Quantum Mechanics, Hermitian Operators and Square Integrable Functions

June 5, 2012

Quantum Mechanics, Hermitian Operators and Square Integrable Functions

Filed under: advanced mathematics, analysis, applied mathematics, calculus, improper integrals, integrals, quantum mechanics, science — collegemathteaching @ 8:36 pm

In one dimensional quantum mechanics, the state vectors are taken from the Hilbert space of complex valued “square integrable” functions, and the observables correspond to the so-called “Hermitian operators”. That is, if we let the state vectors be represented by $\psi(x) = f(x) + ig(x)$ and we say $\psi \cdot \phi = \int^{\infty}_{-\infty} \overline{\psi} \phi dx$ where the overline decoration denotes complex conjugation.

The state vectors are said to be “square integrable” which means, strictly speaking, that $\int^{\infty}_{-\infty} \overline{\psi}\psi dx$ is finite.
However, there is another hidden assumption beyond the integral existing and being defined and finite. See if you can spot the assumption in the following remarks:

Suppose we wish to show that the operator $\frac{d^2}{dx^2}$ is Hermitian. To do that we’d have to show that:
$\int^{\infty}_{-\infty} \overline{\frac{d^2}{dx^2}\phi} \psi dx = \int^{\infty}_{-\infty} \overline{\phi}\frac{d^2}{dx^2}\psi dx$ . This doesn’t seem too hard to do at first, if we use integration by parts:
$\int^{\infty}_{-\infty} \overline{\frac{d^2}{dx^2}\phi} \psi dx = [\overline{\frac{d}{dx}\phi} \psi]^{\infty}_{-\infty} - \int^{\infty}_{-\infty}\overline{\frac{d}{dx}\phi} \frac{d}{dx}\psi dx$ . Now because the functions are square integrable, the $[\overline{\frac{d}{dx}\phi} \psi]^{\infty}_{-\infty}$ term is zero (the functions must go to zero as $x$ tends to infinity) and so we have: $\int^{\infty}_{-\infty} \overline{\frac{d^2}{dx^2}\phi} \psi dx = - \int^{\infty}_{-\infty}\overline{\frac{d}{dx}\phi} \frac{d}{dx}\psi dx$ . Now we use integration by parts again:
$- \int^{\infty}_{-\infty}\overline{\frac{d}{dx}\phi} \frac{d}{dx}\psi dx = -[\overline{\phi} \frac{d}{dx}\psi]^{\infty}_{-\infty} + \int^{\infty}_{-\infty} \overline{\phi}\frac{d^2}{dx^2} \psi dx$ which is what we wanted to show.

Now did you catch the “hidden assumption”?

Here it is: it is possible for a function $\psi$ to be square integrable but to be unbounded!

If you wish to work this out for yourself, here is a hint: imagine a rectangle with height $2^{k}$ and base of width $\frac{1}{2^{3k}}$ . Let $f$ be a function whose graph is a constant function of height $2^{k}$ for $x \in [k - \frac{1}{2^{3k+1}}, k + \frac{1}{2^{3k+1}}]$ for all positive integers $k$ and zero elsewhere. Then $f^2$ has height $2^{2k}$ over all of those intervals which means that the area enclosed by each rectangle (tall, but thin rectangles) is $\frac{1}{2^k}$ . Hence $\int^{\infty}_{-\infty} f^2 dx = \frac{1}{2} + \frac{1}{4} + ...\frac{1}{2^k} +.... = \frac{1}{1-\frac{1}{2}} - 1 = 1$ . $f$ is certainly square integrable but is unbounded!

It is easy to make $f$ into a continuous function; merely smooth by a bump function whose graph stays in the tall, thin rectangles. Hence $f$ can be made to be as smooth as desired.

So, mathematically speaking, to make these sorts of results work, we must make the assumption that $lim_{x \rightarrow \infty} \psi(x) = 0$ and add that to the “square integrable” assumption.

Leave a Comment »

No comments yet.

RSS feed for comments on this post. TrackBack URI

College Math Teaching

June 5, 2012