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I Density matrices, pure states and mixed states

  1. Apr 2, 2016 #1
    I got (very) confused about the concept of states, pure states and mixed states.

    Is it correct that a linear combination of pure states is another pure state?

    Can pure (and mixed) states only be expressed in density matrices?

    Is a pure state expressed in a single density matrix, whereas mixed states are expressed in a linear combination of several density matrices? Since a pure state is a single density matrix, a linear combination of density matrices would be a linear combination of pure states, and hence a pure state. I don't understand.

    I hope someone likes to illuminate me on the matter. Thanks.
     
  2. jcsd
  3. Apr 2, 2016 #2
    Let's start with basics. The density matrix is an mathematical object to describe quantum systems that have classical uncertainties to them. How can we do that? Well simple, let me create an operator, which literally is:

    $$ \hat{\rho} = \sum_{i} p_i |\psi_i \rangle \langle \psi_i | $$
    where the $$p_i$$ are the classical probabilities that I left my system in the particular quantum state $$\psi_i$$

    1. Is it correct that a linear combination of pure states is another pure state?

    Correct. Let's set we have a qbit system that I am 100% sure (that's what a pure state means) is in the following linear combination $$\alpha_1 | 0 \rangle + \beta_1 | 1 \rangle$$. Let's add/superpose to that another state I am 100% sure is in the following linear combination $$\alpha_2 | 0 \rangle + \beta_2 | 1 \rangle$$

    Then the final qbit, I'm still 100% sure it's in the following superposition (modulo normalization)

    $$(\alpha_1 + \alpha_2) | 0 \rangle + (\beta_1 + \beta_2) |1 \rangle$$

    2. Can pure (and mixed) states only be expressed in density matrices?

    Well I'm sure there are alternative ways, but what's wrong with density matrices? they are pretty elegant. Check this out https://en.wikipedia.org/wiki/Density_matrix

    3. Is a pure state expressed in a single density matrix, whereas mixed states are expressed in a linear combination of several density matrices? Since a pure state is a single density matrix, a linear combination of density matrices would be a linear combination of pure states, and hence a pure state. I don't understand.

    A linear combination of pure state above (in 1) is talking about quantum superposition. The
    $$\sum_i p_i \rho_{i}^{pure}$$

    represents a system which has classical probability of being in probability $p_i$ in the pure state represented by the density matrix $$\rho_i^{pure}$$
    This has nothing to do with the summing of coefficients $$\alpha_1, \alpha_2$$ which may be complex? (have you seen complex probabilities before lol)

    Anyways, hope it helps.

    Pro-tip:

    If you are unsure about a quantum problem, always consider the case when the matrix is diagonal. What does a pure state look like when you have a orthonormal basis with the state you are 100% sure your pure state is in is one of the basis vector? What does your mixed state look like in a basis where the density matrix is diagonal?
     
    Last edited: Apr 2, 2016
  4. Apr 2, 2016 #3

    Nugatory

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    yes.

    All states can be expressed in density matrices. Only mixed states must be expressed that way; a pure state can always be written as a vector using the bra-ket notation.

    No. Pure and mixed states are both represented by a single density matrix. A linear combination of matrices is itself a matrix, and any matrix can be written as a linear combination of other matrices, so it makes no sense to say that some states are represented by a single matrix and others by a linear combination.

    This is a B level thread, but there is no way of going further without using math suitable for an I-level thread.
     
  5. Apr 2, 2016 #4
    You're right. Sorry. I am familiar with some of the mathematics now, so I don't object labeling this thread as an I-level one.

    <Moderator's note: This is now an I-level thread.>
     
    Last edited by a moderator: Apr 4, 2016
  6. Apr 2, 2016 #5
    So what is the difference between a density matrix describing a pure state and the one describing a mixed state? (eg Is it that the trace has one 1 in it and then all zero's? Is it that density matrix associated with the pure state is a projection operator?)

    Is a density matrix a state? (since it looks like an observable since it is a matrix)
     
  7. Apr 2, 2016 #6
    Hi Entropy,

    There are several ways to distinguish pure vs. mixed state. But you must first have an intuition what a pure state means.

    1. As I said above, a pure state is a state where you are 100% sure what the wavefunction is. So just rotate your basis so one of the vectors is that wavefunction, and you're done, your density matrix is a 1 0, 0, 0, ... diagonal matrix. In other words, a pure state is a state where a change of basis could lead you to that a matrix, and mixed state? everything else. And yes, pure state density matrices are projection operators. How else would you describe a diagonal matrix with a single 1 on the diagonal?

    2. More formaly,

    $$Tr(\rho^2) = 1$$ for pure state, while this is not true for mixed state. Consider the density matrix in its diagonalized form. The pure state look like the diagonal 1, 0, 0... matrix so obviously $$Tr(\rho^2) = 1$$. The mixed state looks like a diagonal matrix with the diagonal being $$p_1, p_2, .... $$

    such that $$\sum_{i = 1}^N p_i = 1$$.

