# What is the qualitative difference between quantum superpostion and mixed states? [duplicate]

As I understand it, if one has a complete knowledge of the state of a quantum system (insofar as one knows the statistical distributions of all the observables associated with the state) then one can represent it as a state vector (or ket) in an associated Hilbert space, and we say that the system is in a pure state. Furthermore, given an (orthonormal) basis for the Hilbert space, one can express this state vector as a linear combination of the basis vectors. This is usual carried out through the use of an eigenbasis induced by an operator acting on the Hilbert space, representing some observable. We say then that the state is in a quantum superposition.

What confuses me is how this differs from a mixed state? I understand that the situation is at least somewhat different since a mixed state arises when we have a lack of knowledge about the state (i.e. we lack all the possible information we could, in principle, have about it, that is, the statistical distributions of all the observables associated with the state). Hence, we must consider a statistical ensemble of possible pure states that the system could be in, each having an associated probability. This is a so-called classical probability as it does not stem from the intrinsic probabilistic nature of a quantum system, but rather the fact that we lack all the knowledge that we could possibly have about the system.

Is it simply, that in the case of a pure state, even though we know the statistical distribution of the observables associated with this state, we do not, a priori, before measuring the given observable, Know which eigenstate the quantum system is in, and hence must consider it to be in a quantum superposition of the available eigenstates? (In this case, it would be a so-called quantum probability, since such an uncertainty does not arise from lack of information about the state of the system, but is intrinsic in the quantum nature of the system).

Apologies for the long-windedness of this post, I just thought I'd write evergthing down on my thoguhts about it, and hopefully someone can correct/ explain it to me.

Edit: I think that perhaps my confusion stems from how to interpret a quantum superposition of states. How should one interpret this physically? (if I have an understanding of this perhaps it'll clear up things a little).

• Possible duplicate of How is quantum superposition different from mixed state? Apr 25 '16 at 12:21
• "This is a so-called classical probability as it does not stem from the intrinsic probabilistic nature of a quantum system, but rather the fact that we lack all the knowledge that we could possibly have about the system". It is untenable in my opinion, see physics.stackexchange.com/q/98703 Apr 25 '16 at 12:26
• There is a viewpoint on QM where pure and mixed quantum states are treated at the same level. The distinction is just due to the fact that pure state are extremal element of the class of the states. See my answer physics.stackexchange.com/q/116595 Apr 25 '16 at 12:30
• @AccidentalFourierTransform Thanks for the link. I read that post before writing my own, as I didn't feel that it answered the question for me. Apr 25 '16 at 12:32
• I actually typed an answer to this, but then checked the accepted answer at physics.stackexchange.com/questions/80434/… and found that I was saying exactly the same thing, so I deleted my answer. You really will find your answer there. Apr 25 '16 at 13:45

A pure state is a linear combination of basis states $|\psi\rangle = \sum_k c_k |b_k\rangle$. A pure state has unit 2-norm; pure states care about squared weight $\sum_k |c_k|^2=1$. Meaning the weights are amplitudes.

A mixed state is a linear combination of adjoint-squared pure states $\rho = \sum_k p_k |\psi_k\rangle\langle\psi_k|$. A mixed state has unit 1-norm; mixed states care about linear weight $\sum_k p_k = 1$. Meaning the weights are probabilities.

Equivalently, a mixed state is a probability distribution of pure states. If you're not sure which pure state the system is in, your ability to predict is described by a mixed state.

Equivalently, a mixed state is what you get when you marginalize a pure state. If you don't have access to some qubits that your system is entangled with, your ability to predict is described by a mixed state.

You can represent a pure state as a mixed state with a single 100%-likely pure state (i.e. as a density matrix with a single non-zero eigenvalue equal to 1).

You can represent a mixed state as a pure state by adding extra qubits (i.e. purification).

Since each can represent the other, people disagree about whether the fundamental thing is mixed states or pure states. Fortunately the math doesn't care.

