# How does one conjugate psi(x,0)

1. Feb 14, 2006

### Noone1982

I am perplexed. My professor told me that I can conjugate:

$\psi \left( x,0 \right)$

However, I am confused. There is no time dependence, there is no i. It is like saying

z = a + 0i
z* = a - 0i

It is rather useless to even say that and it seems also useless to conjugate something with no i in it. Am I missing something fundamental?

2. Feb 14, 2006

### Hurkyl

Staff Emeritus
Just because it's useless doesn't mean you can't do it.

But I don't see why you think $\psi(x, 0)$ is a purely real number. Why can't it be a complex number?

3. Feb 14, 2006

### Noone1982

Why do I think it is a purely real number? It has no i in it. That probably seems like a silly answer as your post implies. Could you please enlighten me?

4. Feb 14, 2006

### Hurkyl

Staff Emeritus
$\psi$ is a function. There's no reason it cannot take complex values.

For example, it might be defined as:

$$\psi(a, b) := 3 + 4i$$

or

$$\psi(a, b) := b + a i$$

or

$$\psi(a, b) := e^{i (b - a)}$$

or any number of other things.

5. Feb 14, 2006

### Noone1982

I realize that, but my professor says this can be conjugated:

$\psi \left( x,0 \right)\; =\; \sqrt{\frac{2}{a}}\sin \left( \frac{2\pi x}{a} \right)$

I see no "i" in that. I therefore see no way to conjugate it.

6. Feb 14, 2006

### Hurkyl

Staff Emeritus
Where a is meant to be positive, real number? (And, I guess, I'm assuming $\psi$ takes two real arguments)

Of course you can conjugate that; you had it right in your opening post. Just because it's "useless" doesn't mean you can't do it.

7. Feb 14, 2006

### Noone1982

Yes, a is positive, real number. So:

$\psi \cdot \psi \; =\; \sqrt{\frac{2}{a}}\sin \left( \frac{2\pi x}{a} \right)\sqrt{\frac{2}{a}}\sin \left( \frac{2\pi x}{a} \right)\; =\; \frac{2}{a}\sin ^{2}\left( \frac{2\pi x}{a} \right)$

8. Feb 14, 2006

### James R

The congugate of a real number, by the way, is the same number.

9. Feb 14, 2006

### Noone1982

James, yes that is my point. Why did my professor act startled and perhaps a bit angry when I told him it seemed useless to conjugate that psi?

10. Feb 14, 2006

### Hurkyl

Staff Emeritus
Because it's not really "useless". Just think: if you weren't allowed to conjugate purely real values, then you'd have to say:

The amplitude of the wave function, $|\psi|^2$, is given by $\psi^* \psi$, except at the values where $\psi$ is purely real, in which case it is given by $\psi^2$.

Just think of having to attach these caveats to just about everything you say... and when doing problems, having to break every step into two cases, one where the value happens to be purely real, and one where it isn't!

Everything would be a horrible nightmare! Aaaah!