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How to prove that d^2/dx^2 is a hermitian operator?

by alsey42147
Tags: d2 or dx2, hermitian, operator, prove
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alsey42147
#1
Jun1-06, 01:04 PM
P: 22
would anyone mind showing me, for example, how to prove that d^2/dx^2 is a hermitian operator? i've tried to work it out from two different books; they both prove that the momentum operator is hermitian, but when i try to apply the same thing to the operator d^2/dx^2 i get lost pretty quick. thanks in advance.

also, does anyone know a book on QM written from a not so mathematical perspective? right now i don't really care for rigorous mathematical proof, i just want to know how to do things and why i am doing them. all this stuff about operators seems very arbitrary (i know that its not, but thats how it appears with my current level of understanding). right now my QM course just seems like i'm learning random mathematical proofs. like someone trying to learn newtonian mechanics without knowing what a body is.
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Perturbation
#2
Jun1-06, 01:24 PM
P: 124
An operator A is Hermitian (or self adjoint) if

[tex]\langle\psi |A|\phi\rangle =(\langle\phi |A|\psi\rangle )^*[/tex]

I.e. in one dimension, as is relevant to you, [tex]\int\psi^* A\phi dx=\left(\int\phi^* A\psi dx\right)^*[/tex]

So just integrate it by parts a couple of times and impose boundary conditions to [itex]\phi[/itex] and [itex]\psi[/itex] so that they vanish at the limits of integration. If the equality holds then your your differential operator will be self adjoint, which it will for your operator. The boundary conditions then mean that the domain upon which A is Hermitian is that subspace of vectors that obey the boundary conditions.

If something is Hermitian then it means that I can swap between acting on the left and right in an inner product if the vectors in the inner product have the right boundary conditions, as above.

In actual physical terms Hermitian operators correspond to observables, i.e. information we get from experiment or by measurement. The choice of Hermitian operators is because their eigenvalues are real, and this fits best with our experience of physical things having real quantities associated with them, sometimes their nondegenerate eigenstates being orthogonal can be useful.
alsey42147
#3
Jun1-06, 01:35 PM
P: 22
thanks, that was simpler than i thought. i was only integrating by parts once.

Perturbation
#4
Jun1-06, 01:57 PM
P: 124
How to prove that d^2/dx^2 is a hermitian operator?

What book are you using?
lalbatros
#5
Jun1-06, 04:03 PM
P: 1,235
Maybe you could enjoy to proof that the square of an hermitian operator is hermitian too.
And some generalisations ...
alsey42147
#6
Jun2-06, 05:46 AM
P: 22
Quote Quote by Perturbation
What book are you using?
mainly Molecular Quantum Mechanics, P Atkins and R Friedman. i can follow it well enough, i just don't know what any of it 'means'.
nitin
#7
Jun13-06, 10:54 AM
P: 35
I have been thinking about putting up something about the (mathematical) requirement in Quantum Mechanics that physical observables be represented in the formalism by Hermitian operators. I understand that Hermiticity guarantees the real-valuedness of the eigenvalues (hence the measured quantities corresponding to these eigenvalues), and the conservation of probabilities (or unitary evolution). However, I believe that real-valuedness is not a necessity, and that this supposed requirement restricts the number of Hamiltonians that we could normally work with. For example, we only need the Hamiltonian to be PT-symmetric, and with well-defined boundary conditions, we can easily show that the eigenvalues are real and the evolution unitary.

I believe that a PT-symmetric Hamiltonian is more a physical requirement as compared to what appears to be a purely mathematical Hermiticity requirement. Now, this has been known for some time, but never have I been told in my undergraduate QM courses that we can forego Hermiticity (I guess the lecturer had his reservations about it, but then I don't see any problem with it). Has anybody heard about it before?

Paper: Must a Hamiltonian be Hermitian? Bender, Carl M., Brody, Dorje C., and Jones, Hugh F.

