# Density matrix help

Jonathan freeman
Hi, I am wanting to confirm my understanding of the density matrix in quantum mechanics. Is it the wave function co-efficients squared - in other words the wave amplitudes squared which in turn are the probabilities and then these turn out to be placed into a matrix form with the squared wave amplitudes being the diagonal elements of the density matrix itself?

## Answers and Replies

Gold Member
Yes.

• bhobba
Mentor
Things are a bit more complicated than what you wrote. First, there isn't a single density matrix that corresponds to a given system, as when you write a density matrix, you must choose a representation for that matrix, that is you need to choose a set of basis states and the order in which you put them. Also, you have to account for the fact that not all density matrices correspond to pure states, so they do not all correspond to a wave function (the entire point of the density matrix formalism is to be able to treat statistical uncertainty, where the wave function of the system is not exactly known).

But in the case where
$$\psi = \sum_i c_i \phi _i$$
with ##{\phi}## some basis, then choosing this basis to represent the density matrix, then the diagonal elements correspond to ##|c_i|^2##.

• aaroman and Jonathan freeman
Staff Emeritus
Hi, I am wanting to confirm my understanding of the density matrix in quantum mechanics. Is it the wave function co-efficients squared - in other words the wave amplitudes squared which in turn are the probabilities and then these turn out to be placed into a matrix form with the squared wave amplitudes being the diagonal elements of the density matrix itself?

A density matrix combines two different types of uncertainty: (1) classical uncertainty, which is due to incomplete knowledge of what the system state is, and (2) quantum uncertainty, where even a "pure" state can only make probabilistic predictions.

So for example, suppose you have a complete set of basis states ##|\psi_n\rangle##. You randomly select one of the states, ##j## with probability ##p_j##. Then the resulting density matrix can be written this way:

##\rho = \sum_j p_j |\psi_j\rangle \langle \psi_j|##

That's a density matrix written in terms of a basis in which it is "diagonal". In a different basis, the density matrix would look more complicated. For instance, let's write that in terms of a different basis, ##|\phi_\alpha\rangle##:

##\rho = \sum_j \sum_\alpha \sum_\beta p_j |\phi_\alpha\rangle \langle \phi_\alpha |\psi_j\rangle \langle \psi_j|\phi_\beta\rangle \langle \phi_\beta |##
## = \sum_\alpha \sum_\beta \rho_{\alpha \beta} |\phi_\alpha \rangle \langle \phi_\beta|##

where ##\rho_{\alpha \beta} = \sum_j p_j \langle \phi_\alpha|\psi_j\rangle \langle \psi_j|\phi_\beta\rangle##

Sometimes when people talk about the density matrix, they mean the coefficients ##\rho_{\alpha \beta}## rather than the operator ##\rho##.

If ##\rho## is diagonal in the basis ##|\phi_\alpha\rangle##, that means that ##\rho_{\alpha \beta} = 0## unless ##\alpha = \beta##.

That's the general case. What you're talking about is a special case in which the way that we randomly pick ##|\psi_j\rangle## is by a quantum-mechanical measurement, starting with an initial state ##|\psi\rangle = \sum_j c_j |\psi_j\rangle## and measuring whatever property corresponds to the index ##j##. After the measurement is complete, but before you look at the result, you could describe the situation with a diagonal density matrix as above where ##p_j = |c_j|^2##.

So, it's not true in general that the density matrix coefficients are squares of amplitudes. It's only true for the density matrix describing the result of a measurement on a pure state.

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• aaroman, bhobba and Jonathan freeman
Jonathan freeman
Things are a bit more complicated than what you wrote. First, there isn't a single density matrix that corresponds to a given system, as when you write a density matrix, you must choose a representation for that matrix, that is you need to choose a set of basis states and the order in which you put them. Also, you have to account for the fact that not all density matrices correspond to pure states, so they do not all correspond to a wave function (the entire point of the density matrix formalism is to be able to treat statistical uncertainty, where the wave function of the system is not exactly known).

But in the case where
$$\psi = \sum_i c_i \phi _i$$
with ##{\phi}## some basis, then choosing this basis to represent the density matrix, then the diagonal elements correspond to ##|c_i|^2##.

Thanks so much.
And then like you said, if not a pure state, you could theoretically if known even STATE probabilities which would be become the diagonal elements of the matrix. For example, if you knew somehow that there was ( for whatever reason ) a certain probability of some observable then you could include that in the overall "catalog" of probabilities so that n expectation value could be calculated from that whole catalog of values that are captured in the density matrix?

Mentor
Thanks so much.
And then like you said, if not a pure state, you could theoretically if known even STATE probabilities which would be become the diagonal elements of the matrix. For example, if you knew somehow that there was ( for whatever reason ) a certain probability of some observable then you could include that in the overall "catalog" of probabilities so that n expectation value could be calculated from that whole catalog of values that are captured in the density matrix?
Yes. If you have no other information than the probability of being in a given state, then you can construct the density matrix by filling the diagonal elements of the density matrix.

