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Problem with density matrix

  1. Apr 19, 2012 #1

    Jano L.

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    How would you define density matrix for an ensemble of identical harmonic oscillators in thermal equilibrium?

    For example, consider N atoms in a crystalline lattice. I would like to find density matrix to calculate the average dipole moment of the ensemble and also its standard deviation.

    Standard references propose that [itex]\rho[/itex] is defined as

    [tex]
    \rho = \sum_{k}p_k |\psi_k\rangle \langle \psi_k|
    [/tex]

    where [itex]p_k[/itex] is the probability (expected frequency) of the quantum state [itex] |\psi_k\rangle[/itex] of the oscillator. The two obvious questions are:

    1) which states [itex]|\psi_k\rangle[/itex] should we choose to enter the sum (it seems we cannot choose all of them) and

    2) what are the probabilities [itex]p_k[/itex]?


    In case of thermal equilibrium, the standard procedure is

    1) to choose the states [itex]|\psi_k\rangle[/itex] as the eigenvectors of the Hamiltonian operator:

    [tex]
    H|\psi_k \rangle = \epsilon_k | \psi_k \rangle,
    [/tex]

    and

    2) the probabilities are according to Boltzmann's formula:

    [tex]
    p_k = \frac{e^{-\epsilon_k/kT}}{Z}.
    [/tex]


    But, what is the reason behind this choice? Why can we forget the other quantum states of the oscillator (or, why does the probability of any other state vanish) ?
     
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  3. Apr 20, 2012 #2

    kith

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    In general, it doesn't vanish. The probability for obtaining an arbitrary state |θ> in a measurement is tr{ρ|θ><θ|}.

    If |θ> is not an eigenstate of the Hamiltonian, you can either write it as a sum of them and calculate the single probabilities. Or you can represent ρ in a basis which includes |θ> and calculate the probability directly from there.
     
  4. Apr 20, 2012 #3

    Jano L.

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    Kith,

    what you wrote makes sense, after we have obtained [itex]\rho[/itex] we can calculate expected frequency of any quantum state _after_the_measurement.

    But the problem is how to define [itex] \rho [/itex] in the first place. "Define" means how to find matrix [itex]\rho_{mn}[/itex] (in at least one basis, no matter which), prior to measurement, just from known laws of physics.

    The standard definition is based upon the ensemble of vectors [itex]|\psi_k\rangle[/itex], which are supposed to be realized with probabilities [itex]p_k[/itex] (in the sense of expected frequency, prior to measurement). These vectors need not and in general case are not orthogonal.

    But in quantum theory of statistical physics, a strange assumption is made that only eigenvectors of the Hamiltonian operator enter the sum definining the density operator-

    What is the reason behind this? Why the other states do not enter the sum ?
     
  5. Apr 20, 2012 #4

    kith

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    I'm not sure what you are after. If you understand the origin of the Boltzmann distribution in classical statistical mechanics, the generalization to QM should be straightforward. There is for example an algebraic approach to statistical mechanics, which covers both the classical and the quantum case.

    But I still think that the fact that every state can be expanded in Hamiltonian eigenstates may be the solution here. If you have an arbitrary density matrix with an arbitrary number of non-orthogonal states, you can always expand these states in eigenstates of the Hamiltonian (or any other oberservable), which yields ρ = Ʃ En|En><En|. This is simple linear algebra and has nothing to do with measurements. The question why these En fulfill the Boltzmann distribution is -as I already mentioned- not specific to quantum mechanics.
     
  6. Apr 20, 2012 #5

    Jano L.

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    Adding projectors [itex]p_k |\psi_k\rangle \langle\psi_k| [/itex] on arbitrary vectors will produce density matrix which has non-zero off-diagonal terms. It will not be diagonal in general.

    The density matrix ρ = Ʃ p_n |En><En| is a special assumption, which can be rephrased as

    Only the eigenvectors of the Hamiltonian operator are present in the ensemble; and they are present with expected frequency p_n.

    Maybe another example can show my point. Consider silver atoms coming out of hot oven. What density matrix should we use to describe the ensemble of spins of these atoms?

