In Section 4.3, it is shown that$\mathit{\Psi }{\mathbf{\text{(x,t)=Ae}}}^{\mathbf{\text{i(kx-ω}}\mathbf{t}\mathbf{\text{)}}}$satisfies the free-particle Schrödinger equation for all$x$and$t$, provided only that the constants are related by${\mathbf{\text{(}}\mathbf{\hslash }\mathbf{\text{k)}}}^{\mathbf{\text{2}}}\mathbf{/}\mathbf{\text{2m}}\mathbf{=}\mathit{\hslash }\mathit{\omega }$. Show that when the function,$\mathit{\Psi }\mathbf{\text{(x,t)=Acos(kx}}\mathbf{-}\mathbf{\text{0x)}}$is plugged into the Schrodinger equation, the result cannot possibly hold for all values of x and t, no matter how the constants may be related.

So, let's take the above function and plug it back into the Schrodinger equation to see if we can come up with a condition on the value of $\mathbf{\text{ω}}$or $\mathit{k}$ that will allow us to get this result.

$-\frac{{\hslash }^2}{2m}\frac{{\partial }^2\Psi }{\partial x^2}=i\hslash \frac{\partial \Psi }{\partial t}$………………(1)

The wavefunction given is

$\Psi =\text{Acos(kx-ωt)}$……………….(2)

Substitute the value of wavefunction from equation (2) into equation (1), and we get,

$\begin{array}{l}\frac{{\partial }^2\Psi }{\partial x^2}=-{\text{Ak}}^{\text{2}}\text{cos(kx-ωt)}\\ \frac{\partial \Psi }{\partial t}=\text{Aωsin(kx-ωt)}\end{array}$

$\begin{array}{l}-\frac{{\hslash }^2}{\text{2m}}\times \left({\text{-Ak}}^{\text{2}}\text{cos(kx-ωt)}\right)=i\hslash \times \text{Aωsin(kx-ωt)}\\ \frac{\hslash {\text{k}}^{\text{2}}}{\text{2m}}=\text{iωtan(kx-ωt)}\end{array}$

Unlike the exponential case, this final expression has a function that depends on both position and time, but the left-hand side is constant. As a result, we can't discover a condition on $k$or $\text{ω}$ that will make the two sides equal, and the sine function as a solution is no different.