I have a wavefunction ($a=1nm$):


for which I already calculated the normalisation factor (in my other topic):

$$A = \sqrt{\frac{2}{a\sqrt{\pi a}}} = 1.06\frac{1}{nm\sqrt{nm}}$$

What I want to know is how to calculate the expectation value for a kinetic energy. I have tried to calculate it analyticaly but i get lost in the integration:

\begin{align} \langle E_k \rangle &= \int\limits_{-\infty}^{\infty} \overline\psi\hat{T}\psi \,dx = \int\limits_{-\infty}^{\infty} Ax \exp \left[{-\tfrac{x^2}{2a}}\right]\left(-\tfrac{\hbar^2}{2m}\tfrac{d^2}{dx^2}Ax \exp \left[{-\tfrac{x^2}{2a}}\right]\right)\,dx =\dots \end{align}

At this point I go and solve the second derivative and will continue after this:

\begin{align} &\phantom{=}\tfrac{d^2}{dx^2}Ax \exp \left[{-\tfrac{x^2}{2a}}\right] = A\tfrac{d^2}{dx^2}x \exp \left[{-\tfrac{x^2}{2a}}\right]= A\tfrac{d}{dx}\left(\exp \left[{-\tfrac{x^2}{2a}}\right]-\tfrac{2x^2}{2a}\exp \left[{-\tfrac{x^2}{2a}}\right]\right)= \\ &=A \left(-\tfrac{2x}{2a}\exp \left[{-\tfrac{x^2}{2a}}\right] - \tfrac{1}{a}\tfrac{d}{dx}x^2\exp \left[{-\tfrac{x^2}{2a}}\right]\right) = \\ &=A \left(-\tfrac{x}{a}\exp \left[{-\tfrac{x^2}{2a}}\right] - \tfrac{2x}{a}\exp \left[{-\tfrac{x^2}{2a}}\right] + \tfrac{x^3}{a^2}\exp \left[{-\tfrac{x^2}{2a}}\right]\right) = \\ &= A \left(-\tfrac{3x}{a}\exp \left[{-\tfrac{x^2}{2a}}\right] + \tfrac{x^3}{a^2}\exp \left[{-\tfrac{x^2}{2a}}\right]\right) \end{align}

Ok so now I can continue the integration:

\begin{align} \dots &= \int\limits_{-\infty}^{\infty} Ax \exp \left[{-\tfrac{x^2}{2a}}\right]\left(-\tfrac{\hbar^2}{2m} A \left(-\tfrac{3x}{a}\exp \left[{-\tfrac{x^2}{2a}}\right] + \tfrac{x^3}{a^2}\exp \left[{-\tfrac{x^2}{2a}}\right]\right)\right)\,dx = \\ &= \int\limits_{-\infty}^{\infty} -\frac{A^2\hbar^2}{2m}x\exp\left[-\tfrac{x^2}{2a}\right] \left(-\tfrac{3x}{a}\exp \left[{-\tfrac{x^2}{2a}}\right] + \tfrac{x^3}{a^2}\exp \left[{-\tfrac{x^2}{2a}}\right]\right) \,dx\\ &= \int\limits_{-\infty}^{\infty} \frac{A^2\hbar^2}{2m}\left(\tfrac{3x^2}{a}\exp \left[{-\tfrac{x^2}{a}}\right] - \tfrac{x^4}{a^2}\exp \left[{-\tfrac{x^2}{a}}\right]\right) \,dx\\ &= \frac{A^2\hbar^2}{2m} \underbrace{\int\limits_{-\infty}^{\infty}\left(\tfrac{3x^2}{a}\exp \left[{-\tfrac{x^2}{a}}\right] - \tfrac{x^4}{a^2}\exp \left[{-\tfrac{x^2}{a}}\right]\right) \,dx}_{\text{How do i solve this?}}=\dots\\ \end{align}

This is the point where I admited to myself that I was lost in an integral and used the WolframAlpha to help myself. Well I got a weird result. My professor somehow got this ($m$ is a mass of an electron) but I don't know how:

\begin{align} \dots = \frac{\hbar^2}{2m}\cdot\frac{3}{2a} = \frac{3\hbar^2}{4ma} = 0.058eV \end{align}

Can anyone help me to understand the last integral? How can I solve it? Is it possible analyticaly (it looks like professor did it, but i am not sure about it)?