    The fact that the atoms are coming from oven suggest that any state vector is equally possible, which leads to definition of [itex]\rho[/itex] as an integral over all spin vectors with equal weight:

    [tex]
    \rho = \int |\psi(\vartheta,\varphi)\rangle \langle \psi (\vartheta,\varphi) | \sin \vartheta \frac{d\vartheta d\varphi}{4\pi}
    [/tex]

    which equals

    [tex]
    \rho = \frac{1}{2} |z+\rangle\langle z+| + \frac{1}{2} |z-\rangle \langle z-|.
    [/tex]

    In this definition all quantum states are taken with equal weight, which is very satisfactory.

    But in case of energy, standard definition throws most of the vectors out and only concentrates probability on points - eigenvectors of the Hamiltonian.
     
    Last edited: Apr 20, 2012
  7. Apr 20, 2012 #6

    fzero

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    What vectors are being thrown out? As kith already explained, the eigenvectors of a Hermitian operator (like the Hamiltonian) form a basis for the Hilbert space of states. Any vector in the Hilbert space can be expanded in terms of energy eigenstates.

    The only subtlety here is whether there are degenerate energy states. In that case we need additional observables, like angular momentum, to further label the complete basis of states.

    Note, we can choose any Hermitian operators that we want to define our basis. The density matrix in a new basis will be related to the one in the energy basis by a unitary transformation.
     
  8. Apr 20, 2012 #7

    Jano L.

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    All vectors in Hilbert space that are not eigenvectors of the Hamiltonian. In case of spins I wrote above, the integral is over entire Hilbert space. In case of canonical density matrix, the sum is over discrete set of eigenvectors.

    The fact that eigenvectors form a basis does not mean that their linear combinations should not be included into the sum. The definition of density matrix does not require that the vectors in the terms entering the sum form an orthogonal basis.
     
  9. Apr 20, 2012 #8

    kith

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    If you think that states like |En>+|Em> are "missing" in the energy representation of the density matrix of the canonical ensemble, why don't you think that states like |x+> are missing in the density matrix above?
     
  10. Apr 20, 2012 #9

    fzero

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    In general, all linear combinations of states are included in the density matrix. It is not necessary to use an orthogonal basis, but if you happen to do so, the density matrix will be diagonal. As I said above, it is always possible to map the density matrix expressed in a particular basis to the energy eigenstate representation.

    As for the details of the density matrix for the canonical ensemble, I would refer you to Sakurai's Sect 3.4, which gives a very complete explanation of the formulas you refer to in your original post.
     
  11. Apr 20, 2012 #10

    Jano L.

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    The fact that the expression [itex] \rho = 1/2 |z+\rangle \langle z+| + 1/2|z-\rangle \langle z-| [/itex] does not contain |x+> explicitly is just because this is not a definition of the density operator, it is just its evaluation in basis |z+>, |z->. The definition of the operator is by means of the above integral. The state |x+> and alike are all included in the integral above defining the density matrix.

    On the other hand, with energy the story is entirely different. The very definition used in statistical physics prefers Hamiltonian eigenstates [itex]\phi_k[/itex]. The fact that other states are superpositions of these eigenstates is not different from what happens with spin. In order to account correctly for all possible quantum states, all of them should enter into the definition of the density matrix with non-trivial probability. It may be that finally the result would be

    [tex]
    \rho =\sum_k p_k |\phi_k\rangle \langle\phi_k|
    [/tex]

    but can you show it? The problem is that it is hard to define corresponding integral containing projectors on all possible vectors.
     
  12. Apr 20, 2012 #11

    Jano L.

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    I think my question can be reformulated in this way: why is the density matrix diagonal in the basis of eigenvectors of the Hamiltonian?
     
  13. Apr 20, 2012 #12

    fzero

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    The density operator is always defined as

    [tex]
    \hat{\rho} = \sum_{k}p_k |\psi_k\rangle \langle \psi_k|.
    [/tex]

    The density matrix is what you get when you evaluate [itex]\hat{\rho}[/itex] between basis states

    [tex] \rho_{ij} = \langle \psi_i | \hat{\rho} | \psi_j \rangle = \sum_k p_k \langle \psi_i |\psi_k\rangle \langle \psi_k| \psi_j \rangle.[/tex]

    In an orthonormal basis, we see that [itex]\rho_{ij} = p_i \delta_{ij}[/itex].
     
