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 off diagonal terms - what do they mean?

  1. Nov 22, 2007 #1
    A superposition of states such as [tex]a_1|\psi_1\rangle+...+a_n|\psi_n\rangle[/tex] represents a single physical state, a state for which the probability of a measurement finding the system in state [tex]|\psi_k\rangle[/tex] is [tex]|a_k|^2[/tex]. The [tex]a_k[/tex] represent "quantum-type" probabilities.

    On the other hand the density matrix [tex]D=p_1|\psi_1\rangle\langle\psi_1|+...+p_n|\psi_n\rangle\langle\psi_n|[/tex] represents a statistical ensemble of states, statistical in the classical sense in which we accept that the system really is in some particular state but we just don't know which one. The probabilities [tex]p_k[/tex] reflect the uncertainty in our knowledge of the system, the kind of ordinary probabilities that would apply to, say, a poker game.

    I wrote the above just to verify that I understand it this far. Ok? (I know there are large ontological and epistemological gray areas in the above statements, but let's just go with the Copenhagen interpretation for the sake of discussion.)

    Now if we switch to some other set of basis states [tex]|\phi_k\rangle[/tex], this same density matrix D will contain "off-diagonal" terms [tex]|\phi_j\rangle\langle\phi_k|, j\ne k[/tex]. If a term [tex]p_k|\psi_k\rangle\langle\psi_k|[/tex] means "[tex]p_k[/tex] is the probability that the system is actually in state [tex]|\psi_k\rangle[/tex]," then what do the terms of the form [tex]q_{jk}|\phi_j\rangle\langle\phi_k|, j\ne k[/tex] represent?
     
    Last edited: Nov 22, 2007
  2. jcsd
  3. Nov 22, 2007 #2

    f95toli

    User Avatar
    Science Advisor
    Gold Member

    The off-diagonal terms are sometimes called the "coherences" because they describe superposition of states; when states "decay" due to dissipation the off-diagonal terms go to zero (mathematicallt this usually means that the off-diagonal terms decay exponentially).
    When all off diagonal states are zero you basically have a classical (product) state.

    The simplest way to learn about this is probably to play around with the Rabi model and then try to map this to the Block sphere (which in the case of a 2x2 matrix is essentially just a a graphical representation of the system with the "north"- and "south" poles describing "up" and "down" states)

    Also, the denisty matrix approach is usefull also when you do NOT have a statistical ensemble. In the Heisenberg/interaction picture it is often more conventient to use a density matrix.
     
  4. Nov 22, 2007 #3
    Thanks. I think I get it. Actually, the book I was looking at covers the Block sphere on the page after where I stopped to type up my question. :-)

    The problem came from the author pointing out that the density matrix you get if you start with [tex]D = 0.5|up\rangle \langle up|+0.5|down\rangle \langle down|[/tex] and switch to a different z-direction the matrix takes the same form, say, [tex]D = 0.5|right\rangle \langle right|+0.5|left\rangle \langle left|[/tex]. I was surprised that there were no off-diagonal terms no matter which direction we measured from and that led to my question. But it's just because the 0.5,0.5 case is special.

    Todd
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Density matrix off diagonal terms - what do they mean?
Loading...