Is It Meaningful to Talk About Pure vs. Mixed States for (Continuous) Position?

Question

I see a lot written about pure and mixed states regarding state vectors and density matrices/operators that contain a finite number of states/elements.

For something like the continuous state vector of position $|\psi\rangle = \int_{-\infty}^{\infty} \psi(x) |x\rangle dx$, where $\psi(x)$ is the position wave function and $\int_{a}^{b} \rho(x) dx=\int_{a}^{b} \psi^*(x)\psi(x) dx$ yields the probability of measuring the object within $[a,b]$. I think that we could also speak of a continuous density matrix/operator $\hat\rho(x) = |\psi\rangle \langle\psi|$ such that $\rho = \langle x|\hat\rho|x \rangle$ and when $\hat\rho$ is inserted into the resolution of the identity integrated over all x the result is $1$, confirming the $100\%$ probability that the object is somewhere in space.

For continuous position, is it meaningful to speak of pure vs. mixed states? I would think that a mixed state is where we have $\rho(x)$ but not $\psi(x)$? That might happen after entanglement? Consider the double slit experiment, using photons, where we put different polarizing filters over each slit and then have a radial $\psi(x)$ emerging from each slit, but not adding together. In that case, we would have $$ \rho(x) = \frac{1}{2}(\psi^*_1(x)\psi_1(x) + \psi^*_2(x)\psi_2(x)). $$ Is it not possible to find some $\psi_3(x)$ such that $$ \frac{1}{2}(\psi^*_1(x)\psi_1(x) + \psi^*_2(x)\psi_2(x)) = \psi^*_3(x)\psi_3(x)? $$ I believe that for QM, we demand that any $\psi(x)$ be a "test function", such as a Gaussian or a smooth function of compact support. Because of the restriction on $\psi(x)$, maybe there are situations where there is no such $\psi_3(x)$...

Answer 1

Yes, it's perfectly meaningful to talk about pure vs mixed states regardless of the dimension of the Hilbert space.

On the other hand, your understanding of what the density matrix is suggests that you're only seeing a small part of the true core of the concept: by "density matrix" we don't just mean density, we also mean that it must be seen as a matrix i.e. as an operator, which can be queried (in the position basis) with different states on either side: thus, for a pure state, you can happily get $$ \rho(x,x') = \langle x| \hat \rho|x'\rangle = \langle x| \psi\rangle \langle \psi|x'\rangle = \psi(x) \psi(x')^*, $$ with different positions on the two factors. The diagonal of the density matrix, $\rho(x,x) = \rho(x)$, encodes the population (density) over the basis states in the chosen representation, much like it does in finite dimension, and (again like it does in finite dimension) the off-diagonal components $\rho(x,x')$ for $x\neq x'$ encode the coherence between the population present at different sites.

This then informs the answer to your subsidiary question: is it possible to find a state $\psi_3(x)$ such that $$ \frac{1}{2}(\psi^*_1(x)\psi_1(x) + \psi^*_2(x)\psi_2(x)) = \psi^*_3(x)\psi_3(x) $$ given states $\psi_1(x)$ and $\psi_2(x)$? Absolutely, just try $\psi_3(x) = \sqrt{\frac{1}{2}(\psi^*_1(x)\psi_1(x) + \psi^*_2(x)\psi_2(x))}$. But that's the wrong question: what you need to be asking is whether, given two linearly-independent states $\psi_1(x)$ and $\psi_2(x)$, is there a state $\psi_3(x)$ such that $$ \frac{1}{2}(\psi^*_1(x)\psi_1(x') + \psi^*_2(x)\psi_2(x')) = \psi^*_3(x)\psi_3(x') $$ for all (independent) real $x$ and $x'$? Then the answer is simple: no. The operator on the left has rank $2$ and the operator on the right has rank $1$, so they can never be equal.

That's probably still insufficient to clear away all of the misconceptions your post hints at, but it's hopefully enough to point you in the right direction.