    For the mixed state,

    $$Tr(\rho^2) = \sum_{i = 1}^N p_i^2 \neq 1$$

    Remember, being mixed means there's at least 2 nonzero $p_i's$
     
    Last edited: Apr 2, 2016
  8. Apr 2, 2016 #7
    I can't find it back, but I recall that some density matrices are proportional to the unit operator I. Is that correct? (might that be the maximally entangled state?)

    And furthermore:

    As I understand it, a mixed state is represented by a sum of projection operators of a particular state, each operator weighed with a particular probability - the probability that the system is in that particular state. Is that correct?

    I'm getting really confused now I think...
     
    Last edited: Apr 2, 2016
  9. Apr 2, 2016 #8

    vanhees71

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    If you have a finite-dimensional Hilbert space, e.g., when you look only at the spin of a particle, where the Hilbert space is ##(2s+1)##-dimensional you can have
    $$\hat{\rho}=\frac{1}{d} \hat{1}.$$
    That means, if you measure a spin component it has with probability ##1/d## any one of the possible values ##\sigma \in \{-s,-s+1,\ldots,s-1,s \}##.

    This is by the way the state of maximum entropy, i.e., the state you should assign to your system if you don't know anything about it. The entropy is defined as the von Neumann entropy
    $$S(\hat{\rho})=-\mathrm{Tr} [\hat{\rho} \ln \hat {\rho})].$$
    If you don't know anything about the system there's no constraint in finding ##\hat{\rho}## except that it must fulfill
    $$\mathrm{Tr} \hat{\rho}=1.$$
    So we introduce a Lagrange multiplier ##\lambda## and make
    $$\delta S+\lambda \mathrm{Tr} \delta \hat{\rho}=-\mathrm{Tr}(\delta \hat{\rho} (\ln \hat{\rho}+1-\lambda)=0.$$
    Since this must hold true for all ##\delta \hat{\rho}##, we find
    $$\ln \hat{\rho}=-1+\lambda \; \Rightarrow \; \hat{\rho}=C=\text{\const},$$
    and because of ##\mathrm{Tr} \hat{\rho}=1## you get ##C=1/d##, which leads to the statistical operator given above.

    The most simple way to understand a general mixed state
    $$\hat{\rho}=\sum_j p_j |\psi_j \rangle \langle \psi_j|,$$
    where the ##|\psi_j \rangle## are arbitrary normalized Hilbert-space vectors and ##p_j \geq 0## with ##\sum_j p_j=1##, is the following gedanken experiment: Suppose Alice is able to prepare a lot of particles in the pure states represented by the ##|\psi_j \rangle##. Now she sends randomly particles in one of these states such that on average a fraction ##p_j## of these particles is in the state represented by ##|\psi_j \rangle##, i.e., Bob knows that each particle is prepared in one of the states represented by ##|\psi_j \rangle## but not in which one, but Alice tells him the probabilities ##p_j## she uses to send the randomly prepared particles. Then the state Bob has to use to describe the situation is given by the ##\hat{\rho}## above.
     
  10. Apr 2, 2016 #9
    After some re-perusing of my material, I think it's clear to me now. Thanks for the help!
     
  11. Apr 2, 2016 #10

    Nugatory

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    It is always possible to find a basis in which the density matrix of a given pure state takes on that form, but it won't necessarily have that form in other bases. It's a good exercise to try writing the density matrix of a spin-1/2 particle in the spin-up state in the spin-up/spin-down basis and again in the spin-left/spin-right basis. Also, take a look at page 9 of http://www.pa.msu.edu/~mmoore/Lect34_DensityOperator.pdf.... althoughg paralleltransport's advice above ("But you must first have an intuition what a pure state means") is good.
    Matrices can be used to represent all sorts of things. There is no inconsistency in using matrices to represent both states and operators.
     
  12. Apr 2, 2016 #11

    DrDu

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    Also mixed states can be expressed as a Hilbert space vector, however in a "larger" Hilbert space. Usually, this is unwieldy, but sometimes one arrives at such a representation and has to recognize it.
    An example may be helpful:
    Let ##\sigma_i## with ##i\in {x,y,z}## be the usual Pauli matrices and ##\sigma_0## be the 2x2 unit matrix. Usually, we would consider the algebra spanned by these operators in the space of 2-vectors. Then mixed states would have to be represented as density matrices, e.g., ##\rho=\sigma_0/2## would be a mixed state. However, the operators ##\sigma_i \otimes \sigma_0## span the same algebra and we could represent the same mixed state as a vector ## (1/\sqrt{2},0,0,1/\sqrt{2})^T##. You can check the expectation values of all operators remain the same.
    We see here one important point: The matrix representation of the operator algebra, where mixed states are represented as vectors is block diagnonal and therefore reducible.

