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What is the motivation behind density matrices? And, what is the difference between the density matrices of pure states and density matrices of mixed states?

This is a self-answered sequel to What's the difference between a pure and mixed quantum state? & How to find a density matrix of a qubit? You're welcome to write alternate answers.

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Motivation

The motivation behind density matrices is to represent a lack of knowledge about the state of a given quantum system, encapsulating within a single description of this system all the possible outcomes of measurement results, given what we know about the system. The density matrix representation has the added advantage of getting rid of any issues associated with global phases because $$ |\phi\rangle\langle\phi|=(e^{i\varphi}|\phi\rangle)(e^{-i\varphi}\langle\phi|). $$ The lack of knowledge might arise in a variety of ways:

  • A subjective lack of knowledge - a referee prepares for you one of a set of states $\{|\phi_i\rangle\}$ with probability $p_i$, but you don't know which. Even if they know which $|\phi_j\rangle$ they prepared, since you do not, you have to describe it based upon what you know of the possible set of states and their corresponding probabilities, $\rho=\sum_ip_i|\phi_i\rangle\langle \phi_i|$.

  • An objective lack of knowledge - if the quantum system is part of a larger entangled state, it is impossible to describe the system as a pure state, but all possible outcomes of measurements are described by the density matrix obtained by $\rho=\text{Tr}_B(\rho_{AB})$.

It is interesting, however, that the objective lack of knowledge can become subjective - a second party can perform operations on the rest of the entangled state. They can know the measurement results etc. but if they don't pass those on, the person holding the original quantum system has no new knowledge, and so describes their system using the same density matrix as before, but it is now a subjective description.

It is also important to note that choosing a particular way of representing the density matrix, for example, $\rho=\sum_ip_i|\phi_i\rangle\langle \phi_i|$, is a very subjective choice. It may be motivated by a particular preparation procedure, but mathematically, any description that gives the same matrix is equivalent. For example, on a single qubit, $\rho=\frac12\mathbb{I}$ is known as the maximally mixed state. Due to the completeness relation of a basis, this can be represented as a 50:50 mixture or two orthogonal states using any 1-qubit basis. $$ \frac12\mathbb{I}=\frac12|0\rangle\langle 0|+\frac12|1\rangle\langle 1|=\frac12|+\rangle\langle +|+\frac12|-\rangle\langle -|. $$

Pure and Mixed States

The difference between the density matrix of a pure state and a mixed state is straightforward - the pure state is a special case which can be written in the form $\rho=|\psi\rangle\langle\psi |$, while a mixed state cannot be written in this form. Mathematically, this means that the density matrix of a pure state has rank 1, while a mixed state has rank greater than 1. The best way of calculating this is via $\text{Tr}(\rho^2)$: $\text{Tr}(\rho^2)=1$ implies a pure state, otherwise it's mixed. To see this, recall that $\text{Tr}(\rho)=1$, meaning that all the eigenvalues sum to 1. Also, $\rho$ is positive semi-definite, so all the eigenvalues are real and non-negative. So, if $\rho$ is rank 1, the eigenvalues are $(1,0,0,\ldots ,0)$, and their sum-square is clearly 1. The sum-square of any other set of non-negative numbers that sum to 1 must be less than 1.

The pure state corresponds to perfect knowledge of the system, although the fun bit about quantum mechanics is that this does not imply full knowledge of the possible measurement outcomes. Mixed states represent some imperfect knowledge, whether that's knowledge of the preparation, or knowledge of a larger Hilbert space.

That the mixed state description is much richer can be seen from the Bloch sphere picture on a single qubit: the pure states are all those on the surface of the sphere, while the mixed states are all those contained within the volume. In terms of parameter counting, instead of two parameters, you need three, the extra one corresponding to the length of the Bloch vector. $$ \rho=\frac{\mathbb{I}+r\underline{n}\cdot\underline{\sigma}}{2}, $$ where $\underline{n}$ is a 3-element unit vector, $\underline{\sigma}$ is a vector of the Pauli matrices, and $r=1$ for a pure state, and $0\leq r<1$ for a mixed state.

