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Density matrix to represent polarization, what is this? References anyone?

  1. Feb 5, 2010 #1
    Hi,

    I have a particle physics exam tomorrow morning (in a few hours from now, in my time zone). I'm trying to figure out the whole reasoning behind pion-nucleon scattering. Please bear with me..

    We write the scattering matrix as

    [tex]S = 1 - iT[/tex]

    where T is given by

    [tex]T = f + i g \boldsymbol{\sigma}\cdot\hat{n}[/tex]

    Here f and g are scalars, functions of the mandelstam variables s and p (total energy in CM frame squared and difference of momenta modulus). [itex]\hat{n}[/itex] is the normal vector perpendicular to the reaction plane. We reasoned that T has this form because the dependence on total momentum P and differential momentum p must vanish for T to be invariant under parity. So far so good.

    Now, what I do not understand follows below. We write the initial state density matrix as

    [tex]\rho^i = \frac{1}{2}(1 + \vec{\sigma}\cdot \vec{P}_{i})[/tex]

    How? Then we write the final state density matrix as

    [tex]\rho^f = \frac{T\rho^{i}T^{\dagger}}{Tr[T\rho^{i}T^{\dagger}]}[/tex]

    How?

    We also define a quantity

    [tex]\rho_{m'm} = \langle s m'|\rho|s m\rangle[/tex]

    and reason that since the state should be invariant under rotation about the z-axis, we must have

    [tex]e^{-iS_{z}\pi}|m\rangle = (-1)^{-im}|m\rangle[/tex]

    and hence

    [tex]\rho_{m'm} = (-1)^{m-m'}\rho_{m'm}[/tex]

    My questions:

    1. What is the motivation behind writing this density matrix?

    2. How was it written in the first place?

    3. What is [itex]\rho_{m'm}[/itex]? Its some kind of matrix element, but what does it signify? Does it signify a transition from state m to m'? (I'm having a bad day here :-|)

    I never studied scattering theory this way, so I would appreciate if someone could give me a heads-up and point to the relevant text(s).

    Thanks!

    EDIT: I just read the section on projection operators and density matrices from Schiff's book. Am I correct in interpreting this as a way of specifying the initial state in terms of its spin? I get this if we have just one particle to begin with, but when we have two -- as is the case here (pion and nucleon) how do we write a composite density matrix?
     
    Last edited: Feb 5, 2010
  2. jcsd
  3. Feb 5, 2010 #2
    Last edited: Feb 5, 2010
  4. Feb 5, 2010 #3

    Fredrik

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    I don't know the stuff that this course is about, but I might be able to help with some of the technical details. The first thing you should know is that any 2x2 traceless self-adjoint matrix can be written as

    [tex]\begin{pmatrix}x_3 & x_1-ix_2\\ x_1+ix_2 & -x_3\end{pmatrix}=x_i\sigma_i[/tex]

    so the set of Pauli spin matrices is just a basis of the (real) vector space of (complex) 2x2 traceless self-adjoint matrices. If you want to include matrices that aren't traceless, you add the 2x2 identity matrix I as a fourth (or zeroth) basis vector. A density matrix has trace 1, so the coefficient of I must be 1/2. This ensures that your density matrix can be written in that form, but the interpretation of [tex]\vec P[/tex] as momentum isn't obvious to me yet. I'll think about it. (Now I wish I had saved my notes from the time Hurkyl said something that forced me to work through these things).

    By the way, is there a sum over s in your definition of [tex]\rho_{m'm}[/tex]?

    Edit: The reason I ask is that this would make it a "reduced density matrix", which is the statistical representation of a subsystem. The property that makes it useful is fairly easy to derive. (I'm going to have to type this up for my personal notes, so I might as well start now). Let [itex]|\mu\alpha\rangle=|\mu\rangle\otimes|\alpha\rangle[/itex] be basis vectors for the tensor product space.