• How is the linear combination (or superposition) interpreted physically though? Is it simply the fact that one cannot know prior to measurement which eigenstate of a given operator the system is in, and hence, if we use the eigenstates of this operator as a basis, we must consider the system to be in a superposition of all possible eigenstates of this operator?... Apr 25 '16 at 16:36
• ...(I understand mathematically that the state can be expressed as a linear combination due to the state being an element of a vector space and thus can be represented as a linear combination of some basis for the vector space, and also that a superposition of pure states as a solution is a consequence of the linearity of the Schrödinger equation but I'm unsure how to interpret this physically?) Apr 25 '16 at 16:38
• @user35305 It's not that we're missing information about pure states, it's that the information you think exists doesn't. If you're walking northeast, is your velocity north, or is it east? This question has no definite answer, just like, "which eigenstate is the system in" has no answer. Apr 25 '16 at 16:56
• @user35305 The physical interpretation is philosophically controversial. What people do agree on is the math; the linear algebra that describes quantum information. Your best bet is to learn that before worrying about issues of interpretation. I recommend this video series. Apr 25 '16 at 17:00

I'll try to add a rather intuitive take on the difference between pure and mixed states.

Take first the simple example of a single spin-1/2 particle. Its pure states may always be written as superpositions of spin-up and spin-down states measured along some particular direction $z$. That is, we write $|\psi\rangle = a |\uparrow_z\rangle + b|\downarrow_z\rangle$ and interpret $|a|^2$, $|b|^2$ as the probabilities that spin measurements along direction $z$ yield either $|\uparrow_z\rangle$ or $|\downarrow_z\rangle$. However, if $|\psi\rangle$ is a pure state, we can always find a certain direction in 3d-space, say ${\vec u}$, such that spin measurements along ${\vec u}$ always yield with $100\%$ certainty the result $|\uparrow_{\vec u}\rangle$ (or $|\downarrow_{\vec u}\rangle$, depending on the chosen orientation). For a spin-1/2 particle this is the actual physical meaning and operational definition of a pure state. In contrast, mixed states $\hat \rho$ are states for which no such direction ${\vec u}$ may be found. Or if you prefer, for a mixed state the statistics of spin measurements along any direction ${\vec u}$ will always display non-zero probabilities, with non-zero uncertainties, for both $|\uparrow_{\vec u}\rangle$ and $|\downarrow_{\vec u}\rangle$.

Note: Historically, the spin-1/2 interpretation was inspired by the concepts of polarization and coherence for electromagnetic waves. In complete analogy, coherent electromagnetic radiation is characterized by perfect polarization along some direction, while for incoherent radiation no sharp polarization direction can be defined.

Let us generalize now to an arbitrary quantum system. First recall that:

1) The corresponding Hilbert space is always generated as the common pure state eigenbasis $\lbrace |\psi_\lambda\rangle \rbrace$ of a complete set of observables $\lbrace {\hat O}_1, {\hat O}_2, \dots {\hat O}_k\rbrace$. For the spin-1/2 particle the complete set was minimal and consisted of a single observable, ${\hat \sigma}_z$ or ${\hat \sigma}_{\vec u}$ (leaving aside total spin as redundant in this case), but in general we assume the (finite) number of independent observables is some $k > 1$.

2) There is actually a continuum of distinct complete sets of observables that are mutually related through unitary transformations. That is, if $\lbrace {\hat O}_1, {\hat O}_2, \dots {\hat O}_k\rbrace$ is a complete set with eigenbasis $\lbrace |\psi_\lambda\rangle \rbrace$ and ${\hat U}$ is a unitary transformation, ${\hat U}{\hat U}^\dagger = {\hat U}^\dagger{\hat U} = {\hat I}$, then $\lbrace {\hat U}{\hat O}_1{\hat U}^\dagger, {\hat U}{\hat O}_2{\hat U}^\dagger, \dots {\hat U}{\hat O}_k{\hat U}^\dagger\rbrace$ is also a complete set and determines an eigenbasis $\lbrace {\hat U} |\psi_\lambda\rangle \rbrace$.