By the way, I'm reading Penrose's "The Road to Reality", where Penrose tells us about his similar reservations about Hermiticity (page 539, Jonathan Cape Edition). I quote:

"In my opinion, this Hermitian requirement on an observable Q is an unreasonably strong requirement, since complex numbers are frequently used in classical physics, such as for the Riemann sphere representation of the celestial sphere, and in many standard discussions of the harmonic oscillator, etc."

According to him, an "essential requirement of an observable is that its eigenvectors, corresponding to distinct eigenvalues, are orthogonal to one another".

Maybe some people have the chance here to talk about this to their lecturers or friends...
Son Goku
#8
Jun13-06, 07:31 PM
P: 89
This could be wrong..............

Anyway using the bra ket notation:
Let's label the second order derivative operator as [tex]D^2[/tex].

It will operate on an arbitrary ket [tex]\left|f\rangle[/tex] and produce the ket
[tex]\left|\frac{d^2f}{dx^2}\rangle[/tex].

So:
[tex] D^2|f\rangle = \left|\frac{d^2f}{dx^2}\rangle[/tex]

Now finding the projection of both sides along the position eigenkets:
[tex] \langle x|D^2|f\rangle = \langle x|\frac{d^2f}{dx^2}\rangle[/tex]

The right hand side is simply:
[tex]\frac{d^2f}{dx^2}[/tex]

So:
[tex] \langle x|D^2|f\rangle = \frac{d^2f}{dx^2}[/tex]

I'll now introduce the dummy ket [tex]\left|x'\rangle[/tex] to sum over and using the resolution of identity:

[tex]\int \langle x|D^2|x'\rangle \langle x'|f\rangle dx'= \frac{d^2f}{dx^2}[/tex]

Using the fact that [tex]\langle x'|f\rangle = f(x')[/tex]

We get:
[tex]\int \langle x|D^2|x'\rangle f(x') dx'= \frac{d^2f}{dx^2}[/tex]

Knowing the properties of the dirac equation we see from the above that:
[tex]\langle x|D^2|x'\rangle = D^2_{xx'} = \delta(x-x')\delta(x-x')\frac{d^2}{dx'^2}[/tex]

So now we have the components of the operator:
[tex]D^2_{xx'} = \delta(x-x')\delta(x-x')\frac{d^2}{dx'^2}[/tex]

Taking the complex conjugate:
[tex]D^2_{xx'}^{*} = \delta(x-x')\delta(x-x')\frac{d^2}{dx'^2}[/tex]

And now taking the transpose:
[tex]D^2_{x'x}^{*} = D^2_{xx'}^{\dagger} = \delta(x'-x)\delta(x'-x)\frac{d^2}{dx'^2} = -\delta(x-x')-\delta(x-x')\frac{d^2}{dx'^2} = \delta(x-x')\delta(x-x')\frac{d^2}{dx'^2}[/tex]

Which is the same as [tex]D^2_{xx'}[/tex]

So [tex]D^2_{xx'} = D^2_{xx'}^{\dagger}[/tex]

then [tex]D^2 = D^2^{\dagger}[/tex]

Which is the definition of Hermitian.
reilly
#9
Jun15-06, 05:30 PM
Sci Advisor
P: 1,082
Quite sometime ago, Tulio Regge spearheaded research that was based on complex angular momenta -- basically by analytic continuation of standard associated legendre function expsions, as was done much earlier by Arnold Sommerfeld in connection with computing radiation patterns of grounded antenna on a "reflecting earth", So called Regge poles are a staple of particle physics -- this stuff is very complicated, but I imagine Google and Wikopedia can shed some light. Also, one could argue that the so-called Wick rotation involves a formally non-hermitian Hamiltonian operator. Also, look up the Optical potential in scattering theory.

In general, what would you really do with non-Hermitian operators, particularly with the awkward boundary values that an happen ?

Regards,
Reilly Atkinson


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