Wavefunction conjugate

  • Thread starter CyberShot
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  • #1
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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?
 

Answers and Replies

  • #2
G01
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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:

[tex](1-e^{ikx})(1-e^{-ikx})=1-(e^{ikx}+e^{-ikx})+ e^{ikx}e^{-ikx}=2-2cos(kx)[/tex]
 
  • #3
133
2
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:

[tex](1-e^{ikx})(1-e^{-ikx})=1-(e^{ikx}+e^{-ikx})+ e^{ikx}e^{-ikx}=2-2cos(kx)[/tex]
Of course!

Thanks, and sorry for being an idiot. -_-
 
  • #4
133
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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
 
  • #5
1,006
105
Almost. In general, for some operator A,

[tex]<A> = \int \psi^* A \psi dx[/tex]

and [tex]p = -i \hbar \frac{d}{dx}[/tex]

so [tex]p^2 = (-i \hbar \frac{d}{dx})(-i \hbar \frac{d}{dx}) = -\hbar^2 \frac{d^2}{dx^2}[/tex]
 
  • #6
133
2
Almost. In general, for some operator A,

[tex]<A> = \int \psi^* A \psi dx[/tex]

and [tex]p = -i \hbar \frac{d}{dx}[/tex]

so [tex]p^2 = (-i \hbar \frac{d}{dx})(-i \hbar \frac{d}{dx}) = -\hbar^2 \frac{d^2}{dx^2}[/tex]
Why isn't it just

[tex]p^2 = (p)(p) = (-i \hbar \frac{d}{dx})(-i \hbar \frac{d}{dx}) = (-i \hbar \frac{d}{dx})^2 [/tex]

?

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

if so, how come [tex]< x^2 > = x \\int \\ psi^* \\psi dx[/tex]

is [tex]x[/tex] not an operator?
 
Last edited:
  • #7
1,006
105
Why isn't it just

[tex]p^2 = (p)(p) = (-i \hbar \frac{d}{dx})(-i \hbar \frac{d}{dx}) = (-i \hbar \frac{d}{dx})^2 [/tex]

?

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 [tex]< x^2 > = x \\int \\ psi^* \\psi dx[/tex]

is [tex]x[/tex] not an operator?
Your tex seems a bit messed up; the correct equation is

[tex]<x^2> = \int \psi^* x^2 \psi dx[/tex]
 

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