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Commutators; matrices? numbers? both?

  1. Aug 21, 2013 #1
    The commutator of two operators A and B, which measures the degree of incompatibility between A and B, is AB - BA (at least according to one textbook I have). But multiplying/substracting matrices just yields matrices! (http://en.wikipedia.org/wiki/Matrix_multiplication).
    So firstly, how can a matrix measure incompatibility?
    Secondly, how can this equation be valid if the commutator of position and momentum is a number i times h-bar?
    Thirdly, how is a weird number like i times h-bar a measure of anything?

    Very confused, please help.
     
  2. jcsd
  3. Aug 21, 2013 #2
    A commutator in general is an operator, so it is fine that the commutator of 2 matrices is a matrix.
     
  4. Aug 21, 2013 #3
    Thanks but how does that answer my questions? The commutator of position and momentum (I times h-bar) is a number, not a matrix. (Though whether it's a matrix, or I-times-h-bar, either way, hard to see how it measures anything let alone degree of incompatibility.)

    Thoughts anyone?
     
  5. Aug 21, 2013 #4
    Well, does seeing the more general uncertainty relation

    [itex]\langle(\Delta A)^2\rangle\langle(\Delta B)^2\rangle\geq\frac{1}{4}|\langle[A,B]\rangle|^2[/itex]

    help you? All quantities are now calculated with expectation values, which are numbers.
     
  6. Aug 21, 2013 #5
    No, I don't know what the <> brackets mean for start. Take the right hand side. You've got the expression for the commutator of A and B (that is, [A, B]) then I guess your brackets <> transform that matrix into some number; then you take absolute value times by a quarter then square it. Your statement says that that resulting number is less than or equal to the result of some other operation on the two operators.

    So no, that doesn't really help. Are you saying AB-BA does not give the commutator? Or that it does in restricted circumstances (matrix problem still remains). Also I-times-h-bar being a measure of the degree of anything remains.
     
  7. Aug 21, 2013 #6

    CompuChip

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    In general, the commutator of two objects is of the same type as those objects, because it is a sum of two of them.

    Have you seen 3D rotation matrices? Because I think they make a good example. If A and B are rotation matrices, then [A, B] tells you something about the "difference" in the final result if you apply them in a different order. For example, if A and B are both rotations around the same axis (say, x-axis) then [A, B] = 0. This means that for any vector v, ABb = BAv. The zero says: the operations commute. In general, for rotations, this is not the case. For example, rotating first around the x-axis and then around the z-axis gives a different result than first performing the rotation around the z-axis and then around the x-axis. In this case, ABb will give a different vector from BAv. Are the results related? Well, obviously yes: there is a vector Cv such that ABb = BAv + Cv, and C = AB - BA.

    The reason that this concept is important, is that in quantum mechanics operations often don't commute. Still, we sometimes want to have a specific operation at a specific position. Since your question is obviously inspired by QM, let me give an example. If the notation is not familiar to you, let me know and I will try to reformulate. For example, if |E> is some eigenstate of the Hamiltonian (##\hat H |E\rangle = E |E\rangle##), and you have an operator ##\hat A## that acts on your state, you can calculate the energy of the new state from ##\hat H \hat A |E\rangle##. However, you would prefer to have the Hamiltonian act first as you already know what it does to the state, so you would like to swap them. In general, this is not just allowed, but if you know the commutator of H and A you can write
    $$\hat H \hat A |E\rangle = \hat A \underbrace{\hat E |E\rangle}_{= E |E\rangle} + \underbrace{[\hat H, \hat A]}_{\text{known}} |E\rangle$$
    where ##\hat A E |E\rangle = E \cdot \hat A |E\rangle## because E is just a number which commutes with A.

    Position and momentum are not really matrices, but they are operators in a more general sense. For example, in the position basis where ##\hat x |x\rangle = x |x\rangle##, ##\hat p = c \frac{\partial}{\partial x}## where c is some complex constant. The way to calculate [x, p] is by letting it act on a "test" or dummy state like ##|x\rangle## above. However, whatever the commutator is, it must be of the same type as ##\hat x## and ##\hat p##.

