# Solutions of of the time-independent Schrodinger eq.

1. Oct 2, 2010

### xago

1. The problem statement, all variables and given/known data

Supposed that $$\psi$$1 and $$\psi$$2 are two different solutions of the TISE with the same energy E.

a) show that $$\psi$$1 + $$\psi$$2 is also a solution with energy E.
b) show that c*$$\psi$$1 is also a solution with energy E.

2. Relevant equations

TISE: (-$$\hbar$$/2m)*$$\nabla$$^2*$$\psi$$(r) +V(r)*$$\psi$$(r) = E*$$\psi$$(r)

3. The attempt at a solution

My first thought was to solve by substitution but since it doesn't give us the 2 solutions as a function I can't do that. Should I just find 2 different solutions and plug them in to the TISE to see if they satisfy? I feel I am missing a rule or a postulate to solve this with.

P.S. I am very new to QM, be gentle

2. Oct 2, 2010

### fzero

If $$\psi_1$$ is a solution of the TISE with energy E, what must

$$\left(-\frac{\hbar}{2m}\nabla^2 + V(r) \right) \psi_1(r)$$

equal? You don't need to know the functional form of $$\psi_1(r)$$ to answer this.

3. Oct 2, 2010

### xago

$$\left(-\frac{\hbar}{2m}\nabla^2 + V(r) \right) \psi_1(r)$$ = E*$$\psi$$1

I don't see what this proves though, isn't the equation just re-arranged?

4. Oct 2, 2010

### fzero

Well it's reiterating what information you're given. The problem wants you to compute

$$\left(-\frac{\hbar}{2m}\nabla^2 + V(r) \right) (\psi_1(r)+\psi_2(r))$$

and

$$\left(-\frac{\hbar}{2m}\nabla^2 + V(r) \right) ( c \psi_1(r)).$$

5. Oct 2, 2010

### xago

$$\left(-\frac{\hbar}{2m}\nabla^2 + V(r) \right) \psi_1(r) = E\psi_1(r)$$
$$\left(-\frac{\hbar}{2m}\nabla^2 + V(r) \right) \psi_2(r) = E\psi_2(r)$$

therefore $$\left(-\frac{\hbar}{2m}\nabla^2 + V(r) \right) (\psi_1(r)+\psi_2(r)) = E(\psi_1(r)+\psi_2(r)) ?$$

It just doesn't feel like I'm proving anything.

Edit*Perhaps I just have to show that E[f(x)] is a linear operator? (where f(x) would be $$\psi$$)

Last edited: Oct 2, 2010
6. Oct 2, 2010

### fzero

It's not a really deep problem. You're just verifying that the wave equation is linear.

E is a scalar, I don't think you'd be expected to prove that multiplication by a scalar is a linear operation.

7. Oct 2, 2010

### vela

Staff Emeritus
You're supposed to show how what's after "therefore" follows from the first two lines.

8. Oct 3, 2010

### xago

$$\left[\left(-\frac{\hbar}{2m}\nabla^2 + V(r) \right) \psi_1(r) = E\psi_1(r)\right]+$$ $$\left[\left(-\frac{\hbar}{2m}\nabla^2 + V(r) \right) \psi_2(r) = E\psi_2(r)\right]$$
$$\left(-\frac{\hbar}{2m}\nabla^2 + V(r) \right) \left(\psi_1(r)+\psi_2(r)\right) = E\psi_1(r) + E\psi_2(r)$$
Since they both have the same E, it can be factored out:
$$=\left(-\frac{\hbar}{2m}\nabla^2 + V(r) \right) \left(\psi_1(r)+\psi_2(r)\right) = E\left(\psi_1(r) + \psi_2(r)\right)$$

??

Last edited: Oct 3, 2010
9. Oct 3, 2010

### vela

Staff Emeritus
It's clear you understand conceptually what's going on, but you want to write it down in a clear, organized way. A lot of students have trouble writing down a proof because it seems so obvious.

To show $\psi_1(\vec{r})+\psi_2(\vec{r})$ is a solution means to show that

$$\left[-\frac{\hbar}{2m}\nabla^2 + V(\vec{r}) \right] \left(\psi_1(\vec{r})+\psi_2(\vec{r})\right) = E(\psi_1(\vec{r})+\psi_2(\vec{r}))$$

For problems like this, it's good to start with one side of the equation, apply concepts one at a time, and show what you get after each step until you end up at the other side of the equation.

\begin{align*} \left[-\frac{\hbar}{2m}\nabla^2 + V(r) \right] \left(\psi_1(r)+\psi_2(r)\right) &= \cdots \\ & = \cdots \\ && \vdots \\ &= E(\psi_1(\vec{r})+\psi_2(\vec{r})) \end{align*}

In the first step, you'd use the linearity of

$$\left[-\frac{\hbar}{2m}\nabla^2 + V(r) \right]$$

In the next step, you'd use the fact that $\psi_1$ and $\psi_2$ are solutions with energy E, and so on.

10. Oct 3, 2010

### xago

I understand now, thanks for your help!