Two examples that come to mind: a spin-1/2 system where either states (spin-up or spin-down) are equally probable, like an unpolarised source of electrons or neutrons; a system at thermal equilibrium where you can calculate the probability of energy eigenstates using Boltzmann factors.

• Jonathan freeman
Jonathan freeman
So, it's not true in general that the density matrix coefficients are squares of amplitudes. It's only true for the density matrix describing the result of a measurement on a pure state.

Thankyou so much for taking time to reply with all you wrote.
So it is comprised of the amplitudes squared in pure states and even could comprise of stated probabilities from classical imperfection in measurement to be included in a mixed state but is it correct to say the the probabilities will always be the diagonal elements of the density matrix?

Regards,

Jon

Staff Emeritus
Thankyou so much for taking time to reply with all you wrote.
So it is comprised of the amplitudes squared in pure states and even could comprise of stated probabilities from classical imperfection in measurement to be included in a mixed state but is it correct to say the the probabilities will always be the diagonal elements of the density matrix?

Regards,

Jon

I think that most of the ideas can be explained using the simplest state space, which is the two-component spin matrices.

Let ##|u\rangle = \left(\begin{array} \\ 1 \\ 0 \end{array} \right)## be the state for spin-up in the z-direction, and let ##|d\rangle = \left(\begin{array} \\ 0 \\ 1 \end{array} \right)## be the state for spin-down. Those are the "kets". The corresponding "bras" are ##\langle u| = \left(\begin{array} \\ 1 & 0 \end{array} \right)## and ##\langle d| = \left(\begin{array} \\ 0 & 1 \end{array} \right)##

If you randomly put a particle into the spin-up state with probability ##p_1## or spin-down with probability ##p_2##, then that corresponds to the density matrix:

##\rho = p_1 |u\rangle \langle u| + p_2 |d\rangle \langle d| = p_1 \left(\begin{array} \\ 1 \\ 0 \end{array} \right) \left(\begin{array} \\ 1 & 0 \end{array} \right) + p_2 \left(\begin{array} \\ 0 \\ 1 \end{array} \right) \left(\begin{array} \\ 0 & 1 \end{array} \right)##

## = \left( \begin{array} \\ p_1 & 0 \\ 0 & p_2 \end{array} \right)##

So that's a mixed state that is mixture of "spin-up" and "spin-down". It is not a pure state. A pure state can be written in the form ##|\psi\rangle \langle \psi|## (with only one term). Being diagonal means that it is NOT a pure state (except in the special case where all the diagonal values are 0 except for one).

Now, let's switch to another basis: ##|x\rangle = \left( \begin{array} \\ \alpha \\ \beta \end{array} \right)## and ##|y\rangle = \left( \begin{array} \\ -\beta^* \\ \alpha^* \end{array} \right)##

The pure state density matrix ##|x\rangle \langle x| = \left( \begin{array} \\ \alpha \\ \beta \end{array} \right) \left( \begin{array} \\ \alpha^* & \beta^* \end{array} \right) = \left( \begin{array} \\ |\alpha|^2 & \alpha \beta^* \\ \alpha^* \beta & |\beta|^2 \end{array} \right)## is not diagonal (in the spin-up, spin-down basis). A check whether a state is a pure state is that ##\rho^2 = \rho##.

If you start with a pure state ##|x\rangle##, corresponding to the density matrix ##\left( \begin{array} \\ \alpha \\ \beta \end{array} \right) \left( \begin{array} \\ \alpha^* & \beta^* \end{array} \right) = \left( \begin{array} \\ |\alpha|^2 & \alpha \beta^* \\ \alpha^* \beta & |\beta|^2 \end{array} \right)##, and you measure the spin, but don't look at the outcome, then you will get spin-up with probability ##|\alpha|^2## and spin-down with probability ##|\beta|^2##, so that would be described by the matrix ##|\alpha|^2 |u\rangle \langle u| + |\beta|^2 |d\rangle \langle d| = \left( \begin{array} \\ |\alpha|^2 & 0 \\ 0 & |\beta|^2 \end{array} \right)##. The measurement produces a diagonal matrix in the spin-up/spin-down basis, which is a mixed state.

Measurement without checking on the result produces a mixed state that is diagonal in the basis of eigenstates of whatever is being measured.

• Jonathan freeman
Jonathan freeman
I think that most of the ideas can be explained using the simplest state space, which is the two-component spin matrices.