  14. Apr 21, 2012 #13

    Jano L.

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    Yes, but these vectors [itex]|\psi_k\rangle[/itex], although they have unit norm, are not orthogonal in general, and do not even have to form a basis. Sakurai comments this on p.180 of his book (2nd ed.).

    Let us agree on this: the set of vectors [itex]|\psi_k\rangle[/itex] is a general set of unit vectors such that any system from the ensemble considered has a state described by one of these vectors, with relative frequency [itex]p_k[/itex].

    The density operator is then defined in the way you indicated above.

    Now it is useful to introduce another set, a set of orthonormal vectors that form a basis in Hilbert space. Let us denote these by [itex]|\phi_m\rangle[/itex]. The density matrix is defined as

    [tex]
    \rho_{mn} = \langle \phi_m| \sum_k p_k |\psi_k \rangle \langle \psi_k | \phi_n\rangle.
    [/tex]

    This equals

    [tex]
    \rho_{mn} = \sum_k p_k c_{k,m} c_{k,n}^*,
    [/tex]

    which is not diagonal. General density matrix is not diagonal, not even in orthonormal basis.

    Now, this procedure works perfectly when the definition contains only a few vectors [itex]|\psi_k\rangle[/itex], or some denotable set of them, like that for spins from the oven (the vectors are designed by spherical angles, see the integral in my post above).

    However, for harmonic oscillator, the set of all possible unit vectors is difficult to denote, because the basis is infinite.

    Instead, the whole procedure, which works perfectly for spins, is thrown out and the density matrix is postulated out from blue as a matrix that is diagonal in the basis of Hamiltonian eigenstates.

    If there was some another definition or justification for this, I would be glad to hear it.

    But, if _this_diagonal_matrix_ is said to be the definition of the density operator, then it is immediate that only eigenstates were ascribed weight and thus the systems in the ensemble can assume only states described by eigenvectors. That would not be right, because superpositions should be present _and_thus_be_in_the_sum as well.
     
  15. Apr 21, 2012 #14

    kith

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    Ok, now I think I understand your posts.

    Your example is about a situation, where we know nothing about the system (at least as far as spin is concerned). You start with an ensemble which contains all states on the Bloch sphere with equal weightings and find that the resulting density operator is ρ = ½|z+><z+| + ½|z-><z-|.

    Let me first add, that this is not the only possible way to define the density operator here. A (pure) quantum state is determined by specifing the eigenvalues of a complete set of commuting observables (CSCO). Knowing nothing corresponds to an ensemble where the different eigenstates of the CSCO occur with equal weightings. So it is perfectly fine to take an arbitrary CSCO (let's say S and Sx) and define the density operator directly as ρ = ½|x+><x+| + ½|x-><x-|.

    When we know something about the system this usually means we know something about an oberservable. Let's say we prepared the z-component of spin such that our ensemble consists of an incoherent mixture of ⅓ spin up and ⅔ spin down. So we define the density matrix as ρ = ⅓|z+><z+| + ⅔|z-><z-|.

    If we look at a system with given temperature T, we know that we know that the energies are distributed according to the Boltzmann distribution. This is why we define ρ in the usual way.
     
  16. Apr 21, 2012 #15

    kith

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    No, it is not. The weight for an arbitrary state can easily be calculated by a change of basis. Simply use the superposition of interest as one of the elements of the new basis.

    /edit: I want to partly revise my statement here. I think you are right that the superpositions aren't ascribed a weight in the definition. What I wanted to stress is that this doesn't mean that they aren't contained in the density operator.
     
    Last edited: Apr 21, 2012
  17. Apr 21, 2012 #16

    kith

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    That's only true for completely incoherent mixtures which means that the density matrix is proportional to the identity matrix.

    In general, the eigenstates of the density operator form the only basis in which it is diagonal.
     
  18. Apr 21, 2012 #17
    Jano,

    like others have already explained the basis you choose for describing a mixture does not restrict the states contained in the mixture.

