# Proving $\psi$ can be real or not, even or odd

1. Feb 28, 2015

### Maylis

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

2. Relevant equations
Equation 1.20
$$\int_{-\infty}^{\infty} \mid\Psi (x,t)\mid^{2} dx = 1$$

Equation 2.5
$$- \frac {\hbar^{2}}{2m} \frac {d^{2} \psi}{dx^{2}} + V \psi = E \psi$$

3. The attempt at a solution
(a)
$$\Psi (x,t) = \psi e^{-i(E_{0} + i \Gamma)t / \hbar}$$
$$\mid\Psi (x,t)\mid^{2} = \Psi^{*} \Psi = \mid \psi \mid^{2} e^{2 \Gamma t/ \hbar}$$
Using equation 1.20
$$e^{2 \Gamma t/ \hbar} \int_{-\infty}^{\infty} \mid \psi \mid^{2} dx = 1$$
This equality only holds true for all time if $\Gamma$ is zero.

(b) & (c) I am having a hard time understanding what to do, but (c) looks a little easier (probably since it's a shorter question) so I will try it first

Using equation 2.5, it seems they want me to plug in
$$- \frac {\hbar^{2}}{2m} \frac {d^{2} \psi(-x)}{dx^{2}} + V \psi(-x) = E \psi(-x)$$

But now what??

2. Feb 28, 2015

### BvU

In both b) and c) they want you to prove what's suggested by "If..." in the hint. So: not plug it in, but show that if $\Psi(x)$ satisfies 2.5, then $\Psi(-x)$ does so too.

3. Feb 28, 2015

### Maylis

How do I begin to show that? I'm not sure what to start with

4. Feb 28, 2015

### BvU

Well, E is a number, V is an even function. What about $d^2\over {dx^2}$ ?

5. Feb 28, 2015

### Maylis

It's an operator

edit: I'm sorry, the hints you are giving are just to me disjointed and I really don't know where to go with this. The best thing I can think of is to solve the schrödinger equation with the argument of -x.

Last edited: Feb 28, 2015
6. Feb 28, 2015

### BvU

Don't go solving. Just swap x and -x (perhaps that's what you meant to say), knowing $\psi(x)$ is a solution.

7. Feb 28, 2015

### Maylis

So I just say

$$- \frac {\hbar^{2}}{2m} \frac {d^{2} \psi(-x)}{dx^{2}} + \psi(-x) [V(-x) - E] = 0$$

So I know $V(-x) = V(x)$, but how does this prove $\psi(x) = \psi(-x)$? I mean, I suppose $V - E$ will have to be the same for either an argument of $x$ or $-x$, so for this equation to be equal to zero, then $\psi(x)$ will have to be the same as $\psi(-x)$ in order to maintain equality with equation 2.5. This seems trivial and hand wavy...

Last edited: Feb 28, 2015
8. Feb 28, 2015

### BruceW

no, you can prove it, not just by handwaving. Start by assuming that $\psi (x)$ is a solution, then prove that $\psi (-x)$ is a solution. Hint: substitute $-z=x$ into the original equation.

9. Feb 28, 2015

### BvU

 ... messy edit - lost a huge part.

You can extend to $\psi(x)+\psi(-x)$ which is definitely even and $\psi(x)-\psi(-x)$ which is definitely odd.
About $\psi(x)$ itself you know nothing, except that it satisfies 2.5.

This make sense ?

 too much midnight oil. It does not make sense. Point is that at x and at -x $\psi$ satisfies the same differential equation. But I forgot $\psi$ is complex. So back to thinking about $\partial^2 \over \partial x^2$ for such a function. Would prefer to delete whole post, but Maylis has already responded.

 apparently the impulsive delete can't be compensated with ^z except for the smal part of the post you see in the edit window. Sorry I made a mess of it.

Last edited: Mar 1, 2015
10. Feb 28, 2015

### Maylis

That is somewhat what I was getting at, I just was not able to make the distinction to add them as a linear combination. But I think my reasoning was along the right lines. I was saying if equation 2.5 is equal to zero, and if $V(x) = V(-x)$, then substituting with $\psi (-x)$ has to be zero as well.

I think I am missing a sublety in your answer, how do you know $\psi(x) + \psi(-x)$ is even and $\psi(x) - \psi(-x)$ is odd?