  • 5
    $\begingroup$ Wolfram/Mathematica definitely can do those integrals, but you missed some signs when entering the input ($e^{x^2/a}$ instead of $e^{-x^2/a}$). If you use Mathematica, it's good advice to teach Mathematica that $a > 0$, i.e. use "$Assumptions = {a>0}" for example. $\endgroup$
    – Vibert
    Aug 1, 2013 at 23:17
  • $\begingroup$ Now that i have fixed the input in Wolfram alpha i get a result $9/4\,\sqrt{\pi a}$ while $A^2=2/a\,\sqrt{\pi a}$ and so I still get a different result than my prof: $\dfrac{A^2\hbar^2}{2m}9/4\,\sqrt{\pi a} = \dfrac{2\hbar^2}{a\sqrt{\pi a}2m}9/4\,\sqrt{\pi a} = \frac{9\hbar^2}{4a m}$ $\endgroup$
    – 71GA
    Aug 1, 2013 at 23:56

2 Answers 2


In your problem, you need integrals of kind :

$I_{2n} = \int x^{2n} e^{- \large \frac{x^2}{a}} ~ dx$

Note first that $I_0 = (\pi)^\frac{1}{2} (\frac{1}{a})^ {-\frac{1}{2}}$

Now, it is easy to see that there is a reccurence relation between the integrals :

$$I_{2n+2} = - \frac{\partial I_{2n}}{\partial (\frac{1}{a}) } $$

For instance, $$I_2 = - \frac{\partial I_{0}}{\partial (\frac{1}{a}) } = \frac{1}{2}(\pi)^\frac{1}{2} (\frac{1}{a})^ {- \large\frac{3}{2}} = \frac{1}{2}(\pi)^\frac{1}{2} ~a^ {\large\frac{3}{2}}$$

$$I_4 = - \frac{\partial I_{2}}{\partial (\frac{1}{a}) } = \frac{3}{2} \frac{1}{2}(\pi)^\frac{1}{2} (\frac{1}{a})^ {- \large\frac{5}{2}} = \frac{3}{2} \frac{1}{2}(\pi)^\frac{1}{2} ~a^ {\large\frac{5}{2}}$$

A general formula is :

$$I_{2n} = I_0 ~(2n-1)!! ~(\frac{a}{2})^n = \frac{(\pi)^\frac{1}{2}}{2^n} ~(2n-1)!! ~a^{n+\frac{1}{2}}$$ where $(2n-1)!! = (2n-1)(2n-3)......5.3.1$


My statistical physics professor call those ones Laplace integrals $I(h)$.


Note that

$$\int_{-\infty}^{\infty}x^{h}e^{-a^2x^2}dx=2I(h) $$

some values

$$I(0)=\frac{\sqrt{\pi}}{2a}, I(1)=\frac{1}{2a^2}, I(2)=\frac{\sqrt{\pi}}{4a^3},I(3)=\frac{1}{2a^4}, I(4)=\frac{3\sqrt{\pi}}{8a^5} $$

You may brute force by integrating by parts to get rid of $x^{h}$ and use $I(0) $ a classical result, or you may use induction over $h$ or some other method.

  • $\begingroup$ So you think i can solve the integral by per partes? $\endgroup$
    – 71GA
    Aug 2, 2013 at 7:28
  • $\begingroup$ Yes, you only have to know $I(0)$ and decrease the term $x^{h}$ until you get $I(0)$, and of course symmetries such as the integral of an odd function in a symmetric interval is zero and so on. $\endgroup$
    – J L
    Aug 2, 2013 at 7:52

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