Normalize the wave function

  • #1
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Homework Statement


Normalize the wave function ,[itex]\psi(x)[/itex], where [tex]\psi(x)=\frac{1}{1+ix}[/tex].

Homework Equations




The Attempt at a Solution


[tex]\langle\psi\mid\psi\rangle= \int_{-\infty}^{\infty}\frac{1-ix}{1+x^2}\frac{1+ix}{1+x^2}dx=\int_{-\infty}^{\infty}\frac{1}{1+x^2}=\left. arctan(x)\right|_{-\infty}^{\infty}=0[/tex] since arctan(x) is an odd function.
Now I know a normalized wavefunction satisfies [tex]\langle\psi\mid\psi\rangle=1[/tex].
I dont know how to manipulate the given wavefunction in order to satisfy the condition.
 

Answers and Replies

  • #2
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The problem is, only when the integrand is odd, you can say the symmetric integral is zero. But in your case, the integrand is even!
Here's how it should be:
[itex]
\int_{-\infty}^\infty \frac{1}{1+x^2}dx=2 \int_{0}^\infty \frac{1}{1+x^2}dx=2 \tan^{-1}(x) |_0^\infty=(2k+1)\pi
[/itex]
I should say I don't know what k to choose or even its important what k to choose or not!!!
 
  • #3
Orodruin
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The integrand is continuous and so must the primitive function be. Use this to fix k.
 
  • #4
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Well, this is what I think about k.
When people want to define [itex] \tan^{-1}(x) [/itex], they restrict the domain of [itex] \tan(x) [/itex] to [itex] (-\frac \pi 2,\frac \pi 2) [/itex] to make it one-to-one. So the range of [itex] \tan^{-1}(x) [/itex] is [itex] (-\frac \pi 2,\frac \pi 2) [/itex] and so [itex]\lim_{x\rightarrow \infty} \tan^{-1}(x)=\frac \pi 2 [/itex].
Also it seems to me that the normalization constant doesn't have that much physical significant!
 
  • #5
Orodruin
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In this particular problem, it is fine to pick any branch of arctan as any branch is a primitive function to the integrand (I could add an arbitrary constant on top to boot!). The only important thing is that you evaluate both limits of the integral in the same branch.

The normalisation constant will be important when you start evaluating expectation values, unless you want to write
$$
\langle x\rangle = \frac{\langle \psi| x |\psi\rangle}{\langle \psi |\psi\rangle}
$$
everywhere ... Better to just have computed ##\langle \psi |\psi\rangle## once and for all.
 
  • #6
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Orodruin said:
In this particular problem, it is fine to pick any branch of arctan as any branch is a primitive function to the integrand (I could add an arbitrary constant on top to boot!). The only important thing is that you evaluate both limits of the integral in the same branch.
Yeah, that's the right way of looking at it. I missed this in my first post. Considering this, [itex] \tan^{-1}(x)|_0^\infty [/itex] is equal to [itex] \frac \pi 2 [/itex] anyway!

Orodruin said:
The normalisation constant will be important when you start evaluating expectation values, unless you want to write
⟨x⟩=⟨ψ|x|ψ⟩⟨ψ|ψ⟩

everywhere ... Better to just have computed ⟨ψ|ψ⟩ once and for all.
But if we choose [itex] |\phi\rangle=\frac{|\psi\rangle}{\sqrt{\langle \psi|\psi\rangle}}[/itex] to be our wave function, then the normalization constant is not important.
 
Last edited:
  • #7
Orodruin
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Well, then you have just normalised it using the normalisation constant ##1/\sqrt{\langle \psi|\psi\rangle}## ... This is the entire point of computing the normalisation constant.
 
  • #8
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Well, then you have just normalised it using the normalisation constant ##1/\sqrt{\langle \psi|\psi\rangle}## ... This is the entire point of computing the normalisation constant.
Yeah, I know what you mean. But I was saying that if it happens that the normalization integral can give multiple values, as I wrongly thought is the case here, then its not important which one to choose.
 

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