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  • $\begingroup$ (+1) Thank you, as per my understanding, we have the state $|\Psi_{AB}\rangle$ and want to know about $|\Psi_A\rangle$, and there is no preexisting way to find it, hence we are defining density matrix, am I correct? Do we have different definitions of density matrix for different purpose? As, you have mentioned for $\rho=\sum_ip_i|\phi_i\rangle\langle \phi_i|$ for subjective lack of knowledge and for objective $\rho=\text{Tr}_B(\rho_{AB})$ , firstly,it is not clear to me what do you mean by lack of knowledge ? $\endgroup$ – tarit goswami Sep 7 '18 at 20:55
  • $\begingroup$ (contd.) Secondly, can you explain with example what do you mean by subjective and objective ? $\endgroup$ – tarit goswami Sep 7 '18 at 21:00
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    $\begingroup$ @taritgoswami objective means that everybody agrees. So, if I make a pure state, and announce it to the world, everybody knows what that state is. It's an objective fact. But, if different people know different things about a state, e.g. they know it's either |0> or |1>, but I've measured it, and know it's |1>, but I've not told anybody else, then everybody describes the state based on what they know about it, so each subject has a different, personal, description of the state. $\endgroup$ – DaftWullie Sep 12 '18 at 8:57
  • $\begingroup$ @taritgoswami If there's a $|\Psi_{AB}\rangle$ that is entangled, there is no notion of $|\Psi_A\rangle$. It's not that we can't find it; it does not exist. The density matrix is the best description of A by itself that can exist because A does not exist in a state by itself, it is merged with that of B. We don't have different definitions of density matrix. The same fundamental properties hold, whatever you're doing, it's just that there are different philosophies by which you can understand the meaning and relevance of the density matrix. $\endgroup$ – DaftWullie Sep 12 '18 at 9:01
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The motivation behind density matrices[1]:

In quantum mechanics, the state of a quantum system is represented by a state vector, denoted $|\psi\rangle$ (and pronounced ket). A quantum system with a state vector $|\psi\rangle$ is called a pure state. However, it is also possible for a system to be in a statistical ensemble of different state vectors. For example, there may be a $50\%$ probability that the state vector is $|\psi_1\rangle$ and a $50\%$ chance that the state vector is $|\psi_2\rangle$. This system would be in mixed state. The density matrix is especially useful for mixed states, because any state, pure or mixed, can be characterized by a single density matrix. A mixed state is different from a quantum superposition. The probabilities in a mixed state are classical probabilities (as in the probabilities one learns in classical probability theory/statistics), unlike the quantum probabilities in a quantum superposition. In fact, a quantum superposition of pure states is another pure state, for example, $\frac{|0\rangle + |1\rangle}{\sqrt{2}}$ . In this case, the coefficients $\frac{1}{\sqrt {2}}$ are not probabilities, but rather probability amplitudes.

Example: light polarization

An example of pure and mixed states is light polarization. Photons can have two helicities, corresponding to two orthogonal quantum states, $|R\rangle$ (right circular polarization) and $|L\rangle$ (left circular polarization). A photon can also be in a superposition state, such as $\frac{|R\rangle + |L\rangle}{\sqrt{2}}$(vertical polarization) or $\frac{|R\rangle - |L\rangle}{\sqrt{2}}$(horizontal polarization). More generally, it can be in any state $\alpha|R\rangle + \beta |L\rangle$ (with $|\alpha|^2+|\beta|^2=1$) corresponding to linear, circular or elliptical polarization. If we pass $\frac{|R\rangle + |L\rangle}{\sqrt{2}}$ polarized light through a circular polarizer which allows either only $|R\rangle$ polarized light, or only $|L\rangle$ polarized light, the intensity would be reduced by half in both cases. This may make it seem like half of the photons are in state $|R\rangle$ and the other in state $|L\rangle$. But this is not correct: Both $|R\rangle$ and $|L\rangle$ are partly absorbed by a vertical linear polarizer, but the $\frac{|R\rangle+|L\rangle}{\sqrt 2}$ light will pass through that polarizer with no absorption whatsoever.