    That's a general feature: We can represent mixed states as state vectors, but at the price of dealing with a reducible representation of our operator algebra. On the other hand, this explains how mixed states can arrise from pure states:
    A typical situation where this happens is when we split a large system into a subsystem and a surrounding. If we only consider the algebra spanned by the observables of the subsystem, the representation of this algebra will be reducible even if we started from an irreducible representation of the algebra of all the operators from the large system.
     
    Last edited by a moderator: Apr 2, 2016
  13. Apr 2, 2016 #12

    A. Neumaier

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    For a density matrix ##\rho## the following are equivalent:
    • ##\rho## represents a pure state.
    • ##\rho^2=\rho##.
    • ##\rho## is a projector.
    • ##\rho## has rank 1
    • All columns of ##\rho## are parallel.
    Its a state, not an observables - unlike observables it transforms with the opposite sign in time.
     
  14. Apr 3, 2016 #13

    blue_leaf77

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    The ideas of density operator is introduced because a quantum state can also be understood as being the result of a given system preparation procedure. Imagine you have two groups of 10 non-interacting and non-correlating electrons.

    In the first group, each electron is in the state ##|\psi \rangle=\frac{1}{\sqrt{2}}(|z;+\rangle + i|z;-\rangle)##, i.e. all electrons are in the same state. The density operator ##\rho = |\psi \rangle \langle \psi |## is therefore representative of this ensemble because all members of the ensemble assume the same state. In this ensemble, if you measure ##S_z##, your chance is 50% of getting ##\hbar/2## and 50% of getting ##-\hbar/2##, which is clear because every single electron in this ensemble are in a superposition state between ##|z;+\rangle## and ##|z;-\rangle## with equal probabilities. If furthermore you also measure ##S_y##, you can be 100% percent sure that you will get ##\hbar/2## all the time because all electrons in this ensemble are in the state ##|\psi\rangle = |y;+\rangle##.

    In the second group, five electrons are in the state ##|z;+ \rangle## and five other are in ##|z;- \rangle##. Clearly, there is no way you can assign a state for this ensemble because there are more than one state mixed within this ensemble. However, there is an alternative to describe the state (here "state" means condition rather than "state" in the usual term in quantum mechanics) of this system, namely by using the density operator ##\rho = \frac{5}{10}|z;+ \rangle \langle z;+ | + \frac{5}{10}|z;- \rangle \langle z;- |##. Now you perform the same measurement as you do in the first group. First measure ##S_z##. The result for this measurement will be identical to the previous one, namely 50% ##\hbar/2## and 50% ##-\hbar/2## because out of 10 electrons, half of them are in the state ##|z;+ \rangle## and the other half in ##|z;- \rangle##. Next measure ##S_y##. Here comes the difference, 50% of the time you will get ##\hbar/2## and 50% of the rest ##-\hbar/2## for ##S_y##, in contrast to the first group where you will always get ##\hbar/2##. This is because both ##|z;+ \rangle## and ##|z;- \rangle###, which the members of this group assume, are a superposition state between ##|y;+\rangle## and ##|y;-\rangle## with the same probabilities.
     
  15. Apr 5, 2016 #14
    That would only be the case if they were prepared in the σy state, wouldn't it?
     
  16. Apr 5, 2016 #15

    blue_leaf77

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    The state
    is indeed the ##S_y## spin-up state.
     
  17. Apr 5, 2016 #16
    I understand, but shouldn't that be ##|\psi \rangle=\frac{i}{\sqrt{2}}(|z;+\rangle-|z;-\rangle)##, following the Pauli matrix
    [tex]
    \begin{pmatrix}
    0 & -i\\
    i & 0
    \end{pmatrix}
    [/tex]?
     
  18. Apr 5, 2016 #17

    blue_leaf77

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    Nope, check this wolframalpha calculation.
     
  19. Apr 5, 2016 #18

    vanhees71

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    To get the eigenstate of ##\hat{s}_y=\hat{\sigma}_y/2## to the eigenvalue 1/2, you have to solve
    $$\hat{\sigma}_y \begin{pmatrix} a \\ b \end{pmatrix}=\begin{pmatrix} a \\ b \end{pmatrix}.$$
    Doing the matrix-vector product on the left-hand side, this leads to
    $$-\mathrm{i} b = a, \quad \mathrm{i} a=b.$$
    These equations are consistent. So you can find a solution ##\neq 0##. The normalization is arbitrary first. So we have
    $$\psi=a \begin{pmatrix} 1 \\ \mathrm{i} \end{pmatrix}.$$
    To normalize the state we calculate
    $$\psi^{\dagger} \psi=2|a|^2 \stackrel{!}{=} 1 \; \Rightarrow \; a=\frac{1}{\sqrt{2}}.$$
    Finally we have
    $$|\psi \rangle=\frac{1}{\sqrt{2}} \left (|\sigma_z=+1/2 \rangle +\mathrm{i} |\sigma_z=-1/2 \rangle \right) ,$$
    as given already in #13.
     
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