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Momentum operator as differentiation of position vector

  1. Dec 13, 2013 #1
    Is it possible to take momentum operator as [itex]dr/dt[/itex] ([itex]r[/itex] is position operator)? If not, why?
     
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  3. Dec 13, 2013 #2

    bhobba

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    Even in classical mechanics that's velocity not momentum. Because, in QM, it has a factor of -i; in units h bar = 1 that is, and its the derivative wrt to x of the wavefunction.

    Why?

    At the beginning level its associated with the wave particle duality and De-Broglie waves:
    http://en.wikipedia.org/wiki/Momentum_operator

    At the more advanced level where the wave-particle duality is seen as not really a good way to view things its associated with symmetries - see Chapter 3 Ballentine - Quantum Mechanics - A Modern Development.

    Thanks
    Bill
     
    Last edited: Dec 13, 2013
  4. Dec 15, 2013 #3

    tom.stoer

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    It depends on the formalism.

    In canonical formalism one eliminates velocity v = dx/dt already in the classical theory. Instead the canonical variables x, p are used. So in this formalism there is nothing like dx/dt as an elementary entity.

    Later one can study Heisenberg equations of motion. There we do not talk about a velocity either, but about the time evolution of the position operator x and its Heisenberg equation of motion. So one studies [H,x].
     
  5. Dec 17, 2013 #4
    Thank you. As [itex][H,x]=(-i\hbar/m) p[/itex] why don't we consider the operator [itex]v=dr/dt[/itex] instead of the operator [itex]p[/itex]?
     
  6. Dec 17, 2013 #5

    bhobba

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    Because the math shows its wrong.

    Again see Chapter 3 Ballentine page 77 where the correct velocity operator is derived from the derivative of the expectation value of position should give the expectation of velocity, and is V=i[H,X] ie if we require d<X>/dt = <V> then V=i[H,X]. The math isn't hard, but understanding it requires stuff in the preceding pages.

    Thanks
    Bill
     
    Last edited: Dec 17, 2013
  7. Dec 17, 2013 #6

    vanhees71

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    One must be careful with the time dependence of operators in quantum mechanics. The mathematical time dependence is arbitrary in a wide range since you can always do time-dependent unitary transformations on the operators that represent observables and states (either represented by normalized Hilbert-space vectors or statistical operators; here I'll use state vectors, but the same arguments also apply to the general case of stat. ops.):
    [tex]\hat{A}' = \hat{U}(t) \hat{A} \hat{U}^{\dagger}(t), \quad |\psi' \rangle =\hat{U}(t) |\psi \rangle.[/tex]
    So you can shuffle the time dependence from the states to the operators and vice versa without changing the physical results.

    In non-relativistic quantum theory of one particle ("1st quantization") with spin 0 (scalar particle) you have the basic observables [itex]\vec{x}[/itex] and [itex]\vec{p}[/itex]. Any other observable is a function of these.

    The dynamics of the system is characterized by a Hamiltonian, [itex]H(\vec{x},\vec{p})[/itex] (where I assumed a Hamiltonian that is not explicitly time dependent, i.e., which only depends on time through its dependence on the fundamental operators [itex]\vec{x}[/itex] and [itex]\vec{p}[/itex]). As stressed above, the mathematical time dependence of the operators and states is arbitrary to a large degree, and you can choose any "picture of time dependence" you like to solve a problem.

    Usually one starts in the Schrödinger picture, where the observables are time-independent (except when you consider explicitly time dependend observables, but this we won't do in this posting for sake of clarity). The entire time dependence is thus put to the state vectors. This means you have
    [tex]\frac{\mathrm{d} \vec{x}}{\mathrm{d} t}=0, \quad \frac{\mathrm{d} \vec{p}}{\mathrm{d} t}=0, \quad \frac{\mathrm{d}}{\mathrm{d} t} |\psi(t) \rangle = -\frac{\mathrm{i}}{\hbar} H |\psi(t) \rangle.[/tex]

    Independently from the choice of the picture of time evolution you may ask, which operator describes the time derivative of an observable. E.g., you may ask, which operator represents the velocity, which in classical physics is the time derivative of the position vector. In the Schrödinger picture, the mathematical time derivative of the position-vector operator is however 0, and thus this time derivative cannot represent velocity. The correct answer is given by the analogy with analytical mechanics (Hamilton formalism), where the time derivative of an observable is given by the Poisson bracket of the observable with the Hamiltonian. The Poisson brackets of classical canonical mechanics map (up to an imaginary factor) to commutators in quantum theory. Thus the correct "physical time derivative" of the position vector, giving the operator, representing velocity, is
    [tex]\vec{v}=\mathrm{D}_t=\frac{1}{\mathrm{i} \hbar} [\vec{x},H].[/tex]
    This holds true in any picture of the time evolution.

    Another picture is the Heisenberg picture, where the state vectors are constant and the entire mathematical time dependence is put to the operators, representing observables. The formal solution of the state-vector equation of motion in the Schrödinger picture is
    [tex]|\psi(t) \rangle_S=\exp \left (-\frac{\mathrm{i}}{\hbar} t H \right ) |\psi(0) \rangle_{S} = U_S(t) |\psi(0) \rangle_S.[/tex]
    The operator [itex]U_S(t)[/itex] obviously is unitary, because [itex]H[/itex] is self-adfjoint. Thus we can do a transformation to the Heisenberg picture (assuming that state vectors and observable operators coincide in both pictures at the initial time [itex]t=0[/itex]) by setting
    [tex]|\psi(t) \rangle_S=U_s(t) |\psi \rangle_{H},[/tex]
    where [itex]|\psi \rangle_H=|\psi(t=0) \rangle_S[/itex] is the time-independent state vector in the Heisenberg picture.

    The observable operators in the Heisenberg picture become time dependent, because we have to set
    [tex]A_S=U_S(t) A_H(t) U_S^{\dagger} \; \Leftrightarrow\; A_H(t)=U_S^{\dagger}(t) A_S U_S(t).[/tex]
    Now we can take the mathematical time derivative of this expression. For a observable [itex]A[/itex] that is not explicitly time dependent we get
    [tex]\frac{\mathrm{d}}{\mathrm{d} t} A_{H}(t)=\frac{\mathrm{i}}{\hbar} U_S^{\dagger} (H A_S-A_S H) U_S=\frac{\mathrm{i}}{\hbar} [H,A_{H}(t)].[/tex]
    Here we have used that
    [tex]H_S=H_H=H,[/tex]
    because [itex]U_S(t)[/itex] commutes with [itex]H[/itex]. From the previous equation we see that in the Heisenberg picture (and only in the Heisenberg picture) the physical time derivative of observable operators coincides with the mathematical time derivative.

    Your original question, is not so much related to this sometimes a bit confusing issue with the time dependence of quantities in quantum theory but with the physical meaning of the momentum operator. That the momentum operator is given as the gradient of the wave function in position representation comes from the fact that momentum is the generator of spatial translations.
     
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