Find Complex Conjugate of Wave Function in QM Mechanics Book

[CyberShot]

I saw in a QM mechanics book the following wave function:


psi(x) = A*[1 - e^(ikx)]

what is the complex conjugate of this wave function?

isnt it just psi*(x) = A*[1 - e^(-ikx)]

but when you multiply psi(x) by psi*(x) shouldn't you get a real value?

How come I don't?

 

[G01]

Check again. You do get a real function. Remember Euler's identity. After you multiply everything out, you can rewrite the remaining exponentials in terms of trig functions:

$$(1-e^{ikx})(1-e^{-ikx})=1-(e^{ikx}+e^{-ikx})+ e^{ikx}e^{-ikx}=2-2cos(kx)$$

 

[CyberShot]

Check again. You do get a real function. Remember Euler's identity. After you multiply everything out, you can rewrite the remaining exponentials in terms of trig functions:

$$(1-e^{ikx})(1-e^{-ikx})=1-(e^{ikx}+e^{-ikx})+ e^{ikx}e^{-ikx}=2-2cos(kx)$$

Of course!

Thanks, and sorry for being an idiot. -_-

 

[CyberShot]

I have another question.

What would the average of the square of the momentum look like?

I know that <p> = integral[ psi*(x) (hbar/i) d/dx (psi(x)) ] dx

how would you determine <p^2> ?

is it just

<p^2> = integral[ psi*(x) (hbar/i) second derivative (psi(x)) ] dx

 

[The_Duck]

Almost. In general, for some operator A,

$$<A> = \int \psi^* A \psi dx$$

and $$p = -i \hbar \frac{d}{dx}$$

so $$p^2 = (-i \hbar \frac{d}{dx})(-i \hbar \frac{d}{dx}) = -\hbar^2 \frac{d^2}{dx^2}$$

 

[CyberShot]

Almost. In general, for some operator A,

$$<A> = \int \psi^* A \psi dx$$

and $$p = -i \hbar \frac{d}{dx}$$

so $$p^2 = (-i \hbar \frac{d}{dx})(-i \hbar \frac{d}{dx}) = -\hbar^2 \frac{d^2}{dx^2}$$

Why isn't it just

$$p^2 = (p)(p) = (-i \hbar \frac{d}{dx})(-i \hbar \frac{d}{dx}) = (-i \hbar \frac{d}{dx})^2$$

?

Edit: Hmm, is it because p is not an ordinary variable in that its an operator?

if so, how come $$< x^2 > = x \\int \\ psi^* \\psi dx$$

is $$x$$ not an operator?

 

[The_Duck]

Why isn't it just

$$p^2 = (p)(p) = (-i \hbar \frac{d}{dx})(-i \hbar \frac{d}{dx}) = (-i \hbar \frac{d}{dx})^2$$

?

Edit: Hmm, is it because p is not an ordinary variable in that its an operator?

Your form is equivalent to my form; I have simply distributed the exponent. Indeed p is an operator and not a number: p^2 is the operator "apply p twice" (and the action of p is to differentiate the wave function and multiply by -i*h-bar). Often one puts hats on operators to distinguish them from numbers.

if so, how come $$< x^2 > = x \\int \\ psi^* \\psi dx$$

is $$x$$ not an operator?

Your tex seems a bit messed up; the correct equation is

$$<x^2> = \int \psi^* x^2 \psi dx$$