In the context of information theory, entropy is a measure of uncertainty of a random variable. In quantum mechanics, the uncertainty principle states that $\Delta x\Delta k \ge 1/2$. The same can be expressed in terms of information entropy as $S_x + S_k \ge 1 + \ln\pi$. The information entropy $S_x$ is defined as:

$$ S_x = \int_{-\infty}^{\infty} {\rho(x) \ln\rho(x) dx} $$

where $\rho(x)$ is the probability density function of the variable $x$.

In continuation[1,2,3] of my study of the infinite square well problem, I'd like to check whether the informational version of uncertainty principle holds. For that, I need to calculate $S_x$ and $S_k$, for various numbers of $n=1,2,3,\ldots$, and check if their sum is bounded, knowing the respective probability density functions for position $x$ and momentum $k$.

Here is my code for $S_x$:

(* The length of the well *)
L = 1;

(* The eigenfunctions, n=1,2,3,... u[n,x] is zero outside of [0,L] *)
u[n_, x_] := Sqrt[2/L] Sin[n π x/L]

(* Probability density function for the x(=particle position) variable *)
(* Again, the domain of ρ(n,x) is the [0,L] interval *)
ρ[n_, x_] := u[n, x]\[Conjugate] u[n, x]

integrand = 
 Simplify[-ρ[n, x] Log[ρ[n, x]], 
  n ∈ Integers && x ∈ Reals]

(* Out= -2 Log[2 Sin[n π x]^2] Sin[n π x]^2 *)

(* Integrate over [0,L] since we haven't defined u[n,x] outside of [0,L] *)
(* We could have defined it though and then we would be integrating from -inf to +inf *)
Integrate[integrand, {x, 0, L}]

(* Out= -(1/(6 n π))(π (6 n - I π + 6 I n^2 π + n Log[64] - 
      12 n Log[1 - E^(2 I n π)] + 6 n Log[Sin[n π]^2]) + 
   6 I PolyLog[2, E^(2 I n π)] - 
   3 (-1 + Log[2 Sin[n π]^2]) Sin[2 n π]) *)

At this point it's already obvious that the value of the integral cannot be determined, since the terms $\sin(n\pi)$ equal zero and $\ln{0}$ is undefined.

FullSimplify[%, n ∈ Integers]

During evaluation of In[72]:= FullSimplify::infd: Expression Log[1-E^(2 I n [Pi])] simplified to -[Infinity]. >>

During evaluation of In[72]:= FullSimplify::infd: Expression Log[Sin[n [Pi]]^2] simplified to -[Infinity]. >>

During evaluation of In[72]:= FullSimplify::infd: Expression Log[2 Sin[n [Pi]]^2] simplified to -[Infinity]. >>

During evaluation of In[72]:= General::stop: Further output of FullSimplify::infd will be suppressed during this calculation. >>

Out[72]= Indeterminate

My book says that $S_x = \ln(2L) - 1$. Any ideas on how to trick Mathematica to calculate the integral ?

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Calculate integral for arbitrary parameter n in infinite square well problem

  • $\begingroup$ I commented out the Out= lines so that it's easier to copy-and-paste your code. $\endgroup$
    – user484
    Nov 10, 2013 at 20:50
  • $\begingroup$ Thanks @RahulNarain for taking care of that! $\endgroup$
    – stathisk
    Nov 10, 2013 at 20:50

2 Answers 2


I managed to teach Mathematica calculate the integral for arbitrary $n$, with a little aid:

$$\int_0^L\rho(n,x)\ln(\rho(n,x))dx = (2/L)\int_{0}^{L}\sin^2(n\pi x/L)\ln\left[(2/L)\sin^2(n\pi x/L)\right]dx$$

Mathematica has trouble, apparently, handling all the parameters $(n,L,x)$, so I resort to the following substitution: $n\pi x/L=u \Rightarrow (n\pi/L) dx = du$ and the new limits of integration are: $ 0 \rightarrow 0, L \rightarrow n\pi$.

Thus, now, the integral becomes:

$$\begin{align} \int_0^L\rho(n,x)\ln(\rho(n,x))dx &= (2/L) \int_0^{n\pi}\sin^2 u\ln\left[(2/L)\sin^2 u\right]\frac{L}{n\pi}du\\ &= \left(\frac{2}{n\pi}\right)\int_0^{n\pi}\sin^2 u\ln\left[(2/L)\sin^2 u\right]du \end{align}$$

The function $\sin^2 u\ln[(2/L)\sin^2 u$ is periodic with a period $T=\pi$:

f[u_] := Sin[u]^2 Log[(2/L) Sin[u]^2]
f[u + π] == f[u]
(* Out= True *)

As it can also be seen from its plot:

Plot[-Sin[u]^2 Log[Sin[u]^2], {u, 0, 4 Pi}, Filling -> Axis]

enter image description here


$$\begin{align} \int_0^L\rho(n,x)\ln(\rho(n,x))dx &= \left(\frac{2}{n\pi}\right)n \int_0^{\pi}\sin^2(u)\ln\left[(2/L)\sin^2 u\right]du \\ &= (2/\pi)\int_0^{\pi}\sin^2{u}\ln\left[(2/L)\sin^2 u\right]du \end{align}$$

-(2/π) Integrate[Sin[u]^2 Log[(2/L) Sin[u]^2], {u, 0, π}, 
  Assumptions -> L > 0]
(* Out= -1 + Log[2] + Log[L] *)
  • $\begingroup$ By the way, how could I get rid of those tiny integral symbols? Is there anything like \bigint or something ? $\endgroup$
    – stathisk
    Nov 11, 2013 at 19:26
  • 2
    $\begingroup$ If you put the math in display mode with $$…$$ instead of $…$, the integral signs will be full-sized. You'll probably want to use align for the multiline equations then. See meta.math.stackexchange.com/q/5020/856 and meta.math.stackexchange.com/a/5024/856 $\endgroup$
    – user484
    Nov 11, 2013 at 19:34

Here's a way to approach this by defining a function for each $n$, which you can then do separately.

L = 1;
u[n_, x_] := Sqrt[2/L] Sin[n π x/L];
ρ[n_, x_] := u[n, x]\[Conjugate] u[n, x];
integrand[n_, x_] := Simplify[-ρ[n, x] Log[ρ[n, x]], n ∈ Integers && x ∈ Reals]

Now calculate the desired integral:

Integrate[integrand[1, x], {x, 0, L}]
-1 + Log[2]

The $n=2$ and $n=3$ cases gives the same answer.

  • $\begingroup$ Thanks @bills. I was hoping to get M calculate it for any arbitrary n. But if impossible, then I will accept it. $\endgroup$
    – stathisk
    Nov 10, 2013 at 19:32

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