3) Pure states can be mapped into each other by means of unitary transformations. That is, if $|\psi\rangle$, $|\phi\rangle$ are two distinct states, then there exists some unitary $\hat U$ such that $|\psi\rangle = {\hat U} |\phi\rangle$.

As a corollary of all this, it follows that any pure state $|\psi\rangle$ can be mapped by a unitary transformation $\hat U$ into an eigenstate $|\psi_\lambda\rangle$ of a complete set of observables $\lbrace {\hat O}_1, {\hat O}_2, \dots {\hat O}_k\rbrace$, that is, $|\psi\rangle = {\hat U} |\psi_\lambda\rangle$. But then it also follows that $|\psi\rangle$ is necessarily an eigenstate of the complete set $\lbrace {\hat {\bar O}}_1 = {\hat U}{\hat O}_1{\hat U}^\dagger, {\hat {\bar O}}_2 = {\hat U}{\hat O}_2{\hat U}^\dagger, \dots {\hat {\bar O}}_k = {\hat U}{\hat O}_k{\hat U}^\dagger\rbrace$.

In other words, for any pure state $|\psi\rangle$ there exists a complete set of observables $\lbrace {\hat {\bar O}}_1, {\hat {\bar O}}_2, \dots, {\hat {\bar O}}_k\rbrace$ that returns with $100\%$ certainty a set of average values with vanishing standard deviations, $\langle \Delta {\hat {\bar O}}_j^2 \rangle = 0$, $j = 1, 2, \dots, k$. This can be taken then as an operational definition of pure states.

Along the same lines, as before, mixed states $\hat \rho$ are then those for which there exists no complete set such that $\langle \Delta{\hat {\bar O}}_j^2 \rangle = 0$ for all $j = 1, 2, \dots, k$.

Everything else that has already been pointed out holds as usual.

• Thanks for your answer. When it comes to representing a state vector as a superposition of eigenstates of an operator (representing an observable of the quantum system), can one interpret this as being a result of the fact that the observable has a statistical distribution of values, which correspond to the eigenvalues of its associated operator.... Apr 25 '16 at 19:47
• ...Thus, before any measurement is made, (unless the state is prepared in a given eigenstate of an observable) the state isn't simply equal to a given eigenstate of the operator in question and instead one must in general consider it as being in a superposition of all possible eigenstates of the operator. Apr 25 '16 at 19:47
• Not sure what you are trying to say. An observable has eigenvalues and eigenstates that are independent of any state on which the observable might be measured. The statistical distribution always pertains to the measured state, be it pure or mixed. Likewise, the information encapsulated in a state is independent of any particular choice of observables that are to be measured on it in the future, but it is not arbitrary, even prior to measurement. We are not talking information that is inaccessible due to the observer's limitations, but the actual information content of a quantum state.
– udrv
Apr 25 '16 at 20:38
• That is, the same pure state can yield a wide statistical distribution for some observables, but sharp values with null deviation on others. Think a $|\uparrow_z\rangle$ state for the spin-1/2 example. It gives a $+1$ value with probability $1$ in a ${\hat \sigma}_z$ measurement, but only a probability of $1/2$ for either eigenstate of ${\hat \sigma}_x$ or ${\hat \sigma}_y$. Or a vertically polarized monochromatic em mode: gives a sharp result for a vertical vs. horizontal polarization measurement, but only a 50-50 distribution for a 45º vs. 135º measurement.
– udrv
Apr 25 '16 at 20:38
• Another easy example: stationary atomic orbitals are eigenstates corresponding to well-defined energy and total angular momentum eigenvalues (probability one), but measurements of the position operator give the more or less complex spatial probability distributions you are familiar with (although there may be objections as to the actual nature of position eigenstates). Am I anywhere close to what you are asking?
– udrv
Apr 25 '16 at 20:39