# Understanding Mixed Basis States in Hilbert Space: Uses and Applications

• I
• kered rettop
kered rettop
We can represent a state as a vector in Hilbert space. The Hilbert space can be spanned by a set of base vectors. The set is not unique: we can choose to work in any convenient basis. The state vector is therefore a superposition in some (most) bases but not all.

The above views everything as pure states. My question concerns mixed states. To be clear, I do mean proper mixed states, probability distributions. Is there any use for a "basis" comprising mixed states?

Thanks.

First of all to be precise also pure states are not exactly represented by a vector in Hilbert space but by a ray or equivalently by a projection operator as the statistical operator. The reason is that for the representation of the state an arbitrary phase factor of the (normalized) vector is irrelevant and unobservable. This is sometimes important (e.g., it's why half-integer spin makes sense as an observable). As you say, you can represent any vector uniquely by its components with respect to an arbitrary complete set of normalized orthogonal vectors (a basis of the Hilbert space). The state itself is entirely independent of the choice of this basis. You can choose any basis that is most convenient to solve some given problem.

A general state is represented by a positive semidefinite self-adjoint operator of trace 1. It's completely independent of any basis, but of course it can be represented by a basis. If ##|n \rangle## (##n \in \mathbb{N}##) is a complete orthonormal basis, then the matrix elements of the statistical operator ##\hat{R}## uniquely represent this statistical operator,
$$R_{mn}=\langle m|\hat{R}|n \rangle.$$
Since by definition the basis is complete,
$$\sum_{n=1}^{\infty} |n \rangle \langle n|=\hat{1}$$
you immediately can reconstruct the statistical operator from its matrix elements wrt. this basis:
$$\hat{R}=\sum_{m,n=1}^{\infty} |m \rangle \langle m|\hat{R}|n \rangle \langle n| = \sum_{m,n=1}^{\infty} \hat{R}_{mn} |m \rangle \langle n|. \qquad (*)$$
A particularly simple basis is of course the eigenbasis of the statistical operator,
$$\hat{R} |u_n \rangle=\rho_n |u_n \rangle.$$
Because of the properties of the statistical operator the ##\rho_n \in [0,1]## and ##\sum_{n} \rho_n=1##.

Now if you choose these eigenvectors as basis the matrix elements get particularly simple, because
$$\rho_{mn}=\langle u_m |\hat{R}|u_n \rangle=\rho_n \langle u_m|u_n \rangle=\rho_n \delta_{mn}.$$
Plugging this into (*) you get
$$\hat{R}=\sum_{n=1}^{\infty} \rho_n |u_n \rangle \langle u_n|.$$
We note that ##\hat{R}## represents a pure state if only one ##\rho_n=1## and all others ##\rho_m=0## for ##m \neq n##, because only then the statistical operator is a projection operator
$$\hat{R}=|u_n \rangle \langle u_n|.$$

Yes, ray not vector!

Well, thanks a lot for your answer. I may have to chew through it a few times when my brain's a bit clearer. Have you shown that my question was meaningless?

vanhees71
No, why should it be meaningless? It's an important topic. Mixed states are needed to make QT complete. It's not only that it's needed to describe situations, where you need a reduced description, like in many-body physics, where it's simply impossible to describe 1 mol of some matter (i.e., about ##6 \cdot 10^{23}## particles) in all detail. This is as in classical statistical mechanics: you cannot describe all details, so you try a "coarse grained" description of some "relevant degrees of freedom" in a statistical way. E.g., to describe a gas you don't need and cannot describe all the trajectories of each particle, but you rather want to describe it in terms of some "macroscopic" properties like the density, the fluid velocity, temperature, pressure etc. Then you use a coarse-grained description. In thermal equilibrium you then use the Maxwell-Boltzmann distribution or the corresponding quantum version, i.e., the microcanonical, canonical or grand-canonical statistical operators.

In addition you also need it to describe parts of a composed quantum system, even if the complete system is in a pure state. That's particularly important if you have to deal with "entangled states", which is a truely quantum property with no analogue in classical physics. Take e.g., the spins of two electrons in the state with total spin 0, i.e., with the state vector
$$|\psi \rangle=\frac{1}{\sqrt{2}} (|1/2,-1/2 \rangle -|-1/2,1/2 \rangle).$$
The true unique description of the state is then the statistical operator
$$\hat{R}=|\psi \rangle \langle \psi|.$$
The question now is, what's the statistical operator for the spin of one of the electrons? That's given by the "reduced statistical operator", defined by a partial trace, i.e.,
$$\hat{R}_1=\sum_{\sigma_1,\sigma_1',\sigma_2=\pm 1/2} |\sigma_1 \rangle \langle \sigma_1,\sigma_2|\hat{R}|\sigma_1',\sigma_2 \rangle \langle \sigma_1'|=\frac{1}{2} \sum_{\sigma_1=\pm 1/2} |\sigma_1 \rangle \langle \sigma_1|=\frac{1}{2} \hat{1}.$$
This means that in this pure state of the two electrons the single electron is unpolarized, i.e., it is not in a pure state. That's characteristic for entangled states.

kered rettop
Ah! Right! We get a mixed state when we have an entanglement and decide to focus on just one of subsystems, which turns out to be a mixture. But then I get lost because my maths is elementary, I'm afraid. It did occur to me to Google "representing a mixed state in Hilbert space" which led to the statement in Wikipedia that "A pure quantum state can be represented by a ray in a projective Hilbert space over the complex numbers, while mixed states are represented by density matrices, which are positive semidefinite operators that act on Hilbert spaces." But my original question - which you can blame on the tail end of a nasty Covid-like bug - was about constructing a state space which incorporates the mixture in its structure instead of having it imposed by an external matrix. I'm still not convinced, despite your kind assurances, that it is a meaningful question but am open to being persuaded. Thanks a lot for taking the trouble to point me in the right direction even if my steering is then unreliable!

vanhees71
I'm not sure, whether I understand your question. You can as well say that all states are represented by a statistical operator. A state is pure if the statistical operator is a projection operator, i.e., if ##\hat{R}^2=\hat{R}##. Proof: Since ##\hat{R}## is self-adjoint there's a complete orthonormal set of eigenvectors, and since its positive semidefinite all eigenvalues ##p_n \geq 0##. Further its trace is 1, i.e., ##\sum_n p_n=1##. This implies that exactly one of the eigenvalues is ##p_1=1## and all other thus must be ##p_n=0## for all ##n \neq 1##. This implies that ##\hat{R}=|u_1 \rangle \langle u_1|##.

Then it's also clear that ##\hat{R}## determines the unit vector ##|u_1 \rangle## only up to a phase factor. Indeed with ##|u_1' \rangle=\exp(\mathrm{i} \varphi) |u_1 \rangle## with ##\varphi \in \mathbb{R}## you have ##|u_1' \rangle \langle u_1'|=\exp(\mathrm{i} \varphi) |u_1 \rangle \langle u_1| \exp(-\mathrm{i} \varphi) = |u_1 \rangle \langle u_1|=\hat{R}##, i.e., if ##\hat{R}## is a pure state, it determines the corresponding state ket ##|u_1 \rangle## only up to an arbitrary phase factor, but that doesn't matter since all physics is contained in ##\hat{R}##, and thus which phase factor you choose for ##|u_1 \rangle## does not affect any physical content of the statistical operator.

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