    I'd like to add a different argument. You want to describe a mixed state in thermal equilibrium, that means the mixed state is stationary. The time derivative of a mixed state is proportional to the commutator of that state with the hamiltonian of the system. Assuming that the hamiltonian is essentially non degenerate all commuting operators can be written as a function of the hamiltonian (that maps the eigenvalues of the hamiltonian to the eigenvalues of the commuting operator). So you know that rho = f(H), and you only have to identify the function f. There are then statistical arguments that show that f has to be an exponential.

    Cheers,

    Jazz
     
  19. Apr 21, 2012 #18

    kith

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    This is a dynamical question. As Jazzdude pointed out, this is necesarry for the density operator to describe an equilibrium state. The very reason for the system to reach such a stationary state is usually decoherence, i.e. interactions with an environment.
     
  20. Apr 21, 2012 #19

    Jano L.

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    Unfortunately, I think the definition of density operator based on CSCO which you propose is not sufficiently complete. Here is why:

    Repeated measurument of [itex]s,s_z[/itex] in one particular direction on the atoms of the ensemble will reveal frequencies 1/2 and 1/2. But these two numbers are not enough to reconstruct the density matrix. The off-diagonal element can be non-zero. For example, if the oven conspired and produced all atoms in state |x+>, the complete density operator would be

    [tex]
    \rho = |x+\rangle \langle x+|
    [/tex]

    and the density matrix in z basis is

    [tex]
    \rho_{mn} = \left( \begin{array}{cc} 1/2 & 1/2 \\ 1/2 & 1/2 \end{array}\right)
    [/tex]


    The fact that the oven does not conspire but produces atoms in all possible spin states can be revealed by measurement only if we measure spin projection in other directions as well (in other words, we have to measure values of other CSCO as well). This procedure will let us to conclusion that all quantum states can be present in the ensemble with equal probability and we have to use the original definition with the integral over the Bloch sphere.

    The definition based on measurement of CSCO has another drawback; it is inapplicable to statistical description of system in thermal equilibrium. The energy of the system cannot be directly measured. It is just a useful theoretical concept. Statistical physics assigns probabilities a priori, not based on measured frequencies of CSCO.

    In a consistent probabilistic description of system in equilibrium, all quantum states should be assigned some probabilities. It should be the same as in classical statistics : all phase points are assigned probability density.



    Jazzdude, your different argument seems nice at first, but now I think it is not very convincing. You have to assume that evolution of the density matrix by H has to conserve its elements. The reason given for this I've heard is that the expected values of all time-independent operators have to be time-independent in thermal equilibrium, which is possible only if [itex]\rho[/itex] is time-independent.

    However, the systems in thermal equilibrium are not evolved by H, but by a slightly different Hamiltonian [itex]H_{total} = H + H'[/itex], where H' accounts for the interaction with environment and makes the passage to equilibrium possible. Evolution by THIS Hamiltonian should make [itex]\rho[/itex] independent of time, after the relaxation processes cease.

    So [itex]H_{total}[/itex] and [itex]\rho[/itex] do commute, but this does no imply that H and [itex]\rho[/itex] commute.
     
  21. Apr 26, 2012 #20

    Jano L.

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    In case this is still interesting to you, this is how I see it now:

    The problem is that there are two distinct definitions of density matrix, for two distinc situations.

    1) In case we have ensemble of known vectors with known probabilities (whether a priori or measured frequencies), we can define density operator

    [tex]
    \rho = \sum_k p_k |\psi_k\rangle\langle\psi_k|,
    [/tex]

    and its matrix in any basis is easy to calculate.

    This is used for description of spins/light polarization experiments.

    2) In case we have a system in interaction with reservoir, we can not do the same thing.

    Firstly, the system does not have a wave function and thus cannot be said to be with certain probability in a state that is described by vectors from Hilbert space.

    Secondly, even if it somehow was in a definite quantum state, the ensemble of such states would have to be the whole Hilbert space. Because it is infinite-dimensional, there does not seem to be any easy way to define probability on such an ensemble.

    The only way to define the density matrix for a system in interaction seems to be

    [tex]
    \rho_{mn}= \sum_L R_{mL;nL},
    [/tex]

    where [itex]R_{mK;nL}[/itex] is the density matrix of the closed system system+reservoir.

    The latter corresponds to pure state of the system + reservoir. However, to show that the reduction leads to canonical density matrix does not seem easy.
     
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