# What is the actual meaning of the density operator?

I am not able to understand the definition of the density operator. I know that if $V$ is a vector space and if I have $k$ states belonging to this vector space, say $|\psi_{i}\rangle$ for $1\le i\le k$, and the probability of the system being in $|\psi_{i}\rangle$ is $p_i$, then the density operator for the system is given by $$\rho=\sum_{i=1}^{i=k}p_i|\psi_i\rangle\langle\psi_i |.$$ Now what I am unable to understand is

1. Does this mean the system is in exactly one of the $|\psi_i\rangle$ states and we don't know in which state it is in and we just know the probability?
2. Or it is in a superposition of these $k$ states with probabilities being interpreted as weights?

Firstly, what is a state?

A state gives you the complete description of a system. Let's label the state of a system $$\lvert \psi \rangle$$. This is a normalised state vector which belongs in the vector space of states. Keep in mind that we are talking about the full state; I haven't decomposed it into basis states, and I will not. This is not what the density matrix is all about.

The state vector description is a powerful one, but it is not the most general. There are some quantum experiments for which no single state vector can give a complete description. These are experiments that have additional randomness or uncertainty, which might mean that either state $$\lvert \psi_1⟩$$ or $$\lvert \psi_2⟩$$ is prepared. These additional randomness or uncertainties arise from imperfect devices used in experiments, which inevitably introduce this classical randomness, or they could arise from correlations of states due to quantum entanglement.

In this case, then, it is convenient to introduce the density matrix formalism. Since in quantum mechanics all we calculate are expectation values, how would you go about calculating the expectation value of an experiment where in addition to having intrinsic quantum mechanical randomness you also have this classical randomness arising from imperfections in your experiment?

Recall that $$Tr(\lvert\phi_1\rangle\langle\phi_2\rvert)=Tr(\lvert\phi_1\rangle\otimes\langle\phi_2\rvert)=\langle\phi_1\mid\phi_2\rangle,$$ and $$\hat{O}\circ(\lvert\phi_1\rangle\langle\phi_2\rvert)=(\hat{O}\lvert\phi_1\rangle)\otimes\langle\phi_2\rvert$$

Now, using the linearity of the trace, we can compute the expectation value as:

$$\langle \hat{O} \rangle = p_1\langle \psi_1 \lvert \hat{O} \lvert \psi_1 \rangle + p_2\langle \psi_2 \lvert \hat{O} \lvert \psi_2 \rangle$$ $${} = p_1Tr(\hat{O} \lvert \psi_1 \rangle \langle \psi_1 \lvert) + p_2Tr(\hat{O} \lvert \psi_2 \rangle \langle \psi_2 \lvert)$$ $${} =Tr(\hat{O} (p_1 \lvert \psi_1 \rangle \langle \psi_1 \lvert) + p_2 \lvert \psi_2 \rangle \langle \psi_2 \lvert)) = Tr(\hat{O} \rho)$$

where $$p_1$$ and $$p_2$$ are the corresponding classical probabilities of each state being prepared, and $$\rho$$ is what we call the density matrix (aka density operator): it contains all the information needed to calculate any expectation value for the experiment.

So your suggestion 1 is correct, but suggestion 2 is not, as this is not a superposition. The system is definitely in one state; we just don't know which one due to a classical probability.

• +1 for the thorough answer. It might be worth pointing out that not only the expectation of $\hat{O}$ but also all the moments of $\hat{O}$ can be computed as $\left<\hat{O}^n\right> = \mathrm{tr}(\rho\,\hat{O}^n)$ and thus, by linearity $\left<\exp(i\,\hat{O})\right> = \mathrm{tr}(\rho\,\exp(i\,\hat{O}))$. This last one is none other than the characteristic function of the pdf for measurements by $\hat{O}$, so not only can you get expectations, but also the whole probability density function for measurements. Jan 8, 2015 at 12:38
• @sasha The comment above was meant for you too. Another time the density matrix is applicable is in the analysis of Wigner's friend thought experiments. Jan 8, 2015 at 12:38
• "The system is definitely in one state, we just don't know which one due to a classical probability." -- Didn't you say yourself that the uncertainty can also arise from entanglement with another system? How can you say in that case that the system is definitely in one state? Jan 8, 2015 at 12:43
• @sasha Should have said $\left<\exp(i\,\omega\,\hat{O})\right> = \mathrm{tr}(\rho\,\exp(i\,\omega\,\hat{O}))$ where $\omega\in\mathbb{R}$ for the characteristic function, whence the inverse Fourier transform yields the pdf. Jan 8, 2015 at 12:45
• @NorbertSchuch the uncertainty arises from entanglement effects messing up your experiment and producing these classical probabilities. Jan 8, 2015 at 13:19

Check the form of your density operator it should be something like $$\rho=\sum_i p_i |\psi_i\rangle\langle\psi_i|$$

I find the most intuitive way to think of the density matrix as follows. Consider an experimenter in his lab who has a "machine" which produces the quantum state $|\psi\rangle$ but the machine doesn't work perfectly and produces some other states which we don't want. Then to carry out predictions about these states we need something which captures our ignorance of the states which we have i.e we could not use state vectors as we do not have pure states but an ensemble or mixture of states. The density matrix is then what we use. Your suggestion 1.

