Infinite square well – triangular initial state

Required math: calculus

Required physics: Schrödinger equation, particle in a box

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Sec 2.2, Problem 2.7.

In the solution of the one-dimensional particle in a box problem we find that there are an infinite number of wave functions that satisfy the spatial part of the Schrödinger equation, and that each such solution corresponds to a discrete energy. To summarize, the problem was to solve the spatial equation

\displaystyle   -\frac{\hbar^{2}}{2m}\frac{d^{2}\psi}{dx^{2}}+V(x)\psi \displaystyle  = \displaystyle  E\psi \ \ \ \ \ (1)

for the potential

\displaystyle  V(x)=\begin{cases} 0 & 0<x<a\\ \infty & \mathrm{otherwise} \end{cases} \ \ \ \ \ (2)

We found that the solutions, properly normalized, are

\displaystyle   \psi_{n}(x) \displaystyle  = \displaystyle  \sqrt{\frac{2}{a}}\sin\frac{n\pi x}{a} \ \ \ \ \ (3)

with {\psi=0} outside the box. The energy corresponding to solution {\psi_{n}} is

\displaystyle   E_{n} \displaystyle  = \displaystyle  \frac{n^{2}\pi^{2}\hbar^{2}}{2ma^{2}} \ \ \ \ \ (4)

But this solves only the spatial part of the Schrödinger equation. The full solution requires bringing back the function containing the time dependence that arises from solving the equation using separation of variables. Thus the full, time-dependent solution for a particular energy is

\displaystyle   \Psi_{n}(x,t) \displaystyle  = \displaystyle  \psi_{n}(x)e^{-iE_{n}t/\hbar} \ \ \ \ \ (5)

where {\psi_{n}} and {E_{n}} are given above.

The probability of finding a particle that is in state {\Psi_{n}(x,t)} at location {x} at time {t} is therefore {|\Psi_{n}|^{2}=|\psi_{n}|^{2}}. That is, if a particle is in one of the states with a definite energy {E_{n}}, its probability density is independent of time, since the only time dependence comes from the complex exponential function whose modulus is always 1. The time dependence always disappears when the square modulus is calculated. For this reason, the states {\Psi_{n}(x,t)} are called stationary states.

This isn’t the end of the story, however. Since the Schrödinger equation is linear, once we have found two or more solutions, any linear combination of these solutions gives another solution. Thus the most general solution of the time-independent Schrödinger equation for the particle in a box is

\displaystyle   \Psi(x,t) \displaystyle  = \displaystyle  \sum_{n=1}^{\infty}c_{n}\psi_{n}(x)e^{-iE_{n}t/\hbar} \ \ \ \ \ (6)

where the {c_{n}} are arbitrary complex constants. Solutions that are linear combinations of two or more of the stationary states are, however, not stationary states themselves, since each term in the sum contains its own dependence on time in the form of a complex exponential, and these exponentials will not cancel out when the square modulus is calculated.

To explore the consequences of this general solution, we need first to demonstrate some of the properties of the {\psi_{n}} stationary states. We know that {\psi_{n}} is normalized, in the sense that

\displaystyle   \int_{0}^{a}|\psi_{n}(x)|^{2}dx \displaystyle  = \displaystyle  1 \ \ \ \ \ (7)

However the set of {\psi_{n}} functions is also orthogonal, in the sense that if {m\ne n}

\displaystyle   \int_{0}^{a}\psi_{n}(x)\psi_{m}(x)dx \displaystyle  = \displaystyle  0 \ \ \ \ \ (8)

(This can be checked by direct integration, using the trigonometric identity {\sin a\sin b=\frac{1}{2}(\cos(a-b)-\cos(a+b))}.) A set of functions that are both orthogonal and normalized is called an orthonormal set.

This result may surprise you, but you might also be thinking ‘so what?’. The normalization is clearly important if {\psi_{n}} is to be used as a probability density, but the orthogonality seems pretty much irrelevant. To see why this property is important, suppose you want a full dynamical description of a particle starting from time {t=0} when the particle is in some initial state {\Psi(x,0)}. In general, we should be able to specify this initial state to be anything at all. In the classical case, an analogous situation might be something like this: standing on the Earth’s surface we have a rock of mass {m} which we can throw from any height above the ground. We can aim the rock in any direction and at any speed. All of these things are independent of the law of gravity which takes over after the rock has been thrown. But the theory of motion that we use to describe the motion of the rock must allow us to specify the initial conditions to be anything we like.

It is a similar situation in quantum mechanics. The Schrödinger equation tells us how the particle behaves, but only after we have specified the initial conditions. In this case, we can ‘throw’ the particle into the potential well by specifying its initial state, and the theory must be able to accept any (within the constraints of the problem) initial state. For the particle in a box, since we have an infinite potential outside the box, there is no way the particle could ever be found outside, so any initial condition must confine the particle to the region {0\le x\le a}, but apart from that, {\Psi(x,0)} can be anything at all.