However, unpolarized light such as the light from an incandescent light bulb is different from any state like $\alpha|R\rangle + \beta|L\rangle$ (linear, circular or elliptical polarization). Unlike linearly or elliptically polarized light, it passes through the polarizer with $50\%$ intensity loss whatever the orientation of the polarizer; and unlike circularly polarized light, it cannot be made linearly polarized with any wave plate because randomly oriented polarization will emerge from a wave plate with random orientation. Indeed, unpolarized light cannot be described as any state of the form $\alpha |R\rangle + \beta |L\rangle$ in a definite sense. However, unpolarized light can be described with ensemble averages, e.g. that each photon is either $|R\rangle$ with $50\%$ probability or $|L\rangle$ with $50\%$ probability. The same behaviour would occur if each photon was either vertically polarized with $50\%$ probability or horizontally polarized with $50\%$ probability.

Therefore, unpolarized light cannot be described by any pure state but can be described as a statistical ensemble of pure states in at least two ways (the ensemble of half left and half right circularly polarized, or the ensemble of half vertically and half horizontally linearly polarized). These two ensembles are completely indistinguishable experimentally, and therefore they are considered the same mixed state. One of the advantages of the density matrix is that there is just one density matrix for each mixed state, whereas there are many statistical ensembles of pure states for each mixed state. Nevertheless, the density matrix contains all the information necessary to calculate any measurable property of the mixed state.

Where do mixed states come from? To answer that, consider how to generate unpolarized light. One way is to use a system in thermal equilibrium, a statistical mixture of enormous numbers of microstates, each with a certain probability (the Boltzmann factor), switching rapidly from one to the next due to thermal fluctuations. Thermal randomness explains why an incandescent light bulb, for example, emits unpolarized light. A second way to generate unpolarized light is to introduce uncertainty in the preparation of the system, for example, passing it through a birefringent crystal with a rough surface, so that slightly different parts of the beam acquire different polarizations. A third way to generate unpolarized light uses an EPR setup: A radioactive decay can emit two photons travelling in opposite directions, in the quantum state $\frac{|R,L\rangle + |L,R\rangle}{\sqrt{2}}$. The two photons together are in a pure state, but if you only look at one of the photons and ignore the other, the photon behaves just like unpolarized light.

More generally, mixed states commonly arise from a statistical mixture of the starting state (such as in thermal equilibrium), from uncertainty in the preparation procedure (such as slightly different paths that a photon can travel), or from looking at a subsystem entangled with something else.

Obtaining the density matrix[2]:

As mentioned before, a system can be in a statistical ensemble of different state vectors. Say there is $p_1$ probability that the state vector is $|\psi_1\rangle$ and $p_2$ probability that the state vector is $|\psi_2\rangle$ are the corresponding classical probabilities of each state being prepared.

Say, now we want to find the expectation value of an operator $\hat{O}$. It is given as:

$$\langle \hat{O} \rangle = p_1\langle \psi_1 \lvert \hat{O} \lvert \psi_1 \rangle + p_2\langle \psi_2 \lvert \hat{O} \lvert \psi_2 \rangle$$

Note that $\langle \psi_1 \lvert \hat{O} \lvert \psi_1 \rangle$ and $p_2\langle \psi_2 \lvert \hat{O} \lvert \psi_2 \rangle$ are scalars, and trace of scalars are scalars too. Thus, we can write the above expression as:

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

Now, using the cyclic invariance and linearity properties of the trace:

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

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

where $\rho$ is what we call the density matrix. The density operator contains all the information needed to calculate an expectation value for the experiment.

Thus, basically the density matrix $\rho$ is

$$p_1 \lvert \psi_1 \rangle \langle \psi_1 \lvert + p_2 \lvert \psi_2 \rangle \langle \psi_2 \lvert$$ in this case.

You can obviously extrapolate this logic for when more than just two state vectors are possible for a system, with different probabilities.

