Show that the Hamiltonian operator is Hermitian

[JD_PM]

Homework Statement

Show that the Hamiltonian operator $$\hat H = -\frac{\hbar}{2m} \frac{d^2}{dx^2} + V(x)$$ is hermitian

Relevant Equations

$<f|\hat H g> = <\hat H f|g>$

$$<f|\hat H g> = \int_{-\infty}^{\infty} f^*\Big(-\frac{\hbar}{2m} \frac{d^2}{dx^2} + V(x) \Big) g dx$$

Integrating (twice) by parts and assuming the potential term is real (AKA $V(x) = V^*(x)$) we get

$$<f|\hat H g> = -\frac{\hbar}{2m} \Big( f^* \frac{dg}{dx}|_{-\infty}^{\infty} - \frac{df^*}{dx}|_{-\infty}^{\infty} g + \int_{-\infty}^{\infty} \frac{d^2 f}{dx^2}g dx \Big) + \int_{-\infty}^{\infty} V^*(x) f^* g dx $$

In order to get the desired I had to assume that

$$f^* \frac{dg}{dx}|_{-\infty}^{\infty} = 0$$

$$\frac{df^*}{dx}|_{-\infty}^{\infty} g = 0$$

Then we get

$$<f|\hat H g> = -\frac{\hbar}{2m} \int_{-\infty}^{\infty} \frac{d^2 f}{dx^2}g dx + \int_{-\infty}^{\infty} V^*(x) f^* g dx = <\hat H f|g>$$

Checking the solution, they say that these terms indeed vanish 'because both f and g live on Hilbert space'.

But what property of Hilbert space makes this true?

Thanks.

 

[Jaime_mc2]

I would better say that, if $f$ and $g$ are valid wavefunctions, then both belong to the complex space $L_2(\mathbb{R})$, which is a Hilbert space if we consider the scalar product defined as
$$
\langle f | g \rangle = \int_{-\infty}^{\infty} f^*(x)g(x)\ dx\ .
$$

By definition, the complex space $L_2(\mathbb{R})$ is the set of all functions $f$ such that
$$
\int_{-\infty}^{\infty} |f(x)|^2\ dx < +\infty\ ,
$$
and this can only be true if $f$ vanishes at $+\infty$ and $-\infty$. This is why you get that those terms are null when they are evaluated at infinity.

 

[PeroK]

In order to get the desired I had to assume that

$$f^* \frac{dg}{dx}|_{-\infty}^{\infty} = 0$$

$$\frac{df^*}{dx}|_{-\infty}^{\infty} g = 0$$But what property of Hilbert space makes this true?

Thanks.

In general, those terms do not necessarily vanish for square integrable functions. You need a stronger condition here. The argument is that any physical wave-functions will obey these conditions and functions that do not are "pathological" and unphysical.

To quote Griffiths: any decent maths student can furnish you with a counterexample.

PS more fundamentally, the differential operator is not closed acting on $L_2(\mathbb R)$. A counterexample is:
$$f(x) = \frac 1 x \sin(x^3)$$
Which is square integrable on $[1, \infty)$, but its derivative is not.

 

[JD_PM]

In general, those terms do not necessarily vanish for square integrable functions. You need a stronger condition here. The argument is that any physical wave-functions will obey these conditions and functions that do not are "pathological" and unphysical.

To quote Griffiths: any decent maths student can furnish you with a counterexample.

Griffiths indeed states that 'there exist pathological functions that are square-integrable but do not go to zero at infinity'.

He suggests that if we are worried about that issue we simply 'restrict the domain of our operators to exclude them'. What does he mean, that we should not integrate over the whole infinity line?

He also states that such problematic functions do not arise in Physics; but why?

 

[PeroK]

Griffiths indeed states that 'there exist pathological functions that are square-integrable but do not go to zero at infinity'.

He suggests that if we are worried about that issue we simply 'restrict the domain of our operators to exclude them'. What does he mean, that we should not integrate over the whole infinity line?

He also states that such problematic functions do not arise in Physics; but why?

They must all go to zero at infinity. That's true for all square integrable functions. (Although you can mess around with having a sequence of discontinuities.)

What these functions have in common generally is unboundedness of derivatives, especially as the x-coordinate tends to infinity. Remember that in many ways real things are finite and considering a function defined on $(-\infty, \infty)$ is something of an approximation. It can't really go on for ever physically. What you can't have physically is a function that behaves more and more extremely as you go further away from the centre of the action. That's what makes them unphysical. This is true for the example I gave.

The restriction is on the set of functions you consider. One simple idea is that they must go to zero faster than any power of $x$. I.e. eventually they must go to zero exponentially. Otherwise, your physical system is not contained.

 

[vanhees71]

I'd not consult Griffiths's QM textbook on such subtle issues. He is pretty sloppy in the foundations and mathematics. For a nice didactical introduction into these problems, which you can summarize to the conclusion that an operator that should represent an observable should not only be "Hermitian" but must even be "essentially self-adjoint", see

https://arxiv.org/abs/quant-ph/9907069
https://arxiv.org/abs/quant-ph/0103153

 

[PeroK]

@JD_PM I remembered there was a thread recently on a similar issue. The position operator also may map a square integrable function to one that is not. See the concept of the nuclear space of functions here:

https://www.physicsforums.com/threads/nuclear-space-in-qm.980977/

 

[JD_PM]

Thank you all, I now have a better understanding :)

I'd not consult Griffiths's QM textbook on such subtle issues. He is pretty sloppy in the foundations and mathematics. For a nice didactical introduction into these problems, which you can summarize to the conclusion that an operator that should represent an observable should not only be "Hermitian" but must even be "essentially self-adjoint", see

https://arxiv.org/abs/quant-ph/9907069
https://arxiv.org/abs/quant-ph/0103153

Thank you for sharing these links vanhees71 :)

I am actually reviewing Griffiths' chapter 3 (formalism in QM), which contains:

3.1 Introduction to Hilbert space.

3.2 Observables.

3.3 Eigenfunctions of a Hermitian operator.

3.4 Generalized statistical interpretation.

3.5 Heisenberg's uncertainty principle.

3.6 Dirac notation.

I will complement both PDFs with Griffiths.