Symmetry – The Particle in a Box

The particle in the box poses special properties which will be useful in determining the wave solutions.

A. Both $ψ (x)$ and $ψ (- x)$ are solutions of the Schrodinger Equation for potentials in which $V (x) = V (- x)$ .

For a particle of mass $m$ with a total energy $E$ and subject to a potential $V (x)$ , the Time-Independent Schrodinger equation in one dimension is as follows:

$\begin{matrix} (1) & - \frac{ℏ^{2}}{2 m} \frac{d^{2} ψ (x)}{d x^{2}} + V (x) ψ (x) = E ψ (x) \end{matrix}$

If the potential $V (x)$ is an even function of $x : V (- x) = V (x)$ , then replacing $x$ by $- x$ in Eq. $(1)$ gives

$\begin{matrix} (2) & - \frac{ℏ^{2}}{2 m} \frac{d^{2} ψ (- x)}{d x^{2}} + V (x) ψ (- x) = E ψ (- x) \end{matrix}$

Defining

$\begin{matrix} (3) & ϕ (x) \equiv ψ (- x) \end{matrix}$

gives the following relationships

$\begin{matrix} \frac{d}{d x} ϕ (x) = (- 1) \cdot \frac{d}{d x} ψ (- x) \end{matrix}$

$\begin{matrix} (4) & \frac{d^{2}}{d x^{2}} ϕ (x) = \frac{d^{2}}{d x^{2}} ψ (- x) \end{matrix}$

Substituting Eq. $(3)$ and Eq. $(4)$ into Eq. $(2)$ gives

$- \frac{ℏ^{2}}{2 m} \frac{d^{2} ϕ (x)}{d x^{2}} + V (x) ϕ (x) = E ϕ (x)$

thereby showing that $ϕ (x) = ψ (- x)$ is a solution as well.

B. Properties of the linear coefficients $A$ and $B$ for potentials in which $V (x) = V (- x)$ .

For a potential that is an even function of position $x$ , for example when $V (- x) = V (x)$ , if $ψ_{1} = ψ (x)$ is a solution of Eq. $(1)$ , $ψ_{2} = ψ (- x)$ is a solution as well (see above).

The probability $P (a, b)$ of finding a particle in interval $[a, b]$ given by $\int_{a}^{b} ψ^{*} ψ d x$ . Since both $ψ (x)$ and $ψ (- x)$ are valid solutions, then the same value for $P (a, b)$ must be obtained for either $ψ_{1}$ and $ψ_{2}$ . Therefore, for an infinitesimal region $d x$ ,

$\begin{matrix} (5) & ψ^{*} (x) ψ (x) d x = ψ^{*} (- x) ψ (- x) d x \end{matrix}$

Since Eq. $(1)$ is a second order differential equation with a characteristic equation resulting in complex roots, two solutions result and are given by

$f (x) = A e^{ı k x} and g (x) = B e^{- ı k x}$

The general solution is given by a linear combination of $f (x)$ and $g (x)$ . Therefore,

$\begin{matrix} (6) & ψ_{1} \equiv ψ (x) = A e^{ı k x} + B e^{- ı k x} \end{matrix}$

and

$\begin{matrix} (7) & ψ_{2} \equiv ψ (- x) = A e^{- ı k x} + B e^{ı k x} \end{matrix}$

Substituting Eq. $(6)$ and Eq. $(7)$ into Eq. $(5)$ gives

$\begin{matrix} A^{*} A + A^{*} B e^{- ı 2 k x} + A B^{*} e^{ı 2 k x} + B^{*} B = A^{*} A + A^{*} B e^{ı 2 k x} + A B^{*} e^{- ı 2 k x} + B^{*} B \end{matrix}$

Cancelling like terms and rearranging gives

$ı 2 (A^{*} B - A B^{*}) (\cos 2 k x) = 0$

This relationship must hold true for all $x$ , and therefore

$\begin{matrix} (8) & A^{*} B = A B^{*} \end{matrix}$

It should be noted that Eq. $(8)$ holds true regardless of the position of the well.

C. Symmetry of the probability about the midline of the well.

Once again, the well is placed with its left-hand edge at the $x_{L} = - L / 2$ (See Figure 1.). The probability of finding the particle in interval $[- L / 2, 0]$ is given by $P (- L / 2, 0) = \int_{- L / 2}^{0} ψ^{*} (x) ψ (x) d x$ . Using Eq. $(6)$ , the relationship in Eq. $(8)$ (i.e., $A^{*} B = A B^{*}$ ) and noting that $k = n π / L$ (see Quantization of Energy) results in the following

$\begin{matrix} P (- L / 2, 0) = \int_{- L / 2}^{0} (A^{*} A + A^{*} B e^{- ı 2 k x} + A B^{*} e^{ı 2 k x} + B^{*} B) d x \end{matrix}$

$\begin{matrix} = \int_{- L / 2}^{0} [A^{*} A + B^{*} B + 2 A^{*} B \cos (k x)] d x \end{matrix}$

$\begin{matrix} = [(A^{*} A + B^{*} B) x + \frac{1}{k} A^{*} B \sin (k x)] |_{- L / 2}^{0} \end{matrix}$

$\begin{matrix} (9) & = (A^{*} A + B^{*} B) \frac{L}{2} + \frac{1}{k} A^{*} B \sin (\frac{L}{2}) \end{matrix}$

The probability of finding the particle in interval $[0, L / 2]$ is given by $P (0, L / 2) = \int_{0}^{L / 2} ψ^{*} (x) ψ (x) d x$ which yields

$\begin{matrix} P (0, L / 2) = \int_{0}^{L / 2} (A^{*} A + A^{*} B e^{- ı 2 k x} + A B^{*} e^{ı 2 k x} + B^{*} B) d x \end{matrix}$

$\begin{matrix} = \int_{0}^{L / 2} [A^{*} A + B^{*} B + 2 A^{*} B \cos (k x)] d x \end{matrix}$

$= [(A^{*} A + B^{*} B) x + \frac{1}{k} A^{*} B \sin (k x)] |_{0}^{L / 2}$

$\begin{matrix} (10) & = (A^{*} A + B^{*} B) \frac{L}{2} + \frac{1}{k} A^{*} B \sin (\frac{L}{2}) \end{matrix}$

Note that Eq. $(9)$ is equal to Eq. $(10)$ . As to be expected due to the symmetry of the well, the probability of finding the particle in the left half of the well is equal to finding it in the right half of the well. Consequently, since the probability of the particle being found in the well as equal to unity, then $P (- L / 2, 0) = P (0, L / 2) = 1 / 2$ .

More importantly, it needs to be recognized that the symmetry between either half of the box is maintained regardless of the position of the box. In other words, an observer’s frame of reference does not change the probability of finding the particle in either half of the box. Viewed in a different fashion, the position of the box doesn’t alter the probability either. In order to maintain a logical consistency, all observer’s must agree on the same probability throughout a given interval regardless of whether the well is found in the symmetric position, asymmetric position or generic position.