Calculating the density matrix:

Let's take an example, as follows.

enter image description here

In the above image, the incandescent light bulb $1$ emits completely random polarized photons $2$ with mixed state density matrix.

As mentioned before, an unpolarized light can be explained with an ensemble average i.e. say each photon is either $|R\rangle$ or $|L\rangle$ with $50%$ probability for each. Another possible ensemble average is: each photon is either $\frac{|R\rangle+|L\rangle}{\sqrt 2}$ or $\frac{|R\rangle - |L\rangle}{\sqrt 2}$ with $50\%$ probability for each. There are lots of other possibilities too. Try to come up with some yourself. The point to note is that the density matrix for all these possible ensembles will be exactly the same. And this is exactly the reason why density matrix decomposition into pure states is not unique. Let's check:

Case 1: $50\%$ $|R\rangle$ & $50\%$ $|L\rangle$

$$\rho_{\text{mixed}} = 0.5 |R\rangle \langle R| + 0.5 |L\rangle \langle L|$$

Now, in the basis $\{|R\rangle, |L\rangle\}$, $|R\rangle$ can be denoted as $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$ and $|L\rangle$ can be denoted as $\begin{bmatrix} 0 \\ 1 \end{bmatrix}$

$$\therefore 0.5 \left(\begin{bmatrix} 1 \\ 0 \end{bmatrix} \otimes \begin{bmatrix} 1 & 0 \end{bmatrix}\right) + 0.5 \left(\begin{bmatrix} 0 \\ 1 \end{bmatrix} \otimes \begin{bmatrix} 0 & 1 \end{bmatrix}\right)$$

$$= 0.5 \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} + 0.5\begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}$$

$$= \begin{bmatrix} 0.5 & 0 \\ 0 & 0.5 \end{bmatrix}$$

Case 2: $50\%$ $\frac{|R\rangle + |L\rangle}{\sqrt 2}$ & $50\%$ $\frac{|R\rangle - |L\rangle}{\sqrt 2}$

$$\rho_{\text{mixed}} = 0.5 \left(\frac{|R\rangle + |L\rangle}{\sqrt 2}\right)\otimes \left(\frac{\langle R| + \langle L|}{\sqrt 2}\right) + 0.5 \left(\frac{|R\rangle - |L\rangle}{\sqrt 2}\right)\otimes \left(\frac{\langle R| - \langle L|}{\sqrt 2}\right)$$

In the basis $\{\frac{|R\rangle + |L\rangle}{\sqrt 2}, \frac{|R\rangle - |L\rangle}{\sqrt 2}\}$, $\frac{|R\rangle + |L\rangle}{\sqrt 2}$ can be denoted as $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$ and $\frac{|R\rangle - |L\rangle}{\sqrt 2}$ can be denoted as $\begin{bmatrix} 0 \\ 1 \end{bmatrix}$

$$\therefore 0.5 \left(\begin{bmatrix} 1 \\ 0 \end{bmatrix} \otimes \begin{bmatrix} 1 & 0 \end{bmatrix}\right) + 0.5 \left(\begin{bmatrix} 0 \\ 1 \end{bmatrix} \otimes \begin{bmatrix} 0 & 1 \end{bmatrix}\right)$$

$$= 0.5 \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} + 0.5\begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}$$

$$= \begin{bmatrix} 0.5 & 0 \\ 0 & 0.5 \end{bmatrix}$$ Thus, we can clearly see that we get the same density matrices in both case 1 and case 2.

