Linear equations and superposition of wavefunctions

[dyn]

Hi. I have read many times that the Schrödinger equation is a linear equation and so if Ψ₁ and Ψ₂ are both solutions to the equation then so is Ψ₁ + Ψ₂. Is this use of the word linear the same as generally used for differential equations ? As the Schrödinger equation is also an eigenvalue equation for the Hamiltonian.

My main confusion is why a superposition of wavefunctions such as e^ikx + e^-ikx is not a solution to the momentum eigenvalue equation as this also looks like a linear equation ?
Thanks

 

[S.G. Janssens]

Is this use of the word linear the same as generally used for differential equations ?

Yes.

My main confusion is why a superposition of wavefunctions such as eikx + e-ikx is not a solution to the momentum eigenvalue equation as this also looks like a linear equation ?

Because the summands correspond to different eigenvalues.

 

[dyn]

Thanks. So the condition that the superposition of wavefunctions also solves the eigenvalue equation is that it is both linear and the wavefunctions also have the same eigenvalues ?

 

[S.G. Janssens]

Thanks. So the condition that the superposition of wavefunctions also solves the eigenvalue equation is that it is both linear and the wavefunctions also have the same eigenvalues ?

Yes. (Eigenvalue equations are usually linear, by the way. There does exist something called a "nonlinear eigenvalue problem", but that is not relevant here.)

The SE is a partial differential equation, but for your understanding perhaps consider the setting of a linear ordinary differential equation. In the simplest case (constant coefficients, homogeneous) a linear ODE can be written as
$$
\frac{dx}{dt}(t) = Ax(t) \qquad (\ast)
$$
(Think, for example, about the simple harmonic oscillator.) If $x_1$ and $x_2$ are both solutions, then their sum is also a solution and hence satisfies $(\ast)$. This is superposition.

Regarding the corresponding eigenvalue problem, if $\lambda$ and $\mu$ are distinct eigenvalues of $A$ with eigenvectors $u$ and $v$, then $u + v$ is not an eigenvector of $A$. (Try it out.) In contrast, if $\lambda = \mu$ then it is true that
$$
A(u + v) = \lambda(u + v)
$$
In case of the SE similar remarks hold, but the matrices are replaced by (unbounded) operators that act on function spaces instead of $\mathbb{R}^n$ or $\mathbb{C}^n$. Does this clarify matters a bit?

 

[dyn]

Yes. Thanks for your help.