The Solutions – The Particle in a Box

A set of solutions for which observables behave as expected among reference frames can be derived by using a new set of conditions to be imposed on the general solution to the Schrodinger Equation. Interestingly, this requires removal of the traditional boundary conditions. In order to obtain a unique solution, the following conditions need to be met:

1. The values of observable properties need to be consistent from a physical perspective regardless of the coordinate system chosen. For example, the probability of finding a particle in an interval needs to be invariant. Just as in classical mechanics, the expectation value of the momentum needs to be invariant as well with regards to different reference frames.

2. On the other hand, not all observable quantities are invariant. However, all quantities must obey the mathematical rules governing their expression in different coordinate systems. For example, if the expectation value of a particle’s position is given by $x_{avg}$ in one reference frame and by ${\underset{―}{x}}_{avg}$ in a second reference frame (underlined frame) whose origin is shifted to the left by $Δ x$ , the two expectation values need to obey the following relationship ${\underset{―}{x}}_{avg} = x_{avg} + Δ x$ .

3. The solutions must be able to be normalized with the probability of finding the particle within the confines of the well being equal to unity regardless of coordinate system.

4. The energy $E$ of particle within the well must be invariant with respect to coordinate system chosen by the observer.

5. The wave functions inside the well are allowed to be discontinuous with those outside the well. As shown previously ( see Boundary Condition), a well is a special example in which the continuity of the solutions is required only inside the well. The lack of continuity at the boundary of the well poses no physical or mathematical inconsistency.

The Generic Well and its Solutions

Once again, a well of length $L$ is placed with its left-hand edge at an arbitrary position $x_{0}$ resulting in the right-hand edge being at position $x = x_{0} + L$ .

with the potential within the well being given by

$\begin{matrix} (1) & V (x) = {\begin{cases} \infty, if x < x_{0} \\ 0, if x_{0} \leq x \leq x_{0} + L \\ \infty, if x > x_{0} + L \end{cases} \end{matrix}$

Irregardless of the position of the well, the general solutions to the Time-Independent Schrodinger equation for the potential given by Eq. $(1)$ are

$\begin{matrix} (2) & ψ (x) = A e^{ı k x} + B e^{- ı k x} A, B \in C \end{matrix}$

with

$\begin{matrix} (3) & k = \sqrt{\frac{2 m E}{ℏ^{2}}} \end{matrix}$

Eq. $(2)$ needs to be normalized, therefore

$\begin{matrix} P (x_{0}, x_{0} + L) = 1 \end{matrix}$

$\begin{matrix} \int_{x_{0}}^{x_{0} + L} ψ^{*} ψ d x = 1 \end{matrix}$

$\begin{matrix} \int_{x_{0}}^{x_{0} + L} (A^{*} A + B B^{*} + A B^{*} e^{ı 2 k x} + A^{*} B e^{- ı 2 k x}) = 1 \end{matrix}$

$\begin{matrix} (A^{*} A + B B^{*}) L + \frac{A B^{*}}{ı 2 k} e^{ı 2 k x} |_{x_{0}}^{x_{0} + L} - \frac{A^{*} B}{ı 2 k} e^{- ı 2 k x} |_{x_{0}}^{x_{0} + L} = 1 \end{matrix}$

$\begin{matrix} (4) & (A^{*} A + B B^{*}) L = 1 \end{matrix}$

Due to symmetry, the observer in the new coordinate system must calculate the same probability for the particle being in the left half as well as in the right-half of the well as does the observer in which the well is symmetrically placed (See Symmetry). In short, the probability must be invariant between these two coordinate systems. Therefore,

$\begin{matrix} (5) & P (x_{0}, x_{0} + \frac{L}{2}) = \frac{1}{2} \end{matrix}$

Substituting Eq. $(2)$ into Eq. $(5)$ and using the relationship given in Eq. $(4)$ results in the following

$P (x_{0}, x_{0} + \frac{L}{2}) = \frac{1}{2}$

$\begin{matrix} \int_{x_{0}}^{x_{0} + (1 / 2) L} ψ^{*} ψ d x = \frac{1}{2} \end{matrix}$

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

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

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

The relationship

$k_{n} = \frac{n π}{L} n = \pm 1, \pm 2, \pm 3, \dots$

was previously derived from the Bohr-Sommerfeld theorem (see Quantization of Energy). Making this substitution results in

$\begin{matrix} \frac{A B^{*}}{ı 2 k} e^{ı (2 k x_{0} + n π)} - \frac{A^{*} B}{ı 2 k} e^{- ı (2 k x_{0} + n π)} = 0 \end{matrix}$

$\begin{matrix} (6) & - A B^{*} e^{ı 2 k x_{0}} + A^{*} B e^{- ı 2 k x_{0}} = 0 \end{matrix}$

For any arbitrary $x_{0}$ , Eq. $(6)$ is satisfied only when

$\begin{matrix} (7) & A = 0 or B = 0 \end{matrix}$

Therefore, the general solution Eq. $(2)$ reduces to two separate solutions given as

$\begin{matrix} (8) & ψ_{1} (x) = A e^{ı k x} or ψ_{2} (x) = B^{- ı k x} \end{matrix}$

Examining the two solutions in Eq. $(8)$ reveals what is intuitively expected, the particle may be thought of as being a “free particle” represented by a complex wave traveling either to the left or right within the confines of the well.

Finally, using the normalization relationship given in Eq. $(4)$ results in the following the following two solutions:

$ψ_{1_{n}} (x) = A e^{ı k x} | A | = \sqrt{\frac{1}{L}}$

$ψ_{2_{n}} (x) = B^{- ı k x} | B | = \sqrt{\frac{1}{L}}$

with

$k_{n} = \frac{n π}{L} n = \pm 1, \pm 2, \pm 3, \dots$