    The most important thing is that this weird number is non-zero: whenever you use momentum operators and position operators together you must be very careful not to accidentally swap them without the proper bookkeeping. That a ##\hbar## pops up is irrelevant - you can choose your units such that ##\hbar = 1##. Actually you can consider it lucky (it's not, technically, but that would go too far) that the commutator is just the operator that multiplies by a constant. In principle, we could have had
    $$ [\hat x, \hat p] f = \frac{\partial^2 f}{\partial x^2} - x^2 f $$
    and QM would have been (even) much more unpleasant.
     
  8. Aug 21, 2013 #7
    The angle brackets simply mean the expectation value


    AB-BA is the commutator. No, what I'm saying is that the formula gives a way to express the uncertainty related to the operators not using the commutator, but using the expectation value of the commutator. The commutator is an operator (and can be written as a matrix), for example the canonical xp-commutator is actually written for each component as [itex][x_i,p_j]=i\hbar\delta_{ij}[/itex], where the Kronecker delta is zero when i≠j, 1 when i=j: It's a two-indiced object, a matrix (the identity matrix).

    Like I said, the expression above gives a way to interpret iℏ in terms of the uncertainty.
     
  9. Aug 21, 2013 #8
    Thanks Compuchip! I'm still thinking a bit about your answer.

    Deldeal, I know that <> mean expectation value, I just don't know what it means to flank them around a matrix! When ever I use them, I put them around a term for a state and a property, for example, <a|B=bi> is the expectation value for finding a particle in state |a> to be in state bi when I do a B measurement on it.

    Okay that second bit helps, so [x,p] is NOT equal to the number I-times-h-bar but is equal to a matrix whose elements are proportional to I-times-h-bar? That makes a bit more sense.

    Then the question is how a matrix could possibly be a measure of anything. I take it this has something to do your use of expectation value to convert the matrix to a number. I am not familiar with such uses of <> as I say, I've only used <> on two vectors, not a single matrix.
     
  10. Aug 21, 2013 #9

    CompuChip

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    It is really equal to the operator that multiplies by the constant ##i\hbar##. Remember that these operators act on functions. In a way, they are the continuous analog of matrices: if the domain of the function were countably infinite then in principle you could write down the function as an infinite vector from which any function value can be read off, and then the operator x would be an infinite matrix. In this view, the commutator of x and p would be ##i\hbar## times the infinite identity matrix. There are a few problems with this view though, most notably that function domains a uncountable and writing down uncountably many values in matrix form is not feasible.
     
  11. Aug 21, 2013 #10
    The commutator between momentum and position simply means that:

    ##\hat x \hat p ψ = \hat p \hat x ψ + i\hbar ψ##

    If ##ψ## is a continuous function, the operator ##i\hbar## is really just a multiplication by a constant (a mere number). But it still tells you how much the result is different when you switch the 2 operators.
     
    Last edited: Aug 21, 2013
  12. Aug 21, 2013 #11

    kith

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    That's not the expectation value, that's the probability (amplitude). The expectation value of an observable A is defined as <A> = <ψ|A|ψ>. It gives the value you expect after averaging over many measurement outcomes.

    Kind of. Writing the commutator as a number is a sloppy kind of saying that it is the identity operator (resp. matrix) times the number. For x and p, we run into some unrelated technical difficulties if we want to write down a corresponding matrix.
     
    Last edited: Aug 21, 2013
  13. Aug 21, 2013 #12

    kith

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    This is a bit misleading. The commutator [x1,p1] for example is an operator. The reasoning above suggests that it would be only a matrix element aka a number.
     
  14. Aug 21, 2013 #13
    You're right, it was quite poorly worded. I tried to make it clear to OP that the commutators [A,B] are indeed operators, and only quickly included that because he seemed confused about that specific commutator being "a number".
     
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