    [tex]\langle A\otimes I\rangle=\mbox{Tr}(\rho(A\otimes I))=\sum_{\mu\alpha}\langle\mu\alpha|\rho(A\otimes I)|\mu\alpha\rangle=\sum_{\mu\alpha}\sum_{\nu\beta}\underbrace{\langle\mu\alpha|\rho|\nu\beta\rangle}_{=\rho_{\mu\alpha,\nu\beta}}\underbrace{\langle\nu\beta|A\otimes I|\mu\alpha\rangle}_{=\langle\nu|A|\mu\rangle\delta_{\beta\alpha}}[/tex]

    [tex]=\sum_{\mu\nu}\bigg(\underbrace{\sum_\alpha\rho_{\mu\alpha,\nu\alpha}}_{=\rho'_{\mu\nu}}\bigg)\langle\nu|A|\mu\rangle=\sum_{\mu\nu}\langle\mu|\rho'|\nu\rangle\langle\nu|A|\mu\rangle=\mbox{Tr}(\rho'A)[/tex]

    The standard abuse of notation [itex]A=A\otimes I[/itex] turns this into

    [tex]\mbox{Tr}(\rho A)=\mbox{Tr}(\rho'A)[/itex]

    where the left-hand side is a trace over an operator on the tensor product space and the right-hand side is a trace over an operator on the Hilbert space corresponding to one of the subsystems. What we're doing when we define the reduced density matrix by

    [tex]\rho'_{\mu\nu}=\sum_\alpha\rho_{\mu\alpha,\nu\alpha}[/tex]

    is often described as "tracing out" the degrees of freedom of the subsystem that we're not interested in.
     
    Last edited: Feb 5, 2010
  5. Feb 5, 2010 #4

    Fredrik

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    The most general normalized qubit state can be written as

    [tex]|\psi\rangle=\cos\frac\theta 2|0\rangle+\sin\frac\theta 2 e^{i\phi}|1\rangle[/tex]

    This isn't too hard to prove. The corresponding density operator is

    [tex]\rho=|\psi\rangle\langle\psi|=\cos^2\frac\theta 2|0\rangle\langle 0|+...[/tex]

    I'm too lazy to LaTeX it all right now. The operators that appear on the right can be represented by 2x2 matrices. The representation depends on what basis we choose (in the way described here). If we choose [itex]|0\rangle,|1\rangle[/itex] to be the basis vectors, then

    [tex]|0\rangle\langle 0| \leftrightarrow \begin{pmatrix}1 & 0\\ 0 & 0\end{pmatrix}=\frac{I+\sigma_3}{2}[/tex]

    where I'm using the leftrightarrow as "corresponds to". I'll let you do the other ones for yourself. If you go through with this, you should find that

    [tex]\rho\leftrightarrow\frac 1 2(I+\vec r\cdot\vec \sigma)[/tex]

    where [itex]\vec r=(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta)[/itex]. To get your momentum variable in there, I'd have to know how it appears in the state you start with. In other words, I would need your a and b. I'm assuming that you have that information and can work out the details for yourself.
     
    Last edited: Feb 5, 2010
  6. Feb 5, 2010 #5

    Fredrik

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    I don't have a complete answer to the first question, but this might help a little (if you don't know it already). A measurement of an observable B (with non-degenerate discrete spectrum) changes a pure state density matrix as described by

    [tex]\rho=|\psi\rangle\langle\psi|\rightarrow\sum_b|\langle b|\psi\rangle|^2|b\rangle\langle b|=\sum_b|b\rangle\langle b|\psi\rangle\langle\psi|b\rangle\langle b|=\sum_b P_b\rho P_b[/tex]

    where [itex]P_b=|b\rangle\langle b|[/itex] is the projection operator of the 1-dimensional subspace of eigenvectors of B with eigenvalue b. If the state vector [itex]|\psi\rangle[/itex] isn't normalized, everything to the right of the arrow needs to be divided by

    [tex]\sum_b|\langle b|\psi\rangle|^2=\sum_b\langle b|\psi\rangle\langle\psi|b\rangle=\mbox{Tr }\rho[/tex]

    This looks similar enough to the formula you posted to be worth mentioning.

