Solved Example #1: The Infinite Square Well

Solved Example #1: The Infinite Square Well is attached.

11. Example #1 (Application of Schrödinger equation): The Infinite Square Well

Consider a particle with mass m, subjected in a potential:

V (x) = 0 for 0 ≤ x ≤ a, and V (x) → ∞ elsewhere.

The particle is completely free, except at the two ends x = 0 and x = a where an infinite force prevents the particle from escaping. In this case, we would like to find the general solution to the Schrödinger equation for this particular potential.

 Figure 11.1. Infinite square well potential.

The probability of finding the particle outside the well is zero, i.e. ψ (x) = 0. While inside the well, the time-independent Schrödinger equation is:

$-\frac{{{\hbar }^{2}}}{2m}\frac{{{d}^{2}}\psi }{d{{x}^{2}}}=E\psi $                                                    (11.1)

Or:

$\frac{{{d}^{2}}\psi }{d{{x}^{2}}}=-{{k}^{2}}\psi $, with $k=\frac{\sqrt{2mE}}{\hbar }$                                        (11.2)

Equation (11.2) is the simple harmonic oscillator equation, and the general solution is given by:

$\psi (x)=A\sin kx+B\cos kx$                                           (11.3)

With A and B are arbitrary constants. These constants are fixed by boundary conditions of the problem. ψ is continuous, when the potential V (x) goes to infinity, i.e. ψ (a) =  ψ (0) = 0.

Substituting the boundary condition ψ (0) = 0 into equation (11.3), we obtain: B = 0, and hence:

$\psi (x)=A\sin kx$                                                (11.4)

The boundary condition ψ (a) = A sin ka = 0, this condition can be satisfied for A = 0, then ψ (x) = 0. However, if A = 0, we get a trivial, “non-normalizable” solution. Therefore, sin ka = 0 which means that:

ka = 0, ±π, ±2π, ±3π, …                                           (11.5)

The negative solution gives nothing new and the minus sign can be absorbed into A. So the distinct solution is given by:

${{k}_{n}}=\frac{n\pi }{a}$  with n = 1, 2, 3, …                                         (11.6)

With the help of boundary condition at x = a, the possible allowed values of energy E_n can be determined:

${{E}_{n}}=\frac{{{\hbar }^{2}}k_{n}^{2}}{2m}=\frac{{{n}^{2}}{{\pi }^{2}}{{\hbar }^{2}}}{2m{{a}^{2}}}$                                                   (11.7)

In order to fix the constant A, we have to normalize ψ:

$\int\limits_{0}^{a}{{}}$|A|² sin² (kx) dx = 1, so that: |A|² = ${}^{2}/{}_{a}$                                (11.8)

The simplest is to pick the positive real root for constant A: A = $\sqrt{{2}/{a}\;}$then, the solutions inside the potential well are:

${{\psi }_{n}}(x)=\sqrt{\frac{2}{a}}\sin \left( \frac{n\pi }{a}x \right)$                                              (11.9)

ψ₁ is the state with the lowest energy, the so-called ground state, and the other states with higher energies are the excited states. Notice that the energy increases in proportion to n² and each successive state has one more node as we go up in energy (see figure 11.2 below where first three initial states are depicted). And this is a general property, that applies for any shape of the potential.

 Figure 11.2. The first three states of infinite square well.

As can be seen in figure 11.2, the states are alternately even and odd, with respect to the center of the potential well. This property applies whenever the potential is an even function.

Another important property of the states ψ_n is that, they are mutually orthogonal. And this property can be shown:

$\int{{{\psi }_{m}}{{(x)}^{*}}{{\psi }_{n}}(x)dx}=0$                                        (11.10)

For m ≠ n. And for m = n, the integral in equation (11.10) is equal to 1. We can also expressed equation (11.10) in terms of Kronecker delta function which is defined by: δ_mn = 0 if m ≠ n and if m = n, δ_mn = 1. Therefore,

$\int{{{\psi }_{m}}{{(x)}^{*}}{{\psi }_{n}}(x)dx={{\delta }_{mn}}}$                                   (11.11)

We say that the ψ’s are orthonormal.