From the general solution above, we can get an expression for {\Psi(x,0)}:

\displaystyle   \Psi(x,0) \displaystyle  = \displaystyle  \sum_{n=1}^{\infty}c_{n}\psi_{n}(x) \ \ \ \ \ (9)

The problem is therefore, given {\Psi(x,0)} and {\psi_{n}(x)}, find the constants {c_{n}}. It is here that the orthonormal nature of the set of stationary solutions comes into play. Suppose we want to find {c_{N}} for some particular value of {N}. If we multiply this expression by {\psi_{N}} and integrate over the range of the box (from 0 to {a}), then all integrals where {N\ne n} are zero because {\psi_{N}} is orthogonal to {\psi_{n}}, so we are left with just the one term we want:

\displaystyle   \int_{0}^{a}\psi_{N}(x)\Psi(x,0)dx \displaystyle  = \displaystyle  \sum_{n=1}^{\infty}c_{n}\int_{0}^{a}\psi_{N}\psi_{n}\;dx\ \ \ \ \ (10)
\displaystyle  \displaystyle  = \displaystyle  \sum_{n=1}^{\infty}c_{n}\delta_{nN}\ \ \ \ \ (11)
\displaystyle  \displaystyle  = \displaystyle  c_{N} \ \ \ \ \ (12)

where the symbol {\delta_{nN}} is called the Kronecker delta and is a shorthand way of writing a quantity that is zero if {n\ne N}, and 1 if {n=N}.

So for any initial condition, we now have a way of writing it as a sum over the spatial part of the stationary states at time {t=0}. The general solution is obtained merely by restoring the time dependence:

\displaystyle   \Psi(x,t) \displaystyle  = \displaystyle  \sum_{n=1}^{\infty}c_{n}\psi_{n}(x)e^{-iE_{n}t/\hbar} \ \ \ \ \ (13)

Note that I am not saying that any function is a solution of the Schrödinger equation – that is clearly absurd. What is true, however, is that any function of {x} can be expressed as a linear combination of the solutions of the spatial part of the Schrödinger equation. It is important to note that any of the {\psi_{n}} functions on its own is not a solution of the Schrödinger equation unless it is multiplied by {e^{-iE_{n}t/\hbar}}, so the linear combination {\Psi(x,0)=\sum_{n=1}^{\infty}c_{n}\psi_{n}(x)} is not in general a solution of the Schrödinger equation. It is only the final form {\Psi(x,t)=\sum_{n=1}^{\infty}c_{n}\psi_{n}(x)e^{-iE_{n}t/\hbar}} that is a solution.

Example: A particle in the infinite square well starts off with a wave function as follows:

\displaystyle  \Psi(x,0)=\begin{cases} Ax & 0\le x\le a/2\\ A(a-x) & a/2\le x\le a \end{cases} \ \ \ \ \ (14)

The initial wave function is thus a triangle with its peak at {x=a/2} and a height of {Aa/2}. We can normalize the wave function first

\displaystyle   \int_{0}^{a}|\Psi|^{2}dx \displaystyle  = \displaystyle  A^{2}\left(\int_{0}^{a/2}x^{2}dx+\int_{a/2}^{a}(a-x)^{2}dx\right)\ \ \ \ \ (15)
\displaystyle  \displaystyle  = \displaystyle  A^{2}\frac{a^{3}}{12}=1\ \ \ \ \ (16)
\displaystyle  A \displaystyle  = \displaystyle  \frac{2\sqrt{3}}{a^{3/2}} \ \ \ \ \ (17)

Next, we can find {\Psi(x,0)} in terms of the {\psi_{n}}. We need to calculate the {c_{n}}, so:

\displaystyle   c_{n} \displaystyle  = \displaystyle  \int_{0}^{a}\Psi(x,0)\psi_{n}dx\ \ \ \ \ (18)
\displaystyle  \displaystyle  = \displaystyle  \frac{2\sqrt{3}}{a^{3/2}}\sqrt{\frac{2}{a}}\left(\int_{0}^{a/2}x\sin\frac{n\pi x}{a}dx+\int_{a/2}^{a}(a-x)\sin\frac{n\pi x}{a}dx\right)\ \ \ \ \ (19)
\displaystyle  \displaystyle  = \displaystyle  \frac{4\sqrt{6}}{n^{2}\pi^{2}}\sin\frac{n\pi}{2} \ \ \ \ \ (20)

(The integral is straightforward, if a bit tedious. I used mathematical software to do it on a computer.) Thus we get

\displaystyle   \Psi(x,0) \displaystyle  = \displaystyle  \frac{4\sqrt{6}}{\pi^{2}}\sum_{n=1}^{\infty}\frac{\sin\left(n\pi/2\right)}{n^{2}}\psi_{n}(x)\ \ \ \ \ (21)
\displaystyle  \displaystyle  = \displaystyle  \frac{4\sqrt{6}}{\pi^{2}}\left[\sum_{n=1,5,9\ldots}\frac{\psi_{n}}{n^{2}}-\sum_{n=3,7,11\ldots}\frac{\psi_{n}}{n^{2}}\right] \ \ \ \ \ (22)

