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I States diagonal in the reference basis

  1. Sep 1, 2016 #1
    can someone give an example for an incoherent State --> a formula is here on page 7 : http://quantumcorrelations.weebly.com/uploads/6/6/5/5/6655648/2016_robustnessofcoherencetalk.pdf

    I know that coherenc is e.g. a superposition of e.g. Spin-Up and Spin-Down [z] or so....

    But i have no idea about an incoherent state. Is it maybe a "clear" state e.g. Spin-Down [z] (if you measuring Spin [z] you get 100% Spin-Down)?


    PS: sorry, bad english. Not my natural language.
  2. jcsd
  3. Sep 1, 2016 #2


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    First a note about terminology: "incoherent states" are more often called mixed states or incoherent mixtures (of states). You'll find a lot if you search the forums for this. In short, they are states which can't be written as a state vectors / wave functions. You need so-called density matrices to describe them.

    From what you write, it seems to me that you don't really understand state vectors in detail yet. For example, do you know that whether a state vector "is" a superposition depends on what basis you are using? Every state vector can be written as a superposition of state vectors and every superposition of state vectors can be written as a single state vector. I recommend to proceed to mixed states only if you feel comfortable with this (use spin-1/2 to picture this).
  4. Sep 1, 2016 #3
    Thats true! ;-) But it sounds interesting what you write and i think i understand it till here.
  5. Sep 1, 2016 #4


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    That's good, I didn't intend to discourage you!

    Your question is just hard to answer if we don't know where you stand, where you want to go and how much effort you are willing to put in. For example, do you know bra-ket notation? If you are not familiar with the basic math of QM, it takes a lot of words to explain these things.

    The basic concept of a mixed state is really simple: it's just a probabilistic mixture of quantum states. You start with known quantum states and end up with a state of, let's say, 50% probability to find the system in one quantum state and 50% probability to find it in another quantum state. This is analogous to putting balls of different colors into an urn. If probability would only show up in this way in QM, it would simply be classical mechanics. So we have the quantum probabilities associated with superposition and in addition to that, we have classical probability which leads to mixed states. (There's a caveat to this if we have entanglement)

    The difficult part is to understand the quantum part, i.e. superpositions and entanglement. After that, mixed states are an easy addition.
  6. Sep 1, 2016 #5
    Thank you much for you answers! Maybe i have later more questions.
  7. Sep 1, 2016 #6
    E.g. i have an single photon or electron --> then it always can "be written as a state vectors / wave functions" & is never a mixed state.
    E.g. i have two photons, not entangled --> then this is always a mixed state?
  8. Sep 1, 2016 #7


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    If you use an oven to produce silver atoms for a Stern-Gerlach-experiment, the probability to measure spin up in any direction is 50%. This is the standard example for a mixed state. At least in principle, you could tune down your particle emission until you get single atoms. Each of them is in the mixed state then.
    No. The most important states for photons are the number states.

    Mixed states arise mostly in open quantum systems where a quantum system of interest interacts with a large environment. If you have perfect control of your system (like in most gedanken experiments) mixed states don't occur. (Again there is a caveat when there's entanglement)
  9. Sep 1, 2016 #8
    I was thinking this is a superpostion of spin up/down [x,y,z]. hmmmmmm........... :cry:
    can there only be a superpostion of e.g. spin up/down [x] but no superpostion of spin up/down [x] AND spin up/down [y] ... (...)
  10. Sep 1, 2016 #9


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    Sorry, I don't have the time to explain this in detail to you. If you are serious about learning these things, I recommend the "Theoretical Minimum" by Leonard Susskind. I don't really know much resources how to learn it at a lower level.
  11. Sep 1, 2016 #10
    States that can be written as state vectors or wave functions are called pure states.

    There is no pure state that has 50% probability to be in a spin up or down state in all directions. The probabilities have to vary by the Born rule.
  12. Sep 2, 2016 #11
    No problem. With --> (incoherent states) "states which can't be written as a state vectors / wave functions. You need so-called density matrices to describe them." - i have the answer i want! :wink:
  13. Sep 2, 2016 #12
    Can i have a "mixed state" with e.g. Superposition Spin Up/Down [x,y,z]?
  14. Sep 2, 2016 #13


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    Well, take
    $$|\psi \rangle=\frac{1}{\sqrt{2}} (|\sigma_z=1/2 \rangle+|\sigma_z=-1/2 \rangle)$$
    and calculate the probability to find ##1/2## and ##-1/2## when you measure the spin-z component :-)).
  15. Sep 2, 2016 #14
    Note that I said "all directions". :)

    To be perfectly clear, there is no pure state ##| \psi \rangle## that gives ## |\langle \psi|n\rangle |^2 =\frac{1}{2} ## independent of the direction ## |n \rangle = \cos (\theta /2 )|\sigma_z = 1/2 \rangle + e^{i \phi} \sin (\theta /2) |\sigma_z = -1/2 \rangle ## for ## 0 \le \theta \le \pi, 0 \le \phi < 2\pi ##.

    What do you mean by spin up[x,y,z]?
  16. Sep 2, 2016 #15
    the direction
  17. Sep 3, 2016 #16
    If I understand your question correctly then, the answer is yes. For instance, a spin-1/2 particle in a mixed state can always be thought as being in a spin up or down state with some probabilities. Contrast this to a superposition of a spin up and down state which gives a spin state pointing in a different direction. (Sometimes people will say the spin in the latter state is in a spin up and down state at the same time, but that's a misuse of the term "state" because a state is supposed to contain all the information one could have. "Being in two states at once" is an oxymoron.)
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