    The second question is easier. It will be in the form

    [tex]\sum_i w_i|\psi_i\rangle\langle\psi_i|[/tex]

    where [itex]|\psi_i\rangle=|\alpha_i\rangle\otimes|\beta_i\rangle[/itex] is a tensor product of pure states of the component systems, and the [itex]w_i[/itex] are non-negative real numbers that add up to 1. The bilinearity of the tensor product allows us to rewrite the density matrix above as

    [tex]\sum_i w_i|\alpha_i\rangle\langle\alpha_i|\otimes|\beta_i\rangle\langle\beta_i|[/tex]
     
  7. Feb 5, 2010 #6
    Thanks for the detailed posts Fredrik. The vector n thats 'dotted' with the Pauli matrices is the normal defined by the incoming and outgoing momenta in the CM frame. There are 3 vectors available to us: the sum of the initial and final momenta, their difference, and their cross product (n). The first two drop out due to parity considerations. What you have left is n.

    My question to you is: can I write the initial state as a density matrix when there is more than one particle in the initial state? In this case, the dimension is (2*0+1)(2*1/2 + 1) => 2 x 2 because of the neutron (and that the pion has 0 spin). What if we had a spin s1 and a spin s2 particle in the initial state?
     
  8. Feb 5, 2010 #7

    Fredrik

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    Then you would have a 2(s1)+1-dimensional Hilbert space and a 2(s2)+1-dimensional Hilbert space, and their tensor product is (2(s1)+1)(2(s2)+1)-dimensional, with basis vectors of the form [itex]|\mu\alpha\rangle=|\mu\rangle\otimes|\alpha\rangle[/itex], so you would need (2(s1)+1)(2(s2)+1)×(2(s1)+1)(2(s2)+1) matrices.
     
  9. Feb 5, 2010 #8
    Hello guys,

    It is quite understandable why you can write your final density matrix in this form
    [tex]
    \rho ^{'}= \frac{T\rhoT^{\dag}}{Tr[T\rhoT^{\dag}]
    [/tex]

    The sense of your T operator is to "create" final states [tex]\Psi = T v[/tex].

    Lets say we have scattering of relativistic electrons in external field. Imagine that you solved Dirac equation for this case and found wave functions [tex]\Psi[/tex]. Then you have final density matrix

    [tex]\rho^{'}_{\alpha \beta} = \Psi^*_\alpha \Psi_{\beta} = v^*_\gamma T^\dag_{\alpha \gamma} T_{\sigma\beta}v_{\sigma} = T^\dag \rho T[/tex].

    I'm too lazy to write indexes.:rofl: Don't forget that [tex]\rho_{\alpha \beta} = v^*_\alpha v_\beta[/tex],

    And of course [tex]Sp \rho = 1[/tex] that is why you have denominator.

    Regards.
     
    Last edited: Feb 5, 2010
  10. Feb 5, 2010 #9
    What is [itex]Sp[/itex]?
     
  11. Feb 6, 2010 #10
    [tex]Sp = Tr[/tex], Sp from deutsch word spur = trace, I'm surprised that you have never seen this before. For example one of the most famous books on QM by Albert Messia, he uses Sp except Tr there.
     
  12. Feb 6, 2010 #11
    I haven't read Messiah's book. I'm surprised I have never seen this before too :tongue2:. I studied QM from Sakurai, Schiff, Griffiths, Landau/Lifgarbagez, etc.
     
  13. Feb 7, 2010 #12
    I guess it is translation to English. Originally Landau book in Russian also use Sp except Tr.
     
  14. Feb 7, 2010 #13

    Hans de Vries

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    Feynman uses Sp as well in his book Quantum Electro Dynamics but one doesn't
    see it a lot nowadays.

    Regards, Hans
     
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