If two (or more) distinct⁴⁴solutions to the (time-independent) Schrödinger equation have the same energy E . These states are said to be degenerate. For example, the free particle states are doubly degenerate-one solution representing motion to the right. And the other motion to the left. But we have never encountered normalizable degenerate solutions, and this is no accident. Prove the following theorem: In one dimension⁴⁵ there are no degenerate bound states. Hint: Suppose there are two solutions, ${\mathit{\psi }}_{\mathbf{1}}$and ${\mathit{\psi }}_2$with the same energy E. Multiply the Schrödinger equation for ${\mathit{\psi }}_{\mathbf{1}}$by ${\mathit{\psi }}_2$and the Schrödinger equation for ${\mathit{\psi }}_{\mathbf{2}}$by ${\mathit{\psi }}_{\mathbf{1}}$and subtract, to show that $\left({\mathit{\psi }}_2\mathit{d}{\mathit{\psi }}_{\mathbf{1}}\mathbf{/}\mathit{d}\mathit{x}\mathbf{-}{\mathit{\psi }}_{\mathbf{2}}\mathit{d}{\mathit{\psi }}_{\mathbf{1}}\mathbf{/}\mathit{d}\mathit{x}\right)$is a constant. Use the fact that for normalizable solutions $\mathit{\psi }\mathbf{\to }\mathbf{0}\mathbf{}\mathbf{at}\mathbf{}\mathbf{\pm }\mathbf{\infty }$to demonstrate that this constant is in fact zero.Conclude that ${\mathit{\psi }}_2$s a multiple of _{${\mathit{\psi }}_{\mathbf{1}}$}and hence that the two solutions are not distinct.

A differential equation is a foundation for the quantum-mechanical description of matter using the wavelike characteristics of particles in a field. The probability density of a particle in space and time is relevant to its solution.

The time-dependent Schrödinger equation is represented as:

_{$\mathit{i}\mathit{ħ}\frac{\mathbf{d}}{\mathbf{d}\mathbf{t}}\left|\psi \left(t\right)>=\hat{H}\right|\mathit{\psi }\left(t\right)\mathbf{>}$}

The given energy is,

$-\frac{{ħ}^2}{2m}\frac{d^2{\psi }_1}{d{x}^2}+V{\psi }_1=E{\psi }_1$

Multiplying on both sides.

$-\frac{{ħ}^2}{2m}{\psi }_2\frac{d^2{\psi }_1}{d{x}^2}+V{\psi }_1{\psi }_2=E{\psi }_1{\psi }_2$ ...(1)

And

$-\frac{{ħ}^2}{2m}{\psi }_2\frac{d^2{\psi }_1}{d{x}^2}+V{\psi }_2=E{\psi }_2$

Multiplyingon both sides.

role="math" localid="1658213251156" $-\frac{{ħ}^2}{2m}{\psi }_1\frac{d^2{\psi }_1}{d{x}^2}+V{\psi }_1{\psi }_2=E{\psi }_1{\psi }_2$ ...(2)

Now from the equation (1) and (2),

$$\begin{gathered} -\frac{{ħ}^2}{2m}\left[{\psi }_2\frac{d^2{\psi }_1}{d{x}^2}-{\psi }_1\frac{d^2{\psi }_2}{d{x}^2}\right]=0 \\ {\psi }_2\frac{d^2{\psi }_1}{d{x}^2}-{\psi }_1\frac{d^2{\psi }_2}{d{x}^2}=0 \end{gathered}$$

Differentiate the above equation with respect to x.

role="math" localid="1658213785095" $$\begin{gathered} \frac{d}{dx}\left[{\psi }_2\frac{d^2{\psi }_1}{dx}-{\psi }_1\frac{d^2{\psi }_2}{dx}\right]=\frac{d{\psi }_2}{dx}\frac{d{\psi }_1}{dx}+{\psi }_2\frac{d^2{\psi }_1}{d{x}^2}-\frac{d{\psi }_1}{dx}\frac{d{\psi }_2}{dx}-{\psi }_1\frac{d^2{\psi }_2}{d{x}^2} \\ ={\psi }_2\frac{d^2{\psi }_1}{d{x}^2}-{\psi }_1\frac{d^2{\psi }_2}{d{x}^2} \\ {\psi }_2\frac{d^2{\psi }_1}{d{x}^2}-{\psi }_1\frac{d^2{\psi }_2}{d{x}^2}=K\left(\mathrm{a}\mathrm{constant}\right) \end{gathered}$$

But $\psi \to 0\mathrm{at}\infty$at .

So the constant must be zero.

Thus,

$$\begin{gathered} {\psi }_2\frac{d{\psi }_1}{dx}-{\psi }_1\frac{d{\psi }_2}{dx} \\ \frac{1}{{\psi }_1}\frac{d{\psi }_1}{dx}-\frac{1}{{\psi }_2}\frac{d{\psi }_2}{dx} \end{gathered}$$

Then finally

$$\begin{gathered} \mathrm{In}{\psi }_1=\mathrm{In}{\psi }_2+cons\tan t \\ {\psi }_1=\left(cons\tan t\right){\psi }_2 \end{gathered}$$

Therefore, the two solutions are not distinct.