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PhysKid24
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Is the second derivative with respect to position a hermitian operator? (i.e. d^2/dx^2)? Can anyone prove it? I don't think it is. Thanks
Edgardo said:Hello PhysKid24,
Hint: You already know that [itex]\hat p_x = \hat {p_x}^{\dagger} [/itex],
that is
[itex](\phi, \hat{p_x} \psi) = (\hat{p_x}^{\dagger} \phi, \psi) = (\hat{p_x} \phi, \psi)[/itex].
What do you mean? [itex]\langle \psi |\hat p^{\dagger}|\varphi \rangle = \langle \psi | \hat p|\varphi \rangle[/itex] for any vector in the state space and it does imply that the same holds for [itex]p^2[/itex].seratend said:The subset of functions where p is hermitian is not the same as the one where p^2 is hermitian . Therefore, p hermitian on a subset of functions =/=> p^2 hermitan on this subset. We just have a common subset of functions where p and p^2 are both hermitian.
Seratend.
Galileo said:What do you mean? [itex]\langle \psi |\hat p^{\dagger}|\varphi \rangle = \langle \psi | \hat p|\varphi \rangle[/itex] for any vector in the state space and it does imply that the same holds for [itex]p^2[/itex].
I do not understand what you mean.dextercioby said:You haven't shown the operator is at least hermitean on its domain.
Daniel.
Galileo said:Seratend. I read your post. When I mean vectors in the state space, I mean physically realizable states. At the very least they go to zero at infinity and are uniformly continuous. In that case, clearly:
[tex]\lim_{x \to \infty} f^*(x)f'(x)=0[/tex]
A Hermitian operator is a type of linear operator that has the property of being self-adjoint. This means that the operator is equal to its own adjoint, which is the complex conjugate of its transpose. In other words, the operator is symmetric about the main diagonal.
Hermitian operators are important in mathematics because they have real eigenvalues and orthogonal eigenvectors. This makes them useful in solving differential equations and studying quantum mechanics.
The d^2/dx^2 operator, also known as the second derivative operator, is a Hermitian operator. This means that it satisfies the properties of a Hermitian operator and can be used in the same way as any other Hermitian operator.
Yes, the proof for d^2/dx^2 being a Hermitian operator involves complex numbers and complex conjugates. It also involves the use of integration by parts and the properties of complex functions.
Yes, there are some exceptions to d^2/dx^2 being a Hermitian operator. For example, if the function being operated on is not defined on a closed interval, then the operator may not be Hermitian. Additionally, if the function is not continuous or differentiable, the operator may not be Hermitian.