As an aside the density operator can also describe pure states.

Your suggestion number two is incorrect as superpositions and ensembles are physically quite different things. Consider the pure state $$|\psi\rangle=\sum_i \sqrt{p_i}\ |\psi_i\rangle$$ Note that the weights are the square root of the probability which relates to the born rule.

What is new in this answer w.r.t to the other answers. I introduce the density operator by inserting the identity $$1 = \sum \vert i\rangle \langle i \vert$$ in the average, similarly to how one can use the $$\operatorname{tr} \langle A \rangle = \langle A \rangle$$ trick to do the same thing.

Conceptual comments. The density matrix comes whenever your system is part of a bigger system and interacts with it. Since you can't model the total, you average out the effect of the other system on yours. This is equivalent to taking a "marginal distribution/state". It can be interpreted as the system being in one pure state or another with different probabilities.

The mathematical details

Suppose that we have a system which we divide in a subsystem and an environment. We know a basis for the subsystem, $$\vert {\phi_i} \rangle$$, and for the environment, $$\vert {\xi_j} \rangle$$. I assume them to be orthonormal. Then a generic state is: $$\vert {\psi} \rangle = \sum_{ij} c_{ij} \vert {\phi_i} \rangle \otimes \vert {\xi_j} \rangle$$ Suppose we want to perform a measurement on our subsystem only, meaning we have an operator $$O$$ which only acts on $$\vert {\phi_i} \rangle$$: $$(O\otimes {1}) (\vert {\phi_i} \rangle \otimes \vert{\xi_j}) \rangle \equiv (O\vert {\phi_i})\rangle \otimes \vert {\xi_j} \rangle$$ Its expected value is given by \begin{align*} \langle O \rangle = \langle {\psi} \vert O \vert {\psi} \rangle \end{align*} We don't know the total state $$\vert \psi \rangle$$, but we do know is the possible states of our subsystem, $$\lbrace \vert \phi_m \rangle\rbrace$$. What is the best we can do knowing only this?

We can use the full expression of $$\vert {\psi} \rangle$$ and try to "hide" the things we don't know how to calculate, like the expressions involving $$\vert {\xi_j} \rangle$$. We can use that the inner product of tensor products is $$(\langle \phi_i \vert \otimes \langle \xi_j \vert)(\vert \phi_k \rangle \otimes \vert \xi_l \rangle)=\langle \phi_i \vert \phi_k \rangle \langle \xi_j \vert \xi_l \rangle$$ (or alternatively, one can think of the tensor product being defined such that this is true by construction).

\begin{align*} \langle O \rangle = \langle {\psi} \vert O \vert {\psi} \rangle = & \sum_{ij} \sum_{kl}c_{kl} c_{ij}^* (\langle \phi_i \vert \otimes \langle \xi_j \vert) ( O \otimes 1 ) (\vert \phi_k \rangle \otimes \vert \xi_l \rangle) \\ =& \sum_{ij} \sum_{kl} c_{kl}c_{ij}^* \langle \phi_i \vert O \vert \phi_k \rangle \langle \xi_j \vert \xi_l \rangle \\ =& \sum_{ik} \left(\sum_{m} c_{km}c_{im}^* \right) \langle \phi_i \vert O \vert \phi_k \rangle \\ =& \sum_{ik} p_{ki} \langle \phi_i \vert O \vert \phi_k \rangle \end{align*}

So basically so far this is it, because $$p_{ki}$$ gives us all the information we need to compute the average value; $$p_{ki}$$ is the marginal distribution of $$c_{ij}^* c_{kl}$$ (by "marginal" we mean we integrate out or sum the probability over the degrees of freedom of the environment).

Since it has two indices we could think of it as the element of a matrix, and that matrix is easy to construct, we just insert the identity $$1=\sum_m \vert \phi_m \rangle \langle \phi_m \vert$$ and get the density matrix $$\rho$$ by using the commutativity of scalars (we can move them around in the product): \begin{align*} \sum_{ik} p_{ik} \langle \phi_i \vert O \vert \phi_k \rangle = & \sum_{ik} p_{ik} \langle \phi_i \vert \sum_m \vert \phi_m \rangle \langle \phi_m \vert O \vert \phi_k \rangle \\ = & \sum_{ik} \sum_m p_{ik} \langle \phi_m \vert O \vert \phi_k \rangle \langle \phi_i \vert \phi_m \rangle \\ = & \sum_m \langle \phi_m \vert O \left(\sum_{ik} p_{ik} \vert \phi_k \rangle \langle \phi_i \vert\right) \vert \phi_m \rangle = \sum_m \langle \phi_m \vert O \rho \vert \phi_m \rangle \equiv \operatorname{tr} O \rho \end{align*} where $$\rho=\sum_{ik} p_{ki} \vert \phi_k \rangle \langle \phi_i \vert$$.

