Question

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.

Short answer

We can't come up with a condition on $\text{ω}$or k that will make the proposed solution valid in this situation under any circumstances.

Step-by-step solution

Step 1:Schrodinger equation.

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)

Schrodinger equation for the given wavefunction

The wavefunction given is

$\Psi =\text{Acos(kx-ωt)}$……………….(2)

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

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

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

Conclusion.

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.