1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Density matrix, change of basis, I don't understand the basics

  1. Sep 3, 2014 #1

    fluidistic

    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data
    Hello people,
    I am trying to understand a problem statement as well as the density operator, but I still don't get it, desperation is making me posting here.
    The problem comes as
    The problem then continues with other questions but I'm having troubles with the very first one, for now.

    2. Relevant equations

    Diagonalization of the density matrix: ##\hat \rho _{\text{diagonal}} =U^* \rho U## where U is the rotation matrix, it's a unitary operator.

    3. The attempt at a solution
    First, I do not really know how to form the density matrix. The definition I have is "##\hat \rho = \sum _i p_i \left | \phi _i \right > \left < \phi _i \right |##" where ##p_i## is the probability to find the system in the state ##|\phi _i>##.
    I see this as a sum of many operators whose matricial form I am not sure about. So I don't know how to get the matrix representation of the density operator from there.

    Another approach is, ##\hat \rho = \begin{bmatrix} \left < \psi _1 \right | \hat \rho \left | \psi _1 \right > && \left < \psi _1 \right | \hat \rho \left | \psi _2 \right > \\ \left < \psi _2 \right | \hat \rho \left | \psi _1 \right > && \left < \psi _2 \right | \hat \rho \left | \psi _2 \right > \end{bmatrix}## but my problem is that I don't know how the density operator acts on psi_1 and psi_2, the eigenfunctions of ##S_z##.
    I know that written in the basis that diagonalizes ##S_z## and ##S^2##, ##\psi_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix}## and ##\psi_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix}##. But still, this doesn't get me any further.

    Another thoughts: since ## \left | \chi \right >## is arbitrary I can write it as a linear combinations of psi_1 and psi_2 such as ## \left | \chi \right > =\alpha \left | \psi _1 \right > + \beta \left | \psi _2 \right >## where alpha and beta are in principle complex valued.
    Maybe I can also think of ## \left | \chi \right >## as a rotation of ## \left | \psi _1 \right >## by arbitrary angles theta and phi if I work in spherical coordinates.
    The matrix that rotates from an angle phi and theta can be written under the form ##U=\cos \left ( \frac{\theta}{2} \right ) \hat 1 -i \sin \left ( \frac{\theta}{2} \right ) \hat u \cdot \vec \sigma## where ##\hat u## is the unit vector of the axis of rotation and the sigma is the vector of the Pauli matrices.

    But overall I'm still stuck at even going into the right direction to solve the problem.

    The answer (for the diagonal matrix) is, if I'm not mistaken, ##\hat \rho _{\text{diagonalized}}=\begin{bmatrix} 1 &&0 \\ 0&&0 \end{bmatrix}##, that is, if I remember well the solution I've seen.

    Any tip, help and explanation is appreciated! Thanking you.
     
    Last edited: Sep 3, 2014
  2. jcsd
  3. Sep 3, 2014 #2

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Do you have any additional information about the state ##|\chi\rangle##? Are you to assume that it is a general state? What is the density matrix if the particles are all in this state?
     
  4. Sep 3, 2014 #3

    fluidistic

    User Avatar
    Gold Member

    No, I do not have any more information about that state. As I understand it, it's a general state, a linear combination of ##|\uparrow >## and ##|\downarrow >##.
    I don't know the answer to your question about the density matrix. I think the answer will depend upon the chosen basis.
     
  5. Sep 3, 2014 #4

    Matterwave

    User Avatar
    Science Advisor
    Gold Member

    Alright, I'm not sure where this problem is really going, but let's write ##\left|\chi\right>=\alpha\left|z_+\right>+\beta\left|z_- \right>## for complex ##\alpha,\beta## such that ##|\alpha|^2+|\beta|^2=1##, which is the most general pure state one can write down in a 2 state system.

    How would I write down the density matrix in the ##\left|z_+ \right>,\left|z_- \right>## basis?
     
  6. Sep 3, 2014 #5

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Start from the definition of the density matrix and the assumption that the particle is in the state ##|\chi\rangle## with probability 1. What is the density matrix then?

    Since you need to express the state in some way, start by assuming that it is a general linear combination of up and down eigenstates.
     
