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Operators: Do they carry units?

  1. Oct 27, 2008 #1
    1. The problem statement, all variables and given/known data
    Hi all.

    The title says it all: Say we have an operator - e.g. the Hamiltonian. Does this have units? I.e. does the Hamiltonian have units of Joules or nothing?

    Personally I think it is nothing, since it is an operator, but I need confirmation.
     
  2. jcsd
  3. Oct 27, 2008 #2
    When you "apply" the operator to a wavefunction you "get" (some) energy . If the operator has no units then the wavefunction must have units of energy. But if you apply another operator to the same function you'll get another physical quantity... so the wave function cannot have units... the operator must came with the units "in it" :smile:

    You don't need this "deduction" actually. Just look at the expression for Hamiltonian for example. It has Plank's constant and mass and length...all these have units
     
  4. Oct 27, 2008 #3
    The wavefunction \psi(x) is NOT dimensionless, it does have units. In 1-D, when you integrate the absolute value squared of \psi you get 1: so, the dimension of the wavefunction must be 1/sqrt[length]. In 3-D, it similarly must have units of 1/[length]^(3/2).

    Operators have units indeed. The momentum operator has units of momentum, etc. etc.
     
  5. Oct 27, 2008 #4
    You are right, this statement is not true and it does not follow from the previous arguments.
    Thank you.
     
  6. Oct 27, 2008 #5

    gabbagabbahey

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    The wave function only has units when it is expanded into a basis; [itex]\psi(x)=\left<x | \psi \right>[/itex] has the units given by Borgwai, but the abstract wave function [itex]|\psi \rangle [/itex] is generally treated as unitless.
     
  7. Oct 27, 2008 #6
    I'd refer to the ket [itex]|\psi \rangle [/itex] as a state vector, not a wave function (just semantics).

    But indeed, [itex]|\psi \rangle [/itex] can be viewed as dimensionless, whereas [itex]|x \rangle [/itex] has the dimension of [length]^(-1/2). This is the natural choice. I don't think there's a necessity of choosing the dual of a ket to have the same dimension as the ket itself: expressions whose dimension is clear always seem to contain both a ket and a bra (I'll be happy to see if someone knows a counter example!).
     
  8. Oct 28, 2008 #7
    Thanks borgwal

    Another way to realize this (I just thought about this) is to look at Schrödingers time-independent equation: H \psi = E \psi.
     
  9. Oct 28, 2008 #8
    Yes, I think that's what nasu meant also: the units of H follow from H \psi=E\psi.
     
  10. Oct 28, 2008 #9

    D H

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    As others have noted, of course they do. Even quantum mechanics is subject to dimensional analysis and had dang well be dimensionally correct. If we weren't careful about units, we physicists would be little different from (shudder) mathematicians. (D H quickly beats a hasty retreat to don a flame-retardant asbestos suit)
     
  11. Mar 3, 2009 #10
    [itex]
    \sqrt\{\f(x)}= \sqrt\{\g(x)}
    [/itex]
     
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