A wave-function can be written as Ψ=Ae−i(Et−px)/ℏ
where E & p are the energy & momentum of the particle. Now, differentiating Ψ w.r.t. x and t respectively, we get ∂Ψ∂x=iℏpΨ∂Ψ∂t=−iℏEΨ.The above equations can be written in suggestive forms pΨ=(ℏi∂∂x)ΨEΨ=(iℏ∂∂t)Ψ.Evidently the dynamical quantities momentum and energy are equivalent to the operators p=ℏi∂∂xE=iℏ∂∂t.
quoted from Arthur Beiser's Concept of Modern Physics
How can observable quantities be equal to operators? You can't measure an "operator". Can anyone intuitively explain how momentum and energy are equal to operators?
Answer
Not equal, but equivalent, in the sense that they have the same effect on the wavefunction in question.
More precisely, the book is using a slight abuse of terminology. Taking momentum as an example, it's not really the case that the dynamical quantity of momentum is equivalent to the operator ℏi∂∂x, because a number can't actually be equivalent to an operator. A number is a thing all on its own, whereas an operator is something that needs to be applied to something else to have any meaning. But the operation of multiplying the wavefunction by the wave's momentum (a number) is equivalent to the operation of taking the derivative and multiplying by ℏi. In mathematical language: ℏi∂∂xψ=pψ
What we normally do in quantum physics is always think about quantities as operators. That is, if you write just p in an equation, it's implicitly understood that this should be applied to some wavefunction. For example, if you write H=p22m
No comments:
Post a Comment