Let ##|u\rangle = \left(\begin{array} \\ 1 \\ 0 \end{array} \right)## be the state for spin-up in the z-direction, and let ##|d\rangle = \left(\begin{array} \\ 0 \\ 1 \end{array} \right)## be the state for spin-down. Those are the "kets". The corresponding "bras" are ##\langle u| = \left(\begin{array} \\ 1 & 0 \end{array} \right)## and ##\langle d| = \left(\begin{array} \\ 0 & 1 \end{array} \right)##

If you randomly put a particle into the spin-up state with probability ##p_1## or spin-down with probability ##p_2##, then that corresponds to the density matrix:

##\rho = p_1 |u\rangle \langle u| + p_2 |d\rangle \langle d| = p_1 \left(\begin{array} \\ 1 \\ 0 \end{array} \right) \left(\begin{array} \\ 1 & 0 \end{array} \right) + p_2 \left(\begin{array} \\ 0 \\ 1 \end{array} \right) \left(\begin{array} \\ 0 & 1 \end{array} \right)##.

So would it be correct to say that in case of up and down vectors that the co-efficients ( even though not explicitly stated ) could for example in the UP state vector be 1 and 0 which then when squared give the probability of measuring up in the up state 1X1 or certain 100% and in the down state 0x0 never?

And regardless of mixed or pure states will the density matrix elements representing the probabilities always be the diagonal from top left of matrix down to bottomest right?

Staff Emeritus
So would it be correct to say that in case of up and down vectors that the co-efficients ( even though not explicitly stated ) could for example in the UP state vector be 1 and 0 which then when squared give the probability of measuring up in the up state 1X1 or certain 100% and in the down state 0x0 never?

Yes, the density matrix ##\left( \begin{array} \\ 1 & 0 \\ 0 & 0 \end{array} \right)## corresponds to the situation where there is 100% probability of being in the state spin-up.

And regardless of mixed or pure states will the density matrix elements representing the probabilities always be the diagonal from top left of matrix down to bottomest right?

I'm not sure what you mean. The general density matrix for spins would be of the form ##\left( \begin{array} \\ A & B \\ B^* & C \end{array} \right)## where ##A## and ##C## are real, and add up to 1. To be "diagonal" means that ##B=0##.

I think I know what you mean: In that matrix, ##A## and ##C## are indeed the probabilities of measuring ##spin-up## and ##spin-down" for a system described by that matrix.

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• Jonathan freeman
Jonathan freeman
[QUOTE="stevendaryl, post:
I'm not sure what you mean. The general density matrix for spins would be of the form ##\left( \begin{array} \\ A & B \\ B^* & C \end{array} \right)## where ##A## and ##C## are real, and add up to 1. To be "diagonal" means that ##B=0##.[/QUOTE]

Yes, I didn't understand that diagonal meant having B = 0. I meant to ask : in all cases for all density matrices the probabilities will be the left to right diagonal A C in above case and always add to one. They will always have the same form as running from top left entry down to bottom right? In pure states there will be one entry as 1 and rest zero and in mixed states there will be non zero entries < 1 for each entry?

Staff Emeritus
Yes, I didn't understand that diagonal meant having B = 0. I meant to ask : in all cases for all density matrices the probabilities will be the left to right diagonal A C in above case and always add to one. They will always have the same form as running from top left entry down to bottom right? In pure states there will be one entry as 1 and rest zero and in mixed states there will be non zero entries < 1 for each entry?

No, not quite. There are pure states that don't have a 1 anywhere.

What you have to remember is that there is more than one spin operator. Without being explicit, I've been talking about spin in the z-direction. There is also spin in the x-direction, and y-direction, and any other direction you want to measure spin. A pure state is one whose density matrix can be written in the form:

##|\psi\rangle \langle \psi|##

If ##|\psi\rangle## is an eigenvalue of ##s_z##, the spin in the z-direction, then the density matrix will have 0s and 1s in the diagonal. But for a general ##\psi##, the density matrix, as I already said, will have the form:

##\left( \begin{array} \\ |\alpha|^2 & \alpha \beta^* \\ \alpha^* \beta & |\beta|^2 \end{array} \right)##

where ##|\alpha|^2 + |\beta|^2 = 1##. That's a pure state density matrix, it's just not one corresponding to an eigenvalue of ##s_z##.

• Jonathan freeman
Jonathan freeman
No, not quite. There are pure states that don't have a 1 anywhere.
where ##|\alpha|^2 + |\beta|^2 = 1##. That's a pure state density matrix, it's just not one corresponding to an eigenvalue of ##s_z##.

For the above ##|\alpha|^2 + |\beta|^2 = 1## it must mean that each sum term is less than 1 so how can it meet condition of pure state being that tr of p squared has to be equal to 1? Wouldnt squaring terms less than 1 like 1/2 give 1/4 and then eg 1/4 plus 1/4 equal half therefore not meeting definition of pure state ( tr p squared = 1 )? Hope that makes sense.

