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

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 $\psi_3(x)$...

• There seem to be several misconceptions in this question. First off, you can have mixed states for position; there is absolutely nothing different about the math. Second, you're mixing up the density operator (which is an operator) with the position probability distribution (which is a number). Of course you can make a pure state have any position distribution you want. That does not mean every mixed state is pure. – knzhou Apr 20 '18 at 9:16

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.