Let the potential be

$$V = \infty \hspace{3cm}(0>x, x>L)$$ $$V = 0 \hspace{3.7cm}(L>x>0).$$

Now, we measure the position of a particle and discover it is located at $L/4$. What is the probability of finding the particle in each eigenstates of the energy?

So, i guess the wavefunction is $$\psi = \delta(x-L/4)$$

We know the eigenbasis is given by $$\sqrt{\frac{2}{L}} \sin(n \pi x/L).$$


$$\delta(x-L/4) = \sum_n c_n \sqrt{\frac{2}{L}} \sin(n \pi x/L) \\ c_n = \sqrt{\frac{2}{L}} \int \delta(x-L/4) \sin(n \pi x /L)dx = \sqrt{\frac{2}{L}} \sin(\frac{\pi n}{4})$$

But, $\sum |c_n|^2 $ diverges!

So how do i define the probability (generally given by $P_n = \frac{|c_n|^2}{\sum |c_n|^2}$?


1 Answer 1


Lots of things are wrong with $\psi=\delta(x-L/4)$.

First, you cannot normalize it since $$ \int_0^L \vert \psi\vert^2 dx = \int_0^L dx \delta(x-L/4)^2 = \delta(0) $$ is technically infinite.

Next, on dimensional grounds $\psi=\delta(x-L/4)$ does not work. Since $$ \int dx \vert \psi\vert^2 $$ is a probability and thus a dimensionless number, $\psi$ should have dimension of 1/(length)$^{1/2}$, so you $\psi$ does not have the correct dimension.

Another way to see the same thing is that your expansion coefficients $c_n$ do actually have the dimension of (length)$^{-1/2}$ so their modulus square cannot be interpreted as a (dimensionless) probability of finding your initial state in an energy eigenstate.

Sooooo… Your wavefunction $\psi$ is incorrect.

You may want to try instead a normalized Gaussian initial state that is very strongly peaked $x=L/4$, and then take the limit where the width of the Gaussian goes to $0$. Note that such a Gaussian would only approximately satisfy the boundary condition of the problem since the tail of the Gaussian would extend beyond the well, but you might be able to recover something meaningful in the limit of zero width.

Edit: Following the comments of @MichaelSeifert and others, I did some additional work, using $$ \psi(x)=\left\{\begin{array}{ll}1/\sqrt{\epsilon}& \text{if } 1/4-\epsilon/2 < x< 1/4+\epsilon/2\, ,\\ 0&\text{otherwise} \end{array}\right. $$ with $L=1$ for simplicity, and hoping to recover something in the limit where $\epsilon\to 0$ and sharply peaked $\psi$.

It is possible to obtain the expansion coefficients $$ c_n(\epsilon)=\frac{2 \sqrt{2} \sin \left(\frac{\pi n}{4}\right) \sin \left(\frac{\pi n \epsilon }{2}\right)}{\pi n \sqrt{\epsilon }}\, . $$ Expanding the $\sin(n\pi \epsilon/2)$ term yields $$ \vert c_n(\epsilon)\vert^2 = \frac{1}{180} \epsilon \left( 360-30 \pi ^2 n^2 \epsilon ^2+\pi ^4 n^4 \epsilon ^4+\ldots \right) \sin ^2\left(\frac{\pi n}{4}\right)\, . $$ You can easily see how we get in trouble: for any given $\epsilon>0$, there is $n_0$ so that, for any $n>n_0$, $n^2\epsilon^2\pi^2>1$ and the series for $\vert c_n(\epsilon)\vert^2$ stops to make sense as an expansion.

The interesting part is that for finite $\epsilon$, the values of $\vert c_n(\epsilon)\vert^2$ for large $n$, which apparently cause trouble in the expansion of $\vert c_n(\epsilon)\vert^2$, become numerically very small. This is clear from the expression for $c_n(\epsilon)$: for finite $\epsilon$ $c_n(\epsilon)$ scales like $1/n$, and since $\vert\sin(n \pi\epsilon/2)\vert$ is bounded by one, eventually this must decrease.

For instance, for $\epsilon=1/25$ we have the following plots for the values of $\vert c_n(\epsilon)\vert^2$: enter image description here

Summing the probabilities from $n=1$ to $n=500$ gives $0.989873$, so the first $500$ terms capture 99% of the initial state.

For $\epsilon=1/50$, the plot is qualitatively the same, except that the vertical scale is divided by $2$, reflecting the scaling with $\epsilon$ of the leading term $2\epsilon\sin\left(n\pi /4\right)^2$. One needs to now sum to $n=1000$ to get $\sim 99$% of the initial state.

  • $\begingroup$ Infinite square well is nearly as pathological as delta function - if we consider one acceptable, then we could just well settle for the other. E.g., your proposal of Gaussian state (which is usually a foolproof approach) is not viable in an infinite well due to the tails. A possible solution could be taking $|\psi(x)|^2=\delta(x-L/4)$ rather than $\psi(x)=\delta(x-L/4)$. $\endgroup$
    – Roger V.
    Sep 30, 2022 at 7:36
  • $\begingroup$ To avoid the tails of the Gaussian you can use any bump function, e.g. a triangular bump like $\sqrt{\frac{3a}2}(1-|ax|)\theta(1-|ax|),$ where $\theta$ denotes the Heaviside step function. $\endgroup$
    – Ruslan
    Sep 30, 2022 at 10:46
  • 2
    $\begingroup$ @LSS using a rectangle of height $1/\sqrt{\epsilon}$ and width $\epsilon$ as initial state might do the trick: you can likely take the limit $\epsilon\to 0$ “easily”. $\endgroup$ Sep 30, 2022 at 13:40
  • 1
    $\begingroup$ @ZeroTheHero: I tried working through that idea just now and (if I've done it correctly) all of the $c_n$ coefficients go to zero in the $\epsilon \to 0$ limit. This makes a bit of sense to me: we would expect all of the coefficients to be proportional to $\phi_n(L/4)$ in the limit where the particle is found "at $L/4$". But there are an infinite number of eigenstates with $|\phi_n(L/4)| = \sqrt{2/L}$, and so the only way for the sum of the squares of the $c_n$ to go to a finite number is if the proportionality factor is zero. But that's just a seat-of-my-pants post-hoc justification. $\endgroup$ Sep 30, 2022 at 13:58
  • 1
    $\begingroup$ @MichaelSeifert so what you’re saying is that the coefficients $c_n$ are “uniformly distributed” amongst all eigenstates and thus have vanishing amplitudes? One would think ratios of $c_n$’s would be like $\phi_{n}/\phi_m$ at $x=L/4$… $\endgroup$ Sep 30, 2022 at 14:34

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