Please note that any other function f (x), can be expressed as linear combination of state in equation (11.9), this is called Dirichlet’s theorem:

$f(x)=\sqrt{\frac{2}{a}}\sum\limits_{n=1}^{\infty }{{{c}_{n}}\sin \left( \frac{n\pi }{a}x \right)}$                                       (11.12)

This expression for f (x), is in fact the Fourier expansion. The expansion coefficients c_n can be evaluated for a given f (x) and by applying the orthonormality of ψ_n. Thus, the m-th coefficient in the expansion of f(x) is given by:

${{c}_{m}}=\int{{{\psi }_{m}}}{{(x)}^{*}}f(x)dx$                                      (11.13)

The stationary states for the infinite square well are:

${{\Psi }_{n}}(x,t)=\sqrt{\frac{2}{a}}\sin \left( \frac{n\pi }{a}x \right){{e}^{-i({{{n}^{2}}{{\pi }^{2}}\hbar }/{2m{{a}^{2}}}\;)t}}$                              (11.14)

And the most general solution to the time-dependent Schrödinger equation is a linear combination of stationary states:

$\Psi (x,t)=\sum\limits_{n=1}^{\infty }{{{c}_{n}}\sqrt{\frac{2}{a}}\sin \left( \frac{n\pi }{a}x \right){{e}^{-i({{{n}^{2}}{{\pi }^{2}}\hbar }/{2m{{a}^{2}}}\;)t}}}$                           (11.15)

8. Quantum Mechanics: The Wave Function and Statistical Interpretation

Consider a particle of mass m, that can only move by means of specified force F(x,t) along the x-axis. One of the problems of classical mechanics is to determine the position of the particle at any given time t: x(t), this can be determined by applying Newton’s second law: F=ma. For a conservative system, the force F can be expressed in term of potential energy function V: $F=-{}^{\partial V}/{}_{\partial x}$ . And Newton’s law reads: $m{}^{{{d}^{2}}x}/{}_{d{{t}^{2}}}=-{}^{\partial V}/{}_{\partial x}$ . This law together with the initial conditions for instance, the position and velocity at t=0, determine x(t). Once the position of the particle is known, we can determine the velocity ($v={dx}/{dt}\;$), the momentum ($p=mv$), and the kinetic energy ($T={}^{1}/{}_{2}m{{v}^{2}}$) or any other dynamical variables of interest.

In quantum mechanics, we are looking for a wave function Ψ(x,t) of a particle, that is a solution of Schrödinger equation:

$i\hbar \frac{\partial \Psi }{\partial t}=-\frac{{{\hbar }^{2}}}{2m}\frac{{{\partial }^{2}}\Psi }{\partial {{x}^{2}}}+V\Psi $                                              (8.1)

Here, i is $\sqrt{-1}$ and $\hbar ={}^{h}/{}_{2\pi }={{1,054573.10}^{-34}}Js$ i.e. original Planck’s constant h divided by 2π. In analogy to Newton’s law that determines x(t) for all future time, Schrödinger equation – given the initial condition, for instance at t=0  – determines Ψ(x,t) for all future time.

A particle, naturally, is localized at a point in space, whereas the wave function is spread out in space. Therefore we need a statistical interpretation of wave function. This statistical interpretation (provided first by Born) says that: ${{\left| \Psi (x,t) \right|}^{2}}dx$ is the probability of finding the particle between x and (x+dx) at time t. However, the interpretation causes a kind of indeterminacy into quantum mechanics since we cannot predict with certainty the position of the particle through a simple experiment, even though we know everything theoretically about the wave function of the particle. Therefore, quantum mechanics offers only statistical information about the possible results.

Regarding this indeterminacy, there are three positions:

• The realist position, which is advocated by Einstein. This position says that quantum mechanics is incomplete theory, since the theory cannot predict the exact position of the particle. Evidently, the wave function Ψ is not the whole story, some additional information is needed to provide a complete description of the particle.

• The orthodox position or the Copenhagen interpretation, which is suggested by Bohr. Mainly, this position says that the particle was not really anywhere and the act of measurement not only disturb what is to be measured, but also produce it. However, a repeated measurement must return the same value, i.e. the same wave function. The orthodox position says that the wave function ‘collapses’ upon measurement. Therefore, there are two entirely well-defined physical processes: the ‘ordinary’ process in which the wave function evolves ‘smoothly’ under Schrödinger equation and the ‘measurement’ in which Ψ suddenly and discontinuosly collapses.

• The agnostic position suggested by Pauli, that refuses to give any answer, since it doesn’t make any sense to know about the state of particle before a measurement, when the only way to know whether we’re right is to conduct a measurement, in which case the result we get is no longer ‘before the measurement. However, this position is eliminated by Bell’s discovery in 1964. John Bell showed that it makes an observable difference if the position of the particle was precise prior to the measurement.

