# Finding the density matrix of an ensemble

1. May 24, 2005

### Kian

Hi I am just doing an undergraduate degree in physics and currently studying a course in the foundations of QM.

The problem I want to solve is this:

There is an ensemble in a state corresponding to vector (i, 2)
A measurement of Sy (with the operator represented by the 2x2 pauli spin matrix) is made but the values measured are not kept. What is the density matrix that must be used in subsequent measurements of any observable?

I know the denisty matrix is defined as the sum of the classical probability that the vector is in a particular state times by the projection operator of that particular state.
However I am having problems seeing how this is helpful to me.
Any help would be very much appreciated
Many thanks
Kian

2. May 24, 2005

### James Jackson

OK... Say we know the state $|\psi\rangle$ of a quantum system (ie after a measurement of which we know the result), the density operator is then:

$$\rho = |\psi\rangle\langle\psi |$$

This is a pure state.

Now say we can imagine that a quantum system is prepared in a state $\rho_i$ with probability $p_i$. Consider a number of such states, such as the results of a measurement. The density matrix for the (mixed) state is then:

$$\rho=\sum_i p_i \rho_i$$

So, going into operator notation rather than probabilities, we have:

$$\rho=\sum_m p(m) \rho_m$$
$$\rho=\sum_m tr(M_m^t M_m \rho)\frac{M_m \rho M_m^t}{tr(M_m^t M_m \rho)}$$
$$\rho=\sum_m M_m \rho M_m^t$$

which is the result you need.

Edit: Note this is similar to the unitary evolution of a density matrix by the unitary operator U:

$$\rho=\sum_i p_i |\psi_i\rangle\langle\psi_i | \rightarrow^U \sum_i p_i U |\psi_i \rangle\langle \psi_i | \ U^t = U \rho U^t$$

Last edited: May 25, 2005
3. May 25, 2005

### James Jackson

If that's a little too abstract, try expressing the spin vector you have as a linear composition of the Sy eigenvectors. What are the probabilities a measurement of Sy will yield $\frac{\hbar}{2}$ and $\frac{-\hbar}{2}$? What density operator would represent the state now if we threw away the measurement result?

4. May 25, 2005

### Kian

Still not quite there

Thanks for taking the time to reply. However im still unsure how to tackle the problem as I dont understand the meaning of $$\ M_m$$ or what it represents and how I can use it.
In your last reply you have pointed out exactly my problem. I have found the probabilities of a measurement Sy give either 1/10 for $\frac{\hbar}{2}$ or 9/10 for $\frac{-\hbar}{2}$
If we now throw away the result - I assume this is now a mixed state with the state being a linear composition of the Sy eigenvectors (which can be made from my initial spin state)
Is this linear compostion the rho in

$$\sum_m M_m \rho M_m^t$$ ?

Any further help would be greatly appreciated.
Many thanks
Kian

5. May 25, 2005

### James Jackson

You know operator notation, yeah? The $M_m$ is the measurement M leading to result m, where M is an observable operator (i.e. Hermitian and Unitary). The notation could be clearer - the rho in the sumnation is the density operator just before measurement by M, and the rho equal to the sumnation is the density operator after the measurement. Perhaps I should've called them $\rho_{old}$ and $\rho_{new}$

OK, I get the same numbers expressing the spin state in Sy eigenspace. So, you know the probabilities, $p_{up} = \frac{1}{10}$ and $p_{down} = \frac{9}{10}$, and you know the resultant density operators $\rho_{up} = |S_y^{up}\rangle\langle S_y^{up}|$ and $\rho_{down} = |S_y^{down}\rangle\langle S_y^{down}|$. Put these into the first sumnatiion I gave you, and Bob's your father's brother.

This is a very roundabout way of doing things, as we're going from a vector notation, to a density notation by doing it the 'vector' way. This kind of hides the beauty of density operators in that many different ensembles give rise to the same density operator.

Try working through the Stern-Gerlach experiment in density operator terms - you'll get the same results of (say) an Sy measurement followed by an Sz measurement as you would doing it the vector way.

Last edited: May 25, 2005
6. May 27, 2005

### Kian

Thanks

Thanks James
After a long hard think and a bit of maths I think I get it now. I think my understanding of the density matrix was wrong. We can make density matricies of pure states Sy up and down. When the measurements are thrown away all we know is that there are 2 different pure states corresponding to each eigenvector of Sy. We can make a sort of total density matrix from the addition of the 2 pure denisty matrices of Sy eigenvectors but remembering to take into account the probability that these occur at.
I think im on the right track now and the average of an Sx measurement on this new state of pure Sy up and down gives 0 as there is equal probability of measuring Sx up or down from this state.

Many thanks for all the help.
all I need now is some nice questions to come out in my exam next week :)

7. May 27, 2005

### James Jackson

You're clearly getting the right idea when you say that a measurement of Sx will be indeterminate given a measurement of Sy, whether one keeps of throws away the initial result. I hope that you've worked through the maths to see this result from the density operator formulation of QM, rather than going from rote learning of the Stern-Gerlach experiment.

When you say 'We can make a sort of total density matrix...', it's no 'sort of' density matrix, it is a density matrix as illustrated abstractly in my first post. If you have worked through the maths as per the previous paragraph, you'll be beginning see and use the subtle beauty of the density operator formulation. I'm afraid I can't explain it any better than that - I'm no teacher, but all I can say is do problems, work stuff out, set your own problems (this can be very useful) and good luck in your exam!