Staff Emeritus
For the above ##|\alpha|^2 + |\beta|^2 = 1## it must mean that each sum term is less than 1 so how can it meet condition of pure state being that tr of p squared has to be equal to 1?

Every density matrix has a diagonal that sums to 1, whether it's a pure state or a mixed state. The criterion for a pure state is that
##\rho^2 = \rho##.

With the matrix ##\left( \begin{array} \\ |\alpha|^2 & \alpha \beta^* \\ \alpha^* \beta & |\beta|^2 \end{array} \right)##, if you square it you get:

##\left( \begin{array} \\ |\alpha|^2 & \alpha \beta^* \\ \alpha^* \beta & |\beta|^2 \end{array} \right)
\left( \begin{array} \\ |\alpha|^2 & \alpha \beta^* \\ \alpha^* \beta & |\beta|^2 \end{array} \right)##
## = \left( \begin{array} \\ |\alpha|^4 + |\alpha|^2 |\beta|^2 & |\alpha|^2 \alpha \beta^* + |\beta|^2 \alpha \beta^* \\ |\alpha|^2 \alpha^* \beta + |\beta|^2 \alpha^* \beta & |\beta|^4 + |\alpha|^2 |\beta|^2 \end{array} \right)##
## = (|\alpha|^2 + |\beta|^2) \left( \begin{array} \\ |\alpha|^2 & \alpha \beta^* \\ \alpha^* \beta & |\beta|^2 \end{array} \right)##
## = \left( \begin{array} \\ |\alpha|^2 & \alpha \beta^* \\ \alpha^* \beta & |\beta|^2 \end{array} \right)## (since ##|\alpha|^2 + |\beta|^2 = 1##)

Wouldnt squaring terms less than 1 like 1/2 give 1/4 and then eg 1/4 plus 1/4 equal half therefore not meeting definition of pure state ( tr p squared = 1 )? Hope that makes sense.

If you have a matrix of the form ##\left( \begin{array} \\ A & B \\ C & D \end{array} \right)##, the square is ##\left( \begin{array} \\ A^2 + BC & AB + BD \\ AC + DC & D^2 + BC \end{array} \right)##. The diagonal elements are not just ##A^2## and ##D^2##.

• Jonathan freeman, DrClaude and PeterDonis
Jonathan freeman
If you have a matrix of the form ##\left( \begin{array} \\ A & B \\ C & D \end{array} \right)##, the square is ##\left( \begin{array} \\ A^2 + BC & AB + BD \\ AC + DC & D^2 + BC \end{array} \right)##. The diagonal elements are not just ##A^2## and ##D^2##.

Wonderful that makes sense, thank-you.

Ok so would it be correct to think of a density matrix as a way of packaging probabilities ( both classical probabilities in case of say someone setting up a system with classical probability or classical probability due to inefficiency of a measuring device + the probability of quantum uncertainties so that they are all captured in the trace diagonal of the density matrix so that then they can be "operated" on by the observable operator to give expected average?

Thankyou

Jon

Staff Emeritus
Wonderful that makes sense, thank-you.

Ok so would it be correct to think of a density matrix as a way of packaging probabilities ( both classical probabilities in case of say someone setting up a system with classical probability or classical probability due to inefficiency of a measuring device + the probability of quantum uncertainties so that they are all captured in the trace diagonal of the density matrix so that then they can be "operated" on by the observable operator to give expected average?

Thankyou

Jon

Yes, that's exactly what it's good for. If you have any operator ##\hat{O}## and you have a density matrix ##\rho##, then the expectation value of ##\hat{O}## for a system described by that density matrix is: ##\langle O \rangle = Tr(\rho \hat{O})##.

That compact expression on the right means the following: If ##\rho## is written as ##\rho = \sum_{jk} \rho_{ij} |\phi_i\rangle \langle \phi_j|## and you let ##O_{jk} = \langle \phi_j|\hat{O}|\phi_k\rangle##, then

##\langle O \rangle = Tr(\rho \hat{O}) = \sum_{i,j} \rho_{ij} O_{ji}##

If you think of ##\rho## and ##\hat{O}## as square matrices, then you're just multiplying ##\rho## by ##\hat{O}## to get a new square matrix, and then you're summing up the diagonal entries (called taking the trace).

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Mentor
That's why I like to start with density matrices/operators via Gleason ans show the underpinning basis of the Born Rule naturally is expressed using density operators/matrices:
http://kiko.fysik.su.se/en/thesis/helena-master.pdf

And its all important assumption - non-contextuality. Kochen-Specker is actually a simple corollary to Gleason.

Note - although I call it the Born Rue - its not quite what is meant by Born but it has become by many the standard thing to call the trace formula - here is something a bit more careful in the use of the term:
http://www.math.ru.nl/~landsman/Born.pdf

Thanks
Bill