  7. Sep 3, 2014 #6

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    The general rule to get the matrix of components of an operator T, with respect to a particular ordered basis ##(e_i)_{i=1}^n## is ##T_{ij}=(Te_j)_i##. The right-hand side is the ith component of the vector ##Te_j##. This is equal to ##\langle e_i,Te_j\rangle##, so in bra-ket notation, we have ##T_{ij}=\langle i|T|j\rangle##, where ##|j\rangle## is the ket notation for the basis vector ##e_j##. See the https://www.physicsforums.com/showthread.php?t=694922 [Broken] if you need more information about this.

    The problem asks for the density operator of a single particle in the pure state ##|\chi\rangle##. I don't know how to answer that in the form of a hint, so I'll just say it: ##\rho=|\chi\rangle\langle\chi|##.

    That matrix near the end of your post #1 is the matrix of components of ##|\uparrow\rangle\langle\uparrow|## with respect to the ordered basis ##(|\uparrow\rangle,|\downarrow\rangle)##. (You should use the formula in the first paragraph of this post to verify this). So it's also the matrix of components of ##|\chi\rangle\langle\chi|## with respect to the ordered basis ##(|\chi\rangle,|\psi\rangle)##, where ##|\psi\rangle## is some unit vector in the orthogonal complement of ##\{|\chi\rangle\}##.

    I haven't worked out the solution to this problem, so I don't know what exactly will be useful. But the first thing that comes to mind is that every vector in this space is an eigenstate of an operator of the form ##\vec n\cdot\vec S##, where ##\vec n## is some unit vector in ##\mathbb R^3##. This seems relevant, since you're asked to use a rotation operator.

    A rotation around the z axis is probably not going to be sufficient.
     
    Last edited by a moderator: May 6, 2017
  8. Sep 3, 2014 #7

    fluidistic

    User Avatar
    Gold Member

    Thank you very much guys so far.
    Using Fredrik's notation, I do not know how to find the density matrix because it seems to be necessary to know how T acts on ##\left | j \right >##.
    Or using the notation in my first post, how ##\hat \rho## acts on ##\left | \psi _i \right >##.
     
  9. Sep 3, 2014 #8

    Matterwave

    User Avatar
    Science Advisor
    Gold Member

    Frederik gave you the answer in his post basically, but I can expand on that (in my notation since this is a 2 state problem, I think the ijk indices are overboard here). We have ##\rho=\left|\chi\right>\left<\chi\right|##. Given that ##\left|\chi\right>=\alpha\left|z_+\right>+\beta\left|z_-\right>## what do you think ##\rho## equals in terms of ##\left|z_+\right>,\left|z_-\right>##?
     
  10. Sep 3, 2014 #9

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    The ##|\psi_i\rangle## are just ##|\uparrow\rangle## and ##|\downarrow\rangle##, and ##\rho## is just ##|\chi\rangle\langle\chi|## where ##|\chi\rangle## is a linear combination of those two vectors.
     
  11. Sep 3, 2014 #10

    fluidistic

    User Avatar
    Gold Member

    I'm very slow and sloppy with Dirac's notation (our professor never taught it to us).
    Here is an attempt of calculating the first entry of the matrix: ##\left < \uparrow \right | \rho \left | \uparrow \right >##.
    I do it by parts: ##\rho \left | \uparrow \right > = \left | \chi \right > \left < \chi | \uparrow \right > =\left | \chi \right > (\overline \alpha \left < \uparrow \right | +\overline \beta \left < \downarrow \right | ) \left | \uparrow \right >=\left | \chi \right > \overline \alpha \left < \uparrow | \uparrow \right > = \left | \chi \right > \overline \alpha##.
    So that ##\left < \uparrow \right | \rho \left | \uparrow \right > =\left < \uparrow | \chi \right > \overline \alpha =\left < \uparrow \right | (\alpha \left | \uparrow \right > + \beta \left | \downarrow \right > ) \overline \alpha =\overline \alpha ^2 ##.
    Is this correct so far?
    Edit: Can't be correct I guess... the entry has to be real...
     
