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Momentum operator eigenstates/eigenvalues

  1. Mar 30, 2015 #1
    As far as I know, the momentum operator is as follows:

    -iħ(∂/∂x)

    Now lets say that I enact this operator on the famous solution to the 1-D particle in a box example:

    Ψ= squrt(2/L) sin(πnx/L)

    If the momentum operator operates on the above wave function, it yields:

    -iħ * squrt(2/L) * (πn/L) * cos(πnx/L)

    Now, it is true that squrt(2/L) * cos(πnx/L) is actually a solution to the 1D time independent Schrodinger equation that yields the energy eigenvalue E when you enact the Hamiltonian operator on it. However, it is still different from the original wave function (with sine in it). That is why I question whether or not the momentum operator operates on the same eigenstates as the Hamiltonian operator (since by definition of eigenstate, the result of enacting the operator on the wave function must yield an eigenvalue multiplied by the original wave function). So tell me:

    Do the Hamiltonian and momentum operators work on the same eigenstates?
    If not, what are the eigenstates of the momentum operator?
     
  2. jcsd
  3. Mar 30, 2015 #2

    bhobba

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    Generally no. Its only true in the case of a free particle.

    The solutions to the eigenvalue equation P|p> = p|p>. If you expand |p> in terms of position eigenstates you get - well I will leave it as an exercise - its pretty basic and instructive. It is found in many texts eg page 96 of Dirac - Principles of QM which is simply the first text that came to hand.

    If you think a bit deeper about the result you will realise its rather problematical (hint - they are supposed to be elements of a Hilbert space).

    Post back here with the exact problem and we can have a chat about that - best in a separate thread so it can be examined independent of your original question.

    Thanks
    Bill
     
  4. Mar 31, 2015 #3

    vanhees71

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    Be warned, however: On the finite interval there is no momentum operator. The operators, representing observables, must be self-adjoint! Unfortunately, you find misleading statements concerning a momentum operator on a finite interval quite often in textbooks, which is very bad.
     
  5. Mar 31, 2015 #4
    To expand a little bit this comment. Mathematically, you can choose a dense subspace ##D(B) \subset L^2(0,b)## such that the "operator" usually described as:

    ##B: D(B)\subset L^2(0,b) \to L^2(0,b)##

    ##B(f(x)):=-i \frac{d f(x)}{dx}##

    is (an unbounded) self-adjoint operator.

    In fact there exist infinitely many different dense subsets ##D\subset L^2(0,b)## such that the "expression" ##-idf(x)/dx## give rise to a "different" unbounded self-adjoint operator.
     
  6. Mar 31, 2015 #5

    vanhees71

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    This is possible for the finite interval with periodic (or quasi-periodic) boundary conditions,
    $$\psi(x+L)=\exp(\mathrm{i} \varphi) \psi(x), \quad \varphi \in \mathbb{R},$$
    but not for the potential box with infinitely high walls, where the boundary conditions are
    $$\psi(x)=\psi(L)=0.$$
     
  7. Mar 31, 2015 #6
    Exactly.

    Yes. Mathematically you can define ##D(B):=\{ f\in L^2(0,b) : f \in AC(0,b), f'\in L^2(0,b), f(0)=f(b)\}## (or ## f(0) = c f(b)## for some given unit complex number c in the other cases).

    Yes, none of those domains (suitable to define ##-idf(x)/dx## as an unbounded self-adjoint operator) are included in the set ##\{ f\in L^2(0,b) : f(0)=0=f(b)\}##
     
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