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What do vector operators mean?

  1. Aug 17, 2012 #1
    My QM book introduces operators like the momentum operator [itex]\hat{\mathbf{p}}[/itex] which act on their eigenstates to produce new states like [itex]\hat{\mathbf{p}}|\mathbf{p}\rangle = \mathbf{p}|\mathbf{p}\rangle[/itex]. But how can we interpret a state like that? How is multiplication between a Euclidean vector and a vector in Hilbert space defined?
     
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  3. Aug 17, 2012 #2

    atyy

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    It should be [itex]\hat{\mathbf{p}}|\mathbf{p}\rangle = p|\mathbf{p}\rangle[/itex], where the formula is the definition of an eigenstate [itex]|\mathbf{p}\rangle[/itex] of the operator [itex]\hat{\mathbf{p}}[/itex] . An eigenstate of an operator is a vector which is not rotated, but only changed in length by the operator (multiplied by a scalar).
     
  4. Aug 17, 2012 #3

    Fredrik

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    ##\left|\mathbf p\rangle\right.## denotes an eigenstate of all three components of ##\hat{\mathbf p}=\big(\hat{p}_1,\hat{p}_2,\hat{p}_3\big)##. So
    $$\hat{\mathbf p}\left|\mathbf p\rangle\right.=\mathbf p\left|\mathbf p\rangle\right.$$ should be interpreted as
    For all ##i\in\{1,2,3\},## we have
    $$\hat{p}_i\left|\mathbf p\rangle\right.=p_i\left|\mathbf p\rangle\right. .$$​
     
  5. Aug 17, 2012 #4
    No it shouldn't. An eigenstate of the momentum vector operator is a quantum state |px>|py>|pz> with well-defined momentum in the x, y, and z direction, so it has eigenvalues corresponding to all three directions, and that ordered triple of eigenvalues can be written as a vector.
     
  6. Aug 17, 2012 #5

    atyy

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    I see. How do I see that the ordered triple of eigenvalues is a vector?
     
  7. Aug 17, 2012 #6
    Thanks everyone, especially Fredrik.
     
  8. Aug 17, 2012 #7
    Hmm, the question never occurred to me before, but here's a stab at it. Let me start with the position operator, because that's easier to conceptualize. You would have to show that the ordered triple of position eigenvalues corresponding to a given position eigenstate transforms according to the vector transformation law if we apply any rotation operator to the Hilbert space, where the rotation operators are defined in terms of the orbital angular momentum operators (in any given direction). Now to do the same thing for momentum, you would have to use operators that represent rotations in momentum space. (Just due to the relation between position and momentum, I think rotations in momentum space ARE rotations in position space.)
     
  9. Aug 18, 2012 #8

    Fredrik

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    The answer depends on how you intend to associate an ordered triple of real numbers with each coordinate system. It's that association that may or may not be a tensor. A specific triple is just a triple, never a tensor.

    The relevant coordinate systems are in bijective correspondence with the members of SO(3), so the easiest way to define the association is to first assign the triple ##\mathbf p## to the coordinate system we had in mind when we wrote
    $$\hat{\mathbf p}\left|\mathbf p\rangle\right.=\mathbf p\left|\mathbf p\rangle\right. ,$$ and then assign the triple ##R\mathbf{p}## to the coordinate system such that a change of coordinates to this new system is given by R.

    But now there's nothing to prove. What we did is the same thing as saying that we're dealing with "the tensor that has components ##\mathbf{p}## in the coordinate system we're using now", so we have assumed what you wanted to prove.

    Another way to do the association is to first assign a triple of momentum operators to each coordinate system: Consider an arbitrary coordinate system (in addition to the one in which we are already using the notation ##\hat{p}_i## for our momentum operators. Let R be the function (member of SO(3)) that changes coordinates to the new system. Then we associate the operators ##\hat{p}_i'=U(R)\hat{p}_iU(R)^{-1}=R_{ij}\hat{p}_j## with the new coordinate system.

    Now we can associate a triple of real numbers with the new coordinate system, by saying that ##p_i'## denotes the eigenvalue of ##\hat{p}'## associated with the eigenstate ##\left|\mathbf p\rangle\right.##. Then we have
    $$p_i'=\langle\mathbf{p}|\hat{p}_i' |\mathbf{p}\rangle =R_{ij}\langle\mathbf{p}|\hat{p}_i |\mathbf{p}\rangle =R_{ij}p_j=(R\mathbf{p})_i,$$ as desired.
     
  10. Aug 18, 2012 #9
    Projection operation. You have a state vector in momentum space. The eigenstates form a basis for the momentum vector space. The momentum operator takes the state vector and projects it onto the basis vectors. Right?
     
  11. Aug 18, 2012 #10

    Fredrik

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    I don't quite follow you, but momentum operators aren't projection operators. A projection operator only has eigenvalues 0 and 1.

    Momentum space? Momentum vector space? If ##|\psi\rangle## is the state vector, then the function ##\mathbf{p}\mapsto\langle\mathbf{p}|\psi\rangle## can be thought of as a momentum-space wavefunction. Is that what you have in mind?
     
  12. Aug 19, 2012 #11
    The momentum space wavefunction is what I had in mind. That was some rather careless wording on my part. The projection operator is indeed an entirely different beast. What I meant to say, is that only the eigenstates are physically measurable. The eigenstates also form a basis for the momentum vector space. When the momentum operator acts on the general quantum state, it gives a momentum vector in terms of the basis eigenstates...so it's kind of projecting onto the eigenstates and determining how much of that quantum state lies in each basis direction, right?
     
  13. Aug 20, 2012 #12

    atyy

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    Yes, it does seem that momentum eigenvalue triples form a vector space. Can I just check that that isn't true of momentum eigenvectors though (ie. they form the basis of a vector space, but they don't themselves form a vector subspace, since adding two eigenvectors does not produce another eigenvector)?
     
  14. Aug 21, 2012 #13
    If all you wanted to know was that triples of momentum eigenvalues form a vector space, that's trivial, since they're just all the possible triples of real numbers. The more interesting question that me and Fredrik were addressing was whether they transform like vectors under rotations, i.e. whether they satisfy the definition of a rank-1 tensor.
     
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