Finding the density matrix of an ensemble

In summary, the conversation is discussing the concept of density matrices in quantum mechanics and how they can be used to determine the state of a quantum system after a measurement is made. The conversation also touches on the concept of pure and mixed states and how they can be represented by density matrices. The main problem being discussed is how to determine the density matrix after a measurement of a specific observable is made and the result is not kept. The conversation also highlights the beauty of density operator formulation and the importance of working through problems to fully understand the concepts.
  • #1
Kian
3
0
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
 
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  • #2
OK... Say we know the state [itex]|\psi\rangle[/itex] of a quantum system (ie after a measurement of which we know the result), the density operator is then:

[tex]\rho = |\psi\rangle\langle\psi |[/tex]

This is a pure state.

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

[tex]\rho=\sum_i p_i \rho_i[/tex]

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

[tex]\rho=\sum_m p(m) \rho_m[/tex]
[tex]\rho=\sum_m tr(M_m^t M_m \rho)\frac{M_m \rho M_m^t}{tr(M_m^t M_m \rho)}[/tex]
[tex]\rho=\sum_m M_m \rho M_m^t[/tex]

which is the result you need.

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

[tex]\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[/tex]
 
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  • #3
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 [itex]\frac{\hbar}{2}[/itex] and [itex]\frac{-\hbar}{2}[/itex]? What density operator would represent the state now if we threw away the measurement result?
 
  • #4
Still not quite there

Thanks for taking the time to reply. However I am still unsure how to tackle the problem as I don't understand the meaning of [tex]\ M_m[/tex] 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 [itex]\frac{\hbar}{2}[/itex] or 9/10 for [itex]\frac{-\hbar}{2}[/itex]
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

[tex]\sum_m M_m \rho M_m^t[/tex] ?

Any further help would be greatly appreciated.
Many thanks
Kian
 
  • #5
You know operator notation, yeah? The [itex]M_m[/itex] 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 [itex]\rho_{old}[/itex] and [itex]\rho_{new}[/itex]

OK, I get the same numbers expressing the spin state in Sy eigenspace. So, you know the probabilities, [itex]p_{up} = \frac{1}{10}[/itex] and [itex]p_{down} = \frac{9}{10}[/itex], and you know the resultant density operators [itex]\rho_{up} = |S_y^{up}\rangle\langle S_y^{up}|[/itex] and [itex]\rho_{down} = |S_y^{down}\rangle\langle S_y^{down}|[/itex]. 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.
 
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  • #6
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 I am 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
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!
 

1. What is a density matrix?

A density matrix is a mathematical representation of a quantum mechanical system, which describes the statistical behavior of a collection of quantum particles. It is a way of representing the quantum state of an ensemble, which is a group of particles with the same properties.

2. How is the density matrix of an ensemble found?

The density matrix of an ensemble is found by taking the sum of the outer products of all the individual quantum states in the ensemble. This sum is then normalized to ensure that the trace of the density matrix is equal to 1.

3. What information does the density matrix provide?

The density matrix provides information about the statistical behavior of an ensemble, including the average values of observables and the probabilities of different measurement outcomes. It also contains information about the coherence and entanglement of the ensemble.

4. How is the density matrix used in quantum mechanics?

The density matrix is a fundamental tool in quantum mechanics, as it allows for the calculation of physical quantities and the prediction of measurement outcomes. It is used in a variety of applications, including quantum information processing, quantum computing, and quantum simulation.

5. Can the density matrix be used for both pure and mixed ensembles?

Yes, the density matrix can be used for both pure and mixed ensembles. A pure ensemble consists of identical particles with the same quantum state, while a mixed ensemble contains particles with different quantum states. The density matrix is a useful tool for studying both types of ensembles.

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