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Notation question for density matrix

  1. Sep 14, 2010 #1
    This is probably my misunderstanding of the notation...

    The definition of a density matrix is in the attached file. (Sorry, the latex editor is not rendering properly when I preview my post).

    This definition is a sum over only one index 'j', which will invariably lead to a diagonal matrix. But the density matrix is often not diagonal, for instance a basic superposition (pure) state. What am I missing in this definition? Shouldn't it be over indices i and j, for instance, in a two dimensional system?
     

    Attached Files:

  2. jcsd
  3. Sep 15, 2010 #2
    A density matrix of a pure state can be written as the single projector:

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

    The density matrix of a mixed state is a sum over pure states,

    [tex]\rho = \sum_i p_i |\psi_i\rangle\langle\psi_i|[/tex]

    Suppose you have some mixture of particles described by the density matrix of this mixed state. Then the probability that a given particle is in a pure state is given by [tex]p_i[/tex]. This is a different scenario if you are dealing with a single particle which is a superposition of states. It's only in a superposition because you happened to have chosen such a basis for the Hilbert space.

    The index i in the summation does not run over the basis states of your Hilbert space. It runs over the pure states which form mixed state.

    The density matrix of a mixed state is therefore always a sum over projection operators onto pure states. It is diagonal, in this "basis" of projectors.

    Choosing a basis for your Hilbert space means you decompose the projection operators [tex]|\psi_i\rangle[/tex] in terms of the basis vectors you happened to have chosen. Generically, this introduces off-diagonal elements. If you write the density matrix explicitly in terms of some basis for your Hilbert space, then it will look like:

    [tex]\rho = \sum_{n,m} p_{nm} |\phi_n\rangle\langle \phi_m |[/tex]

    This is probably what you were thinking of. Both pure states and mixed states can be decomposed like this. But this notation clouds the fact that a mixed state is a sum over pure states. So it's more important to appreciate the difference between the first two formulas I wrote down.
     
  4. Sep 16, 2010 #3
    Ok, great. So moving to an interpretational question:

    The mixed state doesn't have to be in diagonal form, but often is. (Right?) If it represents a collection of particles, then each diagonal entry (eigenvalue) squared can predict the likelihood of the particle we sample being in that particular basis state.

    Can the density matrix also represent a single particle system? In that case, do the eigenvalues represent the probability of finding the one particle in that particular state?

    If so, then a density matrix fundamentally represents a superposition of state in this case, right? So even when we have a diagonalized matrix, we still may be describing a quantum superposition, so long as it has more than one non-zero eigenvalue.
     
  5. Sep 17, 2010 #4
    Yes.

    Yes. This would be a pure state. Density matrices are just a generalization of these single-particle states.

    Yes.

    Yes. You can cross out "fundamentally".

    Now you're mixing two things again. The eigenvalue of measuring the particle in that particular superposition is always 1: it's a pure state so any measurement will always result in measuring the particle in that superposition. The problem with that statement, however, is that the you also deal with the superposition.

    Let me try to come up with an example here to point at the difference between mixed and pure states.

    Suppose you have an ensemble of particles all prepared in the same state. By that I mean is that you have a bunch of particles, not talking with each other or their environment, and each particle is in the same superposition. That means that each particle is in a pure state. Let's take this to be the superposition of spins relative to the z-axis:

    [tex]|\Psi_1\rangle = \frac{1}{\sqrt{2}} \left(|\uparrow_z\rangle + |\downarrow_z\rangle \right)[/tex]

    The corresponding density matrix describing a particle is:

    [tex]\rho_1 = |\Psi_1\rangle langle\Psi_1|[/tex]

    Now imagine that in some other room we have a mixture of a large number of particles. Half of these particles are in the pure state [tex]|\uparrow_z\rangle[/tex]. The other half are in the state [tex]|\downarrow_z\rangle[/tex].

    So:
    [tex]|\Psi_{2,1}\rangle = |\uparrow_z\rangle[/tex]
    [tex]|\Psi_{2,2}\rangle = |\downarrow_z\rangle[/tex]

    are the two pure state. Together they form the density matrix. The density matrix for this mixture of particles is:

    [tex]\rho_2 = \frac{1}{2}|\uparrow_z\rangle\langle\uparrow_z| + \frac{1}{2}|\downarrow_z\rangle\langle\downarrow_z|[/tex].

    So now the funny part: how do these systems differ? Well, if we measure the spin along the z-axis we can have two outcomes: either we measure spin up or spin down. In both cases either outcome has a probability of 1/2. It doesn't look like there's a difference between a system!

    But look at where these probabilities come from: in the first case we deal with a superposition. When we perform a measurement with respect to some basis in which this superposition is not an eigenstate, then we will get different outcomes with each measurement. You could say that this is a consequence of our choice of basis of the Hilbert space.

    But in the second part this is not the case. The probability of 1/2 comes from the fact that we deal with a mixture of states. There will always be a probability of 1/2 that we measure a particle in either first pure state or the second pure state!

    To illustrate this, imagine the instead of measuring the spin along the z-axis, we measure it along the x-axis. What changes? Well, the superposition is now in an eigenstate of the spin-projector along the x-axis. Namely,

    [tex]|\Psi_1\rangle = |\uparrow_x\rangle[/tex]

    So the density matrix is:

    [tex]\rho_1 = |\uparrow_x\rangle \langle\uparrow_x|[/tex]

    If we measure the spin along the x-axis we will always measure spin up!.

    What happens for the second case. Well, the change of basis has as an effect on the states:

    [tex]
    [tex]|\Psi_{2,1}\rangle = \frac{1}{\sqrt{2}}\left(|\uparrow_x\rangle + |\downarrow_x\rangle\right) [/tex]
    [tex]|\Psi_{2,2}\rangle = \frac{1}{\sqrt{2}}\left(|\uparrow_x\rangle - |\downarrow_x\rangle\right)[/tex]

    The density matrix is still

    [tex]\rho_2 =\frac{1}{2}|\Psi_{2,1}\rangle\langle\Psi_{2,1}| + \frac{1}{2}|\Psi_{2,2}\rangle\langle\Psi_{2,2}|[/tex]

    which you can expand in terms of the x-basis. But, the important thing is: measuring the spin along the x-axis in the mixed state case can give either up or down, with a probability of 1/2 for each outcome! Where did this come from? The reason is that the probability of measuring either "pure state 1" or "pure state 2" does not change when we make a change of basis of our Hilbert space. What changes is the possible outcomes of the measurement we perform. So both pure state 1 and 2 that make up this mixed state are now in a superposition with respect to this basis.
     
  6. Sep 18, 2010 #5
    I am new to the forum, and I decided to answer some questions today in order to demonstrate that the answer to many of the questions being asked here can be found in Landau.

    This question is no exception of course, since Landau first introduced the density matrix. See vol 3, 3rd edition, section 14, "the density matrix". There is also a discussion by Landau of the density matrix in vol 5, and in the relevant paper in his collected works.
     
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