Last edited: Feb 28, 2015
11. Mar 1, 2015

### BvU

Dear Maylis,

Sorry, my post #9 was dumb and I made it even worse when I tried to massage the stupid parts

Nothing subtle about it: + gives the same when swapping x and -x - gives a - sign.

12. Mar 1, 2015

### Maylis

My head hurts, it seems like all I did was just say "it is a solution" because the linear combination is a solution, but how do I know that is a solution? This seems like some sort of circular reasoning to me and nothing has been actually done.

13. Mar 1, 2015

### BruceW

yes. First you must prove that $\psi(-x)$ is a solution. Then after that, you can start thinking about linear combinations. So just start with the first bit. Assuming that $\psi(x)$ is a solution, prove that $\psi(-x)$ is a solution.

14. Mar 1, 2015

### Maylis

I don't understand how to do that. So I will go with your first post and substitute $x = -z$

$$- \frac {\hbar^{2}}{2m} \frac {d^{2} \psi(-z)}{d(-z)^{2}} + [V(-z) - E] \psi(-z) = 0$$

But now what?

Last edited: Mar 1, 2015
15. Mar 1, 2015

### BvU

My head hurts too for a different reason. Feel guilty about both . Fortunately Bruce is here to keep us sane...

Thing to show is that if $\quad \psi(x)$ is a solution $\Rightarrow\quad \psi(-x)$ is a solution too $\quad$, right ? From there, combinations are, so we might as well look at odd and even solutions only when searching for solutions of 5.2. There may well be solutions that are not either odd or even, but those solutions can always be written as linear combinations of odd and even functions.

In other words: if we've shown that if $\psi(x)$ is a solution $\Rightarrow\quad \psi(-x)$ is a solution too, we've shown that for any arbitrary solution the even part and the odd part are both solutions. So in a search for solutions to 2.5 we can confine ourselves to all even and all odd solutions without missing any solution at all.

Now on to showing: perhaps it helps if we swap x for y. Corny. We get $$- \frac {\hbar^{2}}{2m} \frac {d^{2} \psi(y)}{dy^{2}} + V(y) \psi(y) = E \psi(y)$$then we swap y for something else, e.g. -x . Corny too ? We get $$- \frac {\hbar^{2}}{2m} \frac {d^{2} \psi(-x)}{d(-x)^{2}} + V(-x) \psi(-x) = E \psi(-x) \qquad (1)$$and now I hope I have found the right path to show you:

• V is even, so $V(-x) = V(x)$. Then:

• If you remember that ${\partial \over \partial (-x)} = - {\partial \over \partial x}$, then what about
$\partial^2 \over \partial (-x)^2$ in relation to $\partial^2 \over \partial x^2$ ? Right ! Eureka ! So
$$- \frac {\hbar^{2}}{2m} \frac {d^{2} \psi(-x)}{d(x)^{2}} + V(x) \psi(-x) = E \psi(-x)$$​

and this is the same equation for $\psi(-x)$ as 2.5 is for $\psi(x)$.

In spite of the $V(x)$ looking out of place: now you can fill in a value for x to convince yourself that $\psi(-x)$ at x = 5 behaves itself according to the same
differential equation as $\psi(x)$ at x = 5

16. Mar 1, 2015

### Maylis

Okay, so the idea is that if I can manage to get the equation to look the same as 2.5, then it is a solution. One problem I have is that how to know $\frac {d}{d(-x)} = - \frac {d}{dx}$, and thus $\frac {d^{2}}{d(-x)^{2}} = \frac {d^{2}}{dx^{2}}$?

17. Mar 1, 2015

### BvU

You don't have to know it. You only have to see that the equation is unaltered, except for the replacement of $\psi(x)$ by $\psi(-x)$:$$\left [ \frac {\hbar^{2}}{2m} \frac {d^{2} }{dx^{2}} + \bigl ( E-V(x)\bigr ) \right ] \ \psi(-x) = 0$$

18. Mar 1, 2015

### BruceW

yep! specifically, $f(x)$ is a solution exactly in the cases where
$$- \frac {\hbar^{2}}{2m} \frac {d^{2} f(x)}{dx^{2}} + V(x) f(x) = E f(x)$$
And in our case, we are trying to show that $f(x)= \psi(-x)$ is a solution.

try the chain rule.