College Math Teaching

June 5, 2012

Quantum Mechanics, Hermitian Operators and Square Integrable Functions

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.

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