# Kinetic Energy of wavefunction

Suppose we have been provided the form of some wavefunction on a graph, but not the exact mathematical expression of the wavefunction $$\langle x|\psi\rangle$$. Now I'm asked to find the average kinetic energy or the expectation value of the momentum by analyzing nothing except the figure that is given to me. How do I approach problems like this in general.

For example, suppose we have a symmetric wavefunction like this :

We have to find the expectation value of Kinetic energy, or rather the average kinetic energy.

Now my guess is that, since the wavefunction is a constant between $$-a$$ to $$a$$ , the first derivative there will be zero and between $$[-(a+b),-a]$$ and $$[(a+b),a]$$, it's $$\psi(x)=mx$$. Hence the second derivative must vanish as well, the kinetic energy which has the double derivative of the wavefunction inside the integral must therefore be zero. But I feel like I am missing something.

The given answer is $$T=\frac{3\hbar^2}{2mb(3a+b)}.$$

Any help on how to approach this problem would be highly appreciated.

You can start by finding a suitable expression for the function.

$$\psi(x)=\begin{cases} k & |x|\lt a \\ \frac{k}{b}(x+a+b) & -a\lt x\lt -(a+b) \\ \frac{k}{b}(a+b-x) & a\lt x\lt (a+b) \end{cases}$$

Then you need to normalize the wavefunction and find the value of $$k$$. Do it by integrating region by region, in a piecewise manner. You should get $$k^2=\frac{3}{2(3a+b)}$$

Now you need to find $$\langle\psi|\frac{p_{x}^2}{2m}|\psi\rangle$$

This is equivalent to :

$$\frac{-\hbar^2}{2m}\int_{-\infty}^{\infty}\psi^*(x)\frac{\partial^2\psi(x)}{\partial x^2}dx$$

From here, you have two options. You note that, using integration by parts :

$$\int_{-\infty}^{\infty}\psi^*(x)\frac{\partial^2\psi(x)}{\partial x^2}dx=\psi(x)\int\frac{\partial^2\psi(x)}{\partial x^2}dx|_{-\infty}^{\infty}\space -\int_{-\infty}^{\infty}\frac{\partial\psi(x)}{\partial x} (\int_{-\infty}^{\infty}\frac{\partial^2\psi(x)}{\partial x^2}dx)dx$$

The first term on the RHS goes to $$0$$, as $$\psi(x)\rightarrow 0$$ at both the infinities. Hence, you have :

$$\frac{-\hbar^2}{2m}\int_{-\infty}^{\infty}\psi^*(x)\frac{\partial^2\psi(x)}{\partial x^2}dx=\frac{-\hbar^2}{2m}\int_{-\infty}^{\infty}(\frac{\partial\psi(x)}{\partial x})^2dx$$

You can easily calculate $$(\frac{\partial \psi(x)}{\partial x})^2$$ in these intervals. After integrating, you should get $$\frac{\hbar^2 k^2}{mb}$$.

Plugging the value of $$k$$ that you got from normalization, you should get the desired expression.

Another alternative is to note that, the double derivative of a 'kink' is a Dirac delta function. Hence, you should have :

$$\frac{\partial^2\psi(x)}{\partial x^2}=-\frac{k}{b}(\delta(x+a)+\delta(x-a))$$

You can plug this into the integral, and get the same thing out, by noting $$\psi(a)=\psi(-a)=k$$

You can solve the integral in this way too, and reach the final expression.

This is one of the classic subleties of QM where one needs to know whether the wavefunction is in the domain of the unbounded Hamiltonian operator.

A readable discussion is Francois Gieres Mathematical surprises and Dirac’s formalism in quantum mechanics See in particular Example 7 on page 8 of his account. There he displays a function whose fourth derivative is identically zero , but the expectaion of $$H^2\propto \partial^4_x$$ is not zero. The explanation is that the wave function is not in the domain of $$H^2$$. In your case, your wavefunction is not in the domain of $$\partial^2$$.

• I think the two edges at $-a$ and $a$ would give rise to two delta functions if one takes the double derivative. The first derivative of a function at a 'pointed' edge is a discontinuity, and the derivative of the discontinuity is the delta function. I suppose that is the source of the kinetic energy not being $0$ Commented Nov 9, 2021 at 14:53
• Why do you need the second derivative? The momentum operator contains the first derivative and the KE goes like the square of momentum, not like the second derivative. All that question asks is a simple application of the method to calculate average quantities when the wave function is given.
– nasu
Commented Nov 9, 2021 at 16:28
• As @Nakshatra Gangopadhay says the Hamiltonian requires second derivative, and hence delta functions. Delta's are not squre integrable so the given function is not in the domain of the Hamiltonain. Commented Nov 9, 2021 at 18:19