Solutions of of the time-independent Schrodinger eq.

[xago]

Homework Statement

Supposed that \psi₁ and \psi₂ are two different solutions of the TISE with the same energy E.

a) show that \psi₁ + \psi₂ is also a solution with energy E.
b) show that c*\psi₁ is also a solution with energy E.

Homework Equations

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

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 into 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

 

[fzero]

Homework Statement

Supposed that \psi₁ and \psi₂ are two different solutions of the TISE with the same energy E.

a) show that \psi₁ + \psi₂ is also a solution with energy E.
b) show that c*\psi₁ is also a solution with energy E.

Homework Equations

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

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 into 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

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.

 

[xago]

\left(-\frac{\hbar}{2m}\nabla^2 + V(r) \right) \psi_1(r) = E*\psi₁

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

 

[fzero]

\left(-\frac{\hbar}{2m}\nabla^2 + V(r) \right) \psi_1(r) = E*\psi₁

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

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)).

 

[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)

 

[fzero]

\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.

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

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

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

 

[vela]

\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.

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

 

[xago]

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

??

 

[vela]

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*}<br /> \left[-\frac{\hbar}{2m}\nabla^2 + V(r) \right] \left(\psi_1(r)+\psi_2(r)\right) &amp;= \cdots \\<br /> &amp; = \cdots \\<br /> &amp;&amp; \vdots \\<br /> &amp;= E(\psi_1(\vec{r})+\psi_2(\vec{r}))<br /> \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.

 

[xago]

I understand now, thanks for your help!