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Galileo

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That operator is basically [itex]\hat p_x^2[/itex], which we know to be Hermitian.

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Other sources:Ahiezer & Glazman:"The theory of Linear Operators in Hilbert Space".

J.Prugoveçki:"Quantum Mechanics in Hilbert Space".

Birman & Solomjak:"Spectral Theory of Self-Adjoint Operators in Hilbert Space".

Daniel.

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Galileo already gave you a clue. Use the definition of the hermitian operator:

[itex](\phi, \hat{A} \psi) = (\hat{A}^{\dagger} \phi, \psi)[/itex].

So what you have to show is:

[itex]\hat p_x^2 = (\hat p_x^2)^{\dagger} [/itex] ,

that is

[itex](\phi,\hat p_x^2 \psi) = (\hat p_x^2 \phi, \psi) [/itex] .

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].

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Useless,u'd still have to prove the differential operator is at least hermitean...

Daniel.

Daniel.

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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.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].

Seratend.

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Galileo

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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.

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No. P is an unbounded operator (i.e. a non conitnuous operator). In other words, p is not defined on most of the functions of L2(|R,dx) (or a vector of a given separable abstract hilbert space).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].

For example, if f(x) is a continuous derivable and integrable function (subset of L2) , that has a limit 0 as x --> +00, we have (in hbar units):

<f|p|f>= int_{R} if*(x)f'(x)dx= [f*(x)f(x)]_{-oO, +oO} -int_{R} if'*(x)f(x)dx = -int_{R} if'*(x)f(x)dx = <f|-id/dx|f>=<f|p+|f>

where p+ is the hermitian conjugate of p.

now, <f|p^2|f>= -int_{R} f*(x)f''(x)dx=-[f*(x)f'(x)]_{-oO, +oO} +int_{R} if'*(x)f'(x)dx= -[f*(x)f'(x)]_{-oO, +oO} + [f*(x)f(x)]_{-oO, +oO} - int_{R} if''*(x)f(x)dx

= -[f*(x)f'(x)]_{-oO, +oO} - int_{R} if''*(x)f(x)dx

therefore if lim_{x-->+/-oO} f*(x)f'(x)=/=0, p2 is not hermitian ((<f|p^2+)|f>=/= <f|(p^2|f>).

Seratend.

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I think you're diverging from the initial question.

Daniel.

Daniel.

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prego?

Seratend.

Seratend.

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You haven't shown the operator is at least hermitean on its domain.

Daniel.

Daniel.

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I do not understand what you mean.dextercioby said:You haven't shown the operator is at least hermitean on its domain.

Daniel.

I think I have demonstrated that p is an hermitan operator for the vector space of continuous derivable and L2 functions (i.e. the set where p=d/idx is the classical derivation). And that this vector space is not the L2 separable hilbert space.

And In addition, to get a hermitian operator on this subspace (the set where p^2=-d^2/dx^2), we need to restict these functions to the subspace where (lim_{x--> +/-oO}=f(x)f'(x)=a where a is any constant).

However, we may enlarge this domain (extending the classical derivation) but this is not the subject of this thread, I think.

Seratend.

Simply, not understanding what you mean.

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Galileo

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[tex]\lim_{x \to \infty} f^*(x)f'(x)=0[/tex]

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Ok, I understand you now (and for all practicall purposes it is ok).Galileo said:

[tex]\lim_{x \to \infty} f^*(x)f'(x)=0[/tex]

However, for the ones who are interrested, you cannot strictly say that all the physically realizable states (in the sense <psi|psi> <+oO) are uniformly continuous and vanishes to 0 at the infinity vicinity (even if it is sufficient for all practical purposes and with some density results of continuous functions) as this only a subclass of possible states.

We must not forget that all the limits we get are based mainly on the L2 norm and not the usual |x| norm of the limit symbol. In other words, there are lot of points in |R where "[itex]\lim_{x \to \infty} f^*(x)f'(x)\neq 0[/itex]".

Seratend.

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