Due to statistical interpretation, the theory of probability plays an important role in quantum mechanics. We will summarize the theory of probability by means of an example. Suppose there’re 14 people in a room, whose ages are:

14 years old : 1 person

15 years old : 1 person

16 years old : 3 people

22 years old : 2 people

24 years old : 2 people

25 years old : 5 people

Let N(j) be the number of people of age j, then:

$N(14)=1$

$N(15)=1$

$N(16)=3$

$N(22)=2$

$N(24)=2$

$N(25)=5$

The total number of people in the room is given by:

$N=\sum\limits_{j=0}^{\infty }{N(j)}$                                                        (8.2)

For instance, if P(j) is the probability of getting age j, then $P(14)={}^{1}/{}_{14}$, $P(15)={}^{1}/{}_{14}$, $P(25)={}^{5}/{}_{14}$, etc. Generally, the probability P(j) can be expressed by:

$P(j)=\frac{N(j)}{N}$                                                         (8.3)

Please notice that the probability of getting either 14 or 15 is the sum of the individual probabilities and the sum of all probabilities is 1:

$\sum\limits_{j=1}^{\infty }{P(j)=1}$                                                        (8.4)

The most probable age in this case is 25. Generally, the most probable j is the j for which P(j) is maximum. The median age is obviously 23, while the mean or average age is: $\frac{14+15+(3.16)+(2.22)+(2.24)+(5.25)}{14}=\frac{294}{14}=21$. In general, the average value of j is given by:

$\left\langle j \right\rangle =\frac{\sum{jN(j)}}{N}=\sum\limits_{j=0}^{\infty }{jP(j)}$                                                (8.5)

In quantum mechanics, the average is usually to be called the expectation value.

The average of the square of age 14 is: ${{14}^{2}}=196$ with probability ${1}/{14}\;$ and for age 15: ${{15}^{2}}=225$ with probability ${1}/{14}\;$ , etc.Then, the average is:

$\left\langle {{j}^{2}} \right\rangle =\sum\limits_{j=0}^{\infty }{{{j}^{2}}P(j)}$                                                     (8.6)

In general, the average value of some function f is given by:

$\left\langle f(j) \right\rangle =\sum\limits_{j=0}^{\infty }{f(j)P(j)}$                                                  (8.7)

Another important quantity is the variance σ of the distribution or the standard deviation, which is the measure of the spread about $\left\langle j \right\rangle $.

The variance is given by σ²≡<(j-<j>)²>, working this equation out further with (8.5), we obtain: 

${{\sigma }^{2}}=\sum{\left( {{j}^{2}}-2j\left\langle j \right\rangle +{{\left\langle j \right\rangle }^{2}} \right)P(j)}$

      $=\sum{{{j}^{2}}P(j)-2\left\langle j \right\rangle \sum{jP(j)}+{{\left\langle j \right\rangle }^{2}}\sum{P(j)}}$

${{\sigma }^{2}}=\left\langle {{j}^{2}} \right\rangle -2\left\langle j \right\rangle \left\langle j \right\rangle +{{\left\langle j \right\rangle }^{2}}=\left\langle {{j}^{2}} \right\rangle -{{\left\langle j \right\rangle }^{2}}$                                    (8.8)

 

The deviation term $j-\left\langle j \right\rangle $ is usually given by $\Delta j$, i.e. $\Delta j=j-\left\langle j \right\rangle $.

Since ${{\sigma }^{2}}$ is surely non-negative, equation (8.8) implies that $\left\langle {{j}^{2}} \right\rangle \ge {{\left\langle j \right\rangle }^{2}}$, and when both terms are equal, the distribution $\sigma =0$, i.e. the distribution has no spread at all (every member has the same value).

We are dealing so far with discrete variable. While in practice, the probability of getting an age precisely 16 years, 4 hours, 27 minutes and 3,33333 seconds is zero. Therefore, it is more convenient to speak about the probability of getting an age that lies in some interval. Now we are talking about continuous variable.

We introduce therefore a proportionality factor ρ(x), the so-called probability density, which is the probability of getting x. For example, the probability that x lies between a finite interval a and b is given by the integral of ρ(x):

${{P}_{ab}}=\int\limits_{a}^{b}{\rho (x)dx}$                                                          (8.9)

And the rules for discrete variable can be rewritten:

$\int\limits_{-\infty }^{+\infty }{\rho (x)dx}=1$                                                       (8.10)

$\left\langle x \right\rangle =\int\limits_{-\infty }^{+\infty }{x\rho (x)dx}$                                                   (8.11)

$\left\langle f(x) \right\rangle =\int\limits_{-\infty }^{+\infty }{f(x)\rho (x)dx}$                                               (8.12)

${{\sigma }^{2}}\equiv \left\langle {{\left( \Delta x \right)}^{2}} \right\rangle =\left\langle {{x}^{2}} \right\rangle -{{\left\langle x \right\rangle }^{2}}$                                              (8.13)