  12. Sep 3, 2014 #11

    Matterwave

    User Avatar
    Science Advisor
    Gold Member

    You can do it that way, but it seems quite convoluted. Your calculation is correct except for the fact that ##\alpha\bar{\alpha}=|\alpha|^2## so you did get a real number at the end. But not all entries in the density matrix have to be real anyways, it is a Hermitean matrix, so it is in fact a complex matrix. An easier way would be to perhaps just write ##\left|\chi\right>\left<\chi\right|=(\alpha\left|z_+\right>+\beta\left|z_-\right>)(\bar{\alpha}\left<z_+\right|+\bar{\beta}\left<z_-\right|)##.
     
  13. Sep 3, 2014 #12

    fluidistic

    User Avatar
    Gold Member

    The problem is that I do not get any alpha. I only get the complex conjugate of it (twice). The diagonal entries must be real I think. The off-diagonal need not be real as you said.
    And about the suggestion that's basically what I've used. But by parts.


    Nevermind, I do get the modulus of alpha squared. I got confused about the convention of the inner product. Thanks I will proceed with the other entries.
     
  14. Sep 3, 2014 #13

    Matterwave

    User Avatar
    Science Advisor
    Gold Member

    Why do you have a conjugate in the second part of your calculation? I don't see a conjugate there.

    Also, my suggestion was to simply foil it out and then you can basically read off the entries of the matrix, without having to go through these calculations.
     
  15. Sep 3, 2014 #14

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    I suggest that you do it the way Matterwave proposed even if you want to get it in matrix representation rather than in braket notation. Multiplying the parentheses will allow you to simply read off the matrix elements. The next step is to relate the absolute values of alpha and beta upon which you will get a relation they must fulfill. I am sure you know how to parametrize this relation ...
     
  16. Sep 3, 2014 #15

    fluidistic

    User Avatar
    Gold Member

    I just calculated the matrix rho the hard way: ##[\hat \rho ] =\begin{bmatrix} |\alpha | ^2 && \alpha \overline \beta \\ \overline \alpha \beta && |b| ^2 \end{bmatrix}##.
    Orodruin, are you suggesting me to replace ##|\beta |^2## by ##1- |\alpha | ^2##?

    Edit: I've just tried your way MatterWave. Indeed much better! I get rho as a sum of operators and I can see that the coefficients are the entries of the matrix.
     
    Last edited: Sep 3, 2014
  17. Sep 3, 2014 #16

    Matterwave

    User Avatar
    Science Advisor
    Gold Member

    You could make that substitution if you wanted, but I don't see how it will really help you. The next part of the problem is to rotate that matrix so it becomes diagonal. To do that though, one should use a little ingenuity instead of brute force. In fact it is much easier to figure out the result of this rotation than the rotation matrix itself.

    Frederick already gave you the answer in his post earlier to this question, but, in which basis will the density matrix be diagonal? And what will the density matrix be in such a basis? (Hint: a pure state has what kind of a diagonal density matrix?)
     
  18. Sep 3, 2014 #17

    fluidistic

    User Avatar
    Gold Member

    A pure state has a 1 somewhere in the diagonal, all the other entries are 0.
     
  19. Sep 3, 2014 #18

    Matterwave

    User Avatar
    Science Advisor
    Gold Member

    Yeah, so you already know the answer. Do you still need to find the rotation matrix? The problem did not specify what exactly you were looking for.
     
  20. Sep 3, 2014 #19

    fluidistic

    User Avatar
    Gold Member

    Yeah I must find the rotation matrix and apply it to diagonalize rho.

    I've just found the matrix ##\hat n \cdot \vec S##. It's worth ##\frac{\hbar}{2} \begin{bmatrix} \cos \theta && \sin (\theta) e^{- i \phi} \\ \sin (\theta )e^{i \phi} && - \cos \theta \end{bmatrix}## where I've used the spherical coordinates, theta is the zenith angle and phi is the azimuth one.
     
  21. Sep 3, 2014 #20

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Exactly, do you know of two functions of one single parameter that fulfill this relation? To completely parametrize the density matrix you will need this single parameter and the phase of ##\alpha\bar\beta##.

    After that, think about how you can remove the phase by multiplying by a unitary matrix on each side. Once that is done you are almost done.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Density matrix, change of basis, I don't understand the basics
Loading...