# Question about the hermiticity of the momentum operator

by neelakash
Tags: hermiticity, momentum, operator
 P: 511 Is momentum operator $$\frac{\hbar}{i}\frac{\partial}{\partial\ x }$$ is Hermitian only for a normalized wave function?What is the case for the box normalization as done for a free particle? Actually when we prove the Hermiticity of the momentum operator, we do simple by parts integartion and use the scalar products.I never bothered about whether the wave function is normalized or not. Can anyone suggest anything?
 Sci Advisor HW Helper P: 4,738 http://mathworld.wolfram.com/HermitianOperator.html See eq (3) Thats how you can do it. You can use any wave function.
 P: 511 So, hermiticity of the momentum operator is independent of whether the wave function is normalized or not----right? I faced this question in university question book and got astonished.They asked to prove this and also asked what would it be if the wave function is box-normalized.
 Sci Advisor HW Helper P: 4,738 Question about the hermiticity of the momentum operator It should not matter. Look at eq (13) and below, never is it stated that the wave function is normalized. Normalized means that = 1, so it is just some multiplicatible constants that is added. And constants never affects what your operator does with the wave function. A is an operator, b is constant: [A,b] = 0, they commute. etc.
P: 173
 Quote by neelakash Is momentum operator $$\frac{\hbar}{i}\frac{\partial}{\partial\ x }$$ is Hermitian only for a normalized wave function?What is the case for the box normalization as done for a free particle? Actually when we prove the Hermiticity of the momentum operator, we do simple by parts integartion and use the scalar products.I never bothered about whether the wave function is normalized or not. Can anyone suggest anything?
Actually the analysis on P must be carry out in the differnet configurations scheme. I mean that its hermitianity depend upon its domain D(P).
In fact mathematically speaking what is really important is the couple (A,D(A))
to analize all the spectrum of an operator. In some circumstances can happen that P does posses residue spectrum.
Obviously they have no physical meaning.

Normalization on a function doesn't mean anything. f(x)=100g(x) its ok either.
regards
marco