The term {\sin(n\pi/2)} is zero if {n} is even, and {\pm1} for odd values of {n}, giving the sums shown in the last line. The series consists of odd {n} only, with half the terms being positive and the other half negative. The full solution is found by replacing the exponentials containing the time dependence:

\displaystyle   \Psi(x,t) \displaystyle  = \displaystyle  \frac{4\sqrt{6}}{\pi^{2}}\left[\sum_{n=1,5,9\ldots}\frac{\psi_{n}}{n^{2}}e^{-iE_{n}t/\hbar}-\sum_{n=3,7,11\ldots}\frac{\psi_{n}}{n^{2}}e^{-iE_{n}t/\hbar}\right] \ \ \ \ \ (23)

The probability that the energy is {E_{1}} is

\displaystyle  c_{1}^{2}=\left(\frac{4\sqrt{6}}{\pi^{2}}\right)^{2}=\frac{96}{\pi^{4}}=0.9855 \ \ \ \ \ (24)

[Incidentally, since we know that {\sum_{n}\left|c_{n}\right|^{2}=1,} we get from 20 that

\displaystyle  \sum_{n\;odd}\frac{1}{n^{4}}=\frac{\pi^{4}}{96} \ \ \ \ \ (25)

The average energy can be found by using the energy levels for the infinite square well:

\displaystyle  E_{n}=\frac{\left(n\pi\hbar\right)^{2}}{2ma^{2}} \ \ \ \ \ (26)

We then have

\displaystyle   \left\langle H\right\rangle \displaystyle  = \displaystyle  \sum_{n=0}^{\infty}\left|c_{n}\right|^{2}E_{n}\ \ \ \ \ (27)
\displaystyle  \displaystyle  = \displaystyle  \frac{96}{\pi^{4}}\frac{\pi^{2}\hbar^{2}}{2ma^{2}}\sum_{n\;odd}\frac{1}{n^{2}}\ \ \ \ \ (28)
\displaystyle  \displaystyle  = \displaystyle  \frac{96}{\pi^{4}}\frac{\pi^{2}\hbar^{2}}{2ma^{2}}\frac{\pi^{2}}{8}\ \ \ \ \ (29)
\displaystyle  \displaystyle  = \displaystyle  \frac{6\hbar^{2}}{ma^{2}}\ \ \ \ \ (30)
\displaystyle  \displaystyle  \approx \displaystyle  1.216E_{1} \ \ \ \ \ (31)

The sum

\displaystyle  \sum_{n\;odd}\frac{1}{n^{2}}=\frac{\pi^{2}}{8} \ \ \ \ \ (32)

is not commonly found in tables, but a link (live at the time of writing) showing a derivation is here. The derivation requires knowledge of either the Riemann zeta function or the residue theorem from complex variable theory.

To summarize, the situation with the particle in a box is:

  1. Using separation of variables, we can split the solution {\Psi(x,t)} into a spatial function {\psi(x)} and a temporal factor {e^{-iEt/\hbar}}.
  2. When we solve the spatial equation with appropriate boundary conditions and normalization, we get the set of functions {\psi_{n}(x)}. Each of these functions is associated with a specific discrete energy {E_{n}}.
  3. The set of functions {\psi_{n}(x)} is orthonormal.
  4. The full solution of the Schrödinger equation for a particular energy {E_{n}} is {\psi_{n}(x)e^{-iE_{n}t/\hbar}}. The square modulus of this solution is independent of time, so it is a stationary state.
  5. The general solution of the Schrödinger equation is a linear combination of the set of stationary states. The square modulus of a general solution will not, in general, be stationary (time-independent).
  6. The constants {c_{n}} in the general solution can be obtained using the initial condition {\Psi(x,0)} and the orthonormality of the {\psi_{n}(x)}.

6 thoughts on “Infinite square well – triangular initial state

  1. Pingback: Infinite square well with triangular initial state using delta function « Physics tutorials

  2. khaiyichin


    I get that we can obtain the value of pi^4/96 as the sum of odd n’s of 1/n^4, but I’m struggling to see how does that lead to sum_odd_n’s (1/n^2) = pi^2/8 (or maybe it doesn’t?). I wasn’t able to find this particular geometric series, so can you briefly explain the origin for sum_odd_n’s (1/n^2) = pi^2/8?


    1. gwrowe Post author

      A good question. Stupidly, I didn’t record where I’d found the sum, but I’ve now edited the post to give a link to a derivation. I don’t think you can get this from {\sum_{n\;odd}\frac{1}{n^{4}}}, since the derivation involves some rather advanced mathematics.


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