# Density matrix's non-diagonal components

1. Nov 29, 2011

### TheDestroyer

Hello guys,

The diagonal components of a density matrix represent the probability to find a particle in the corresponding state of the matrix; so in a spin +/- 1/2 (+ for up, - for down) system it's the probability to find a particle in either spin +1/2 or spin -1/2 state.

If the state of the system is pure up, then the density matrix is (1,0;0;0). If the state of the system is pure down then the density matrix is (0,0;0,1).

What do the non-diagonal components represent? can you please explain with examples of states in a system, and how the non diagonal components would change with different system situations? I heard it has something to do with correlation. But I can't get the idea.

Thank you for any efforts :)

2. Nov 29, 2011

### A. Neumaier

They represent contributions to the probability of superpositions. The probability of observing psi given rho
is <psi|rho|psi>.

3. Nov 29, 2011

### TheDestroyer

Could you please explain it with an example? how would the change of the average polarisation state affect them?

4. Nov 30, 2011

### A. Neumaier

An unpolarized state has density matrix rho = [0.5 0; 0.5 0] (where ; denotes a new row). The probability of measuring a particular polarization state psi is <psi|rho|psi>=0.5 <psi|psi>=0.5, no matter which state psi you are looking at.

A fully polarized state in the superposition (|left>+|right>)/sqrt(2) has density matrix rho = [0.5 0.5; 0.5.5 0]. The probability of measuring a particular polarization state psi is <psi|rho|psi>=|psi_1|^2, which equals 1 for the prepared state psi=(|left>+|right>)/sqrt(2) , 0 for the complementary state psi= (|left>-|right>)/sqrt(2), and 0.5 for both the left-polarized and the right-polarized state,

A fully left-polarized state has density matrix rho = [1 0;0 0]. The probability of measuring a particular polarization state psi is <psi|rho|psi>=|psi_1|^2, which equals 1 for the left-polarized state, 0 for the right-polarized state, and is somewhere in between for other states.

In general, the probability is
$$\langle \psi|\rho|\psi\rangle=\rho_{11} |\psi_1|^2 = \rho_{22}|\psi_2|^2 + 2 Re \rho_{12} \psi_1^*\psi_2,$$
which shows how much the diagonal terms and the off-diagonal term contribute. In particular,
rho_{12} is nonzero iff the state is not a mixture of the preferred basis states.

5. Dec 1, 2011

### TheDestroyer

I've been investigating, and could alomst understand the use of non-diagonal terms. But I couldn't relate what I learned with the experimental stuff.

So in the book called "Density Matrix, Theory and Applications" by Blum, it says that the density matrix can be written in terms of the polarisation vector as follows:

rho = 0.5*{1+P_z, P_x- i P_y; P_x + i P_y, 1 - P_z},

where P_k is the polarisation vector of the system in the direction k, i is the complex unit number sqrt(-1).

You said that the off-diagonal components could be non-zero if the state is not a mixture of the preferred basis states. This complies with the matrix I wrote, because the quantisation axis is z, and if there are any other components on x or y, then the off-diagonal components will be non-zero. This is how I understood it (is that right?).

The point confusing me now is, how could I get the x and y components in an experiment, while the x and y components of the spin don't commute with the z component?

Let's assume I have a Stern-Gerlach filter. How can I use it to measure the off-diagonal components of the system for a specific beam? I think rotating it about z will only help me in measuring the intensities along 1 direction, since the others don't commute with this direction.

How could this measurement be done?

Last edited: Dec 1, 2011
6. Dec 1, 2011

### A. Neumaier

A factor 1/2 in front of the matrix is missing, since the trace of a density matrix must be 1. Indeed, the diagonal elements are the probabilities for measuring the alternatives defined by the basis states, and these must sum to 1. It is a trivial exercise to show that every Hermitian matrix of trace one can be written in the above (but corrected) form. This defines the polarization vector in terms of the density matrix.
Yes. rho_{12} is nonzero when P_x or P_y is nonzero.
This has nothing to do with commutation. The polarization vector is not an operator, but a state descriptor.

In experiments you make statistics to determine rho, which is related to the vector P by using the above (but corrected) formula. This means that you choose a number of states psi_1, ..., psi_N,(in complex 2-space), perform enough statistics to get reliable probabilities, insert these and the formula expressing rho in terms of P into the formula relating density and probabilities, and solve the resulting equations in a least squares sense to get P. (One can do it a bit more efficiently; the recipe just given gives the basic principle.)
Polarisation is a property of a beam in the direction of the beam, so what you describe is just
what one wants to have. The orientation of the filter tells you which psi you are using.

The directions in the Hilbert space have nothing to do with the direction of the beam (which is not modelled at all at this level of description) but are directions in spin space.

By the way, the second equality in the equation in post #4 should have been a +.
You should rewrite this formula in terms of P using the relation discussed above.

7. Dec 1, 2011

### TheDestroyer

Thank you for your answer. I understood a lot, but the picture isn't yet fully clear for me. I will resume reading and investigating, if I have a question I'll post it. Please subscribe to the thread.

Thank you again.