However, after passing through the vertical plane polarizer (3), the remaining photons are all vertically polarized (4) and have pure state density matrix:

$$\rho_{\text{pure}} = 1 \left(\frac{|R\rangle + |L\rangle}{\sqrt 2}\right)\otimes \left(\frac{\langle R| + \langle L|}{\sqrt 2}\right) + 0 \left(\frac{|R\rangle - |L\rangle}{\sqrt 2}\right)\otimes \left(\frac{\langle R| - \langle L|}{\sqrt 2}\right) $$

In the basis $\{\frac{|R\rangle + |L\rangle}{\sqrt 2}, \frac{|R\rangle - |L\rangle}{\sqrt 2}\}$, $|R\rangle$ can be denoted as $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$ and $|L\rangle$ can be denoted as $\begin{bmatrix} 0 \\ 1 \end{bmatrix}$

$$\therefore 1 \left(\begin{bmatrix} 1 \\ 0 \end{bmatrix} \otimes \begin{bmatrix} 1 & 0 \end{bmatrix}\right) + 0 \left(\begin{bmatrix} 0 \\ 1 \end{bmatrix} \otimes \begin{bmatrix} 0 & 1 \end{bmatrix}\right)$$

$$= 1 \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} + 0\begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}$$

$$= \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$$

The single qubit case:

If your system contains just a single qubit and you're know that its state $|\psi\rangle = \alpha|0\rangle + \beta|1\rangle$ (where $|\alpha|^2+|\beta|^2$) then you are already sure that the 1-qubit system has the state $|\psi\rangle$ with probability $1$!

In this case, the density matrix will simply be:

$$\rho_{\text{pure}} = 1|\psi\rangle \langle \psi|$$

If you're using the orthonormal basis $\{\alpha|0\rangle + \beta|1\rangle,\beta^*|0\rangle - \alpha^*|1\rangle\}$,

the density matrix will simply be:

$$\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$$

This is very similar to 'case 2' above, so I didn't show the calculations. You can ask questions in the comments if this portion seems unclear.

However, you could also use the $\{|0\rangle,|1\rangle\}$ basis as @DaftWullie did in their answer.

In the general case for a 1-qubit state, the density matrix, in the $\{|0\rangle,|1\rangle\}$ basis would be:

$$\rho = 1(\alpha |0\rangle + \beta |1\rangle) \otimes (\alpha^* \langle 0| + \beta^* \langle 1|)$$

$$= \begin{bmatrix} \alpha \\ \beta \end{bmatrix} \otimes \begin{bmatrix} \alpha^* & \beta^* \end{bmatrix}$$

$$= \begin{bmatrix} \alpha\alpha^* & \alpha\beta^* \\ \beta\alpha^* & \beta\beta^* \end{bmatrix}$$

Notice that this matrix $\rho$ is idempotent i.e. $\rho = \rho^2$. This is an important property of the density matrices of a pure state and helps us to distinguish them from density matrices of mixed states.

Obligatory exercises:

1. Show that density matrices of pure states can be diagonalized to the form $\text{diag}(1,0,0,...)$.
2. Prove that density matrices of pure states are idempotent.


Sources & References:

[1]: https://en.wikipedia.org/wiki/Density_matrix

[2]: https://physics.stackexchange.com/a/158290

Image Credits:

User Kaidor on Wikimedia

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  • $\begingroup$ It's a little bit confusing at first what you're considering as your initial situation. Maybe consider switching |L> and |R> to |H> and |V> (with the polarizer set to D)? While technically it's all the same stuff in some basis, I think its more natural to think about polarizers in the H, V basis. $\endgroup$ – Steven Sagona Jun 16 '18 at 21:59
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    $\begingroup$ I think this question misses the most fundamental aspect of the different between pure and mixed, and that is that mixed states do not behave quantum mechanically. You say that states are classical mixtures, but you do not point out how superpositions states behave quantum mechanically (which is nontrivial). For example if you have something in a 1qubit superposition there's also a 50/50 chance of each option. So how is this state different than a classical one. I think showing how we can see "quantum interference" of a superposition state is how to properly illustrate the difference. $\endgroup$ – Steven Sagona Jun 16 '18 at 22:03
  • $\begingroup$ ^This idea is discussed a bit here: physics.stackexchange.com/questions/409205/… $\endgroup$ – Steven Sagona Jun 16 '18 at 22:09
  • $\begingroup$ @StevenSagona Thanks for pointing that out. I will update my answer. $\endgroup$ – Sanchayan Dutta Jun 17 '18 at 18:13

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