |0\rangle $$

But I'm interested in working strictly in the density matrix picture. I want to add the density matrices without having to look at the kets at all. At first glance, I naively thought that I could simiply add them together in the same way as the kets:

$$ \frac{1}{2} (\rho_{\psi} +\rho_{\phi}) = \frac{1}{2}\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$

When I think about the result of this I get confused. I'm adding all of the elements of my density matrix; and looking at the subtraction in the off-diagonals, the coherence terms of each state are certainly talking to each other (which indicates to me that there's quantum interference). But half of me thinks that adding two separate density matrices is simply constructing a mixture (and consequently the quantum coherence isn't interfering).

Doing the math you can see that this is different from the density matrix produced by summing the ket states:

$$ \frac{1}{2} \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$$

I can almost come up with a way to do it, but the answer is a lot uglier than I was expecting, and I'm stuck at the very end.

Looking at our summed states: |summed\rangle = (|ψ\rangle+|φ\rangle)(\langleφ|+\langleψ|) $$ = |φ\rangle \langleφ| +|φ\rangle \langleψ|+ |ψ\rangle \langleφ|+|ψ\rangle \langleψ| \\ = \rho_φ + \rho_ψ +|φ\rangle \langleψ|+ |ψ\rangle \langleφ| $$

$$ |φ\rangle \langleψ| = \frac{1}{\langleφ||ψ\rangle}|φ\rangle \langleφ||ψ\rangle \langleψ| = \frac{1}{\langleφ||ψ\rangle}\rho_ψ \rho_φ$$

$$ \rho_φ + \rho_ψ + \frac{1}{\langleφ||ψ\rangle}\rho_ψ \rho_φ + \frac{1}{\langleψ||φ\rangle}\rho_φ \rho_ψ$$

I'm not sure how to express $\langle \psi || \phi \rangle$ in terms of $\rho\phi$ and $\rho\psi$. I'm assuming its some combo of trace and multiplication, since ${\rm Tr}[\rho\phi\rho\psi] = 2 \langle \psi ||\phi \rangle \langle \phi ||\psi \rangle$.

Answer 1

What you want is not possible: You cannot represent the density operator $\rho$ of $\lvert\phi\rangle+\lvert\psi\rangle$ as a function of $\rho_\phi$ and $\rho_\psi$.

The reason is that $\lvert\phi\rangle+\lvert\psi\rangle$ depends on the global phase of the two states -- e.g., if you change $\lvert\psi\rangle\to e^{i\vartheta}\lvert\psi\rangle$, you will get a different state (not only by a global phase), and thus also a different $\rho$.

Yet, the information about the global phase is not contained in $\rho_\phi$ and $\rho_\psi$. Thus, there is no way to express $\rho$ solely using of $\rho_\phi$ and $\rho_\psi$: You need a way to additionally fix the relative phase of the states $\lvert\phi\rangle$ and $\lvert\psi\rangle$. (The individual global phases are not relevant, only their difference: A joint global phase drops out again in $\rho$.)