We have reduced the computation of the global average to calculating averages over the states of our subsystem of a new object, $$O\rho$$. The object $$\rho$$ has hidden in it both the state of our subsystem and the average effect of the environment. I will make a few further comments about how one can think about $$\rho$$ and the mathematical justification.

First of all, it is easy to prove (see any textbook or Wikipedia) that $$\rho^\dagger = \rho$$, therefore there exists an orthonormal basis in which $$\rho$$ is diagonal: $$\rho = \sum_\lambda p_\lambda \vert \lambda \rangle \langle \lambda \vert \quad \sum p_\lambda = 1$$ What is the significance of the states $$\vert \lambda \rangle$$? They represent the most "classical-like" states. This is so because they are orthonormal states, $$\langle \lambda \vert \lambda^{'} \rangle=\delta_{\lambda \lambda^{'}}$$, there are no correlations between them. They are as mutually exclusive as you can get, meaning that if you know your system is in state $$\vert \lambda \rangle$$, then automatically it cannot be in $$\vert \lambda^{'} \rangle$$ if $$\lambda^{'}\neq \lambda$$ because the projection onto it is zero.

Consider a two level system $$\lbrace \vert 0 \rangle, \vert 1 \rangle \rbrace$$. Then if you are in state $$\vert 0 \rangle$$, you know you cannot be in $$\vert 1 \rangle$$, but you have $$50\%$$ of being in any of the two diagonal states, therefore the system isn't "classical-like" regarding diagonal states. Also, if you are assured all your particles are in one of these states ($$\vert 0 \rangle$$ or $$\vert 1 \rangle$$), you can measure without collapsing provided you have an observable with $$\lbrace \vert 0 \rangle, \vert 1 \rangle \rbrace$$ as eigenstates.

Such an observable always exists because we can just build it ourselves, we only need it to have the same eigenstates as the density matrix, $$\lbrace \vert \lambda \rangle \rbrace$$: $$M = \sum_\lambda m_\lambda \vert \lambda \rangle \langle \lambda \vert$$ (It measures 0 on everything which is outside $$\text{span}\lbrace \vert \lambda \rangle \rbrace$$).

The usefulnes of this observable is that we can use it to prepare the mixed state! We measure a random $$\vert \psi \rangle$$. After measuring, we know the state $$\vert \lambda \rangle$$ to which it collapsed. We will take $$N$$ measurements of $$N$$ different states and after a measurement we will keep the state or discard it according to whether our ratio of states is the one prescribed by the density matrix, $$p_\lambda = N_\lambda / N$$ where $$N_\lambda$$ is the number of states in state $$\vert \lambda \rangle$$. If we now measure $$M$$, then the average value will be

$$\langle M \rangle = \sum \frac{N_\lambda}{N} \langle \lambda \vert M \vert \lambda \rangle = \sum p_\lambda m_\lambda$$

and we see that this is a classical average over $$p_\lambda$$ of the distribution $$m_\lambda$$. It can be shown that this equals $$\text{tr}M\rho$$.

As a curiosity, Von Neumann uses the idea of $$M$$ in his book "Mathematical Foundations of Quantum Mechanics" as a way to implement a filter which separates quantum systems in a way that doesn't collapse them. If you have particles in two different quantum states and all mixed in a box, you can physically separate the particles with a moving wall which acts as a filter only if the possible states of the particles are orthogonal. In that case you can find a measurement which can distinguish between the two states without collapsing them, thus allowing you to create a filter which lets or lets not pass a certain kind of particle, letting you separate the quantum systems.

So to sum up, the first introduction was a top down approach, where we got the density matrix by averaging out the environment, arriving at the fact that mathematically the result is like having an average over pure states. The second approach is taking that "average over pure states" as reality and considering ensembles of pure states. Then, if we have $$N$$ systems and $$N_\lambda$$ in state $$\vert \lambda \rangle$$, then averaging an observable over the $$N$$ systems gives the same result as a mixed state given by $$\rho$$.

The density operator of a quantum system summarises the expectation values of the observables of that system alone. If you have two systems $S_1$ and $S_2$ that are entangled with one another, then there is no pure Schrodinger picture state for either of the entangled systems. However, there are observables of $S_1$ (or $S_2$) and those observables have expectation values. The standard way of writing expectation values in terms of the pure state is given by $\langle\psi|\hat{A}|\psi\rangle$ for some arbitrary observable $\hat{A}$, but this can also be written as $\text{tr}(|\psi\rangle\langle\psi|\hat{A})$. So then the question is whether there is an operator on the Hilbert space of $S_1$ alone that would give you the right expectation values using the second formula. It turns out that the correct choice is given by the partial trace since the partial trace is the only choice consistent with the Born rule, see "Quantum Computation and Quantum Information" by M.A. Nielsen, I.L. Chuang, p. 107.

Does this mean the system is in exactly one of the $|\psi_i\rangle$ states and we don't know in which state it is in and we just know the probabilities?
No. If $S_1$ and $S_2$ can be manipulated jointly in a coherent process then they can undergo quantum interference. During quantum interference, the square magnitude of probability amplitudes do not in general obey the calculus of probabilities, see