Should the Normalization Constant be Positive or Complex?

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SUMMARY

The normalization constant A in the context of the time-independent Schrödinger equation can be complex, represented as A = √(2/a) e^(iφ), where φ is an arbitrary real phase. While Griffiths emphasizes that only the magnitude |A| = √(2/a) is necessary for normalization, the phase does not affect observable quantities, leading to the common practice of selecting φ = 0 for simplicity. The integral normalization condition is established as 1 = ∫ |ψ(x)|² dx, confirming that A's magnitude suffices for calculations involving probabilities and expectation values.

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  • Understanding of the time-independent Schrödinger equation
  • Familiarity with complex numbers and their magnitudes
  • Knowledge of wave functions in quantum mechanics
  • Basic integration techniques in calculus
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In finding solutions to the time independent Schrödinger equation we have to normalize \psi to find the constant A. So we get \int_{0}^{a} |A|^{2} sin^{2}(kx) dx = |A|^2 \frac{a}{2}=1

For A we then get |A|^2 = \frac{2}{a}. Griffiths says that this only determines the magnitude of A but it's simplest to pick the positive real root. I know how to work with complex numbers generally but I'm a little confused as to what the imaginary root would be anyway. It looks to me like A= \pm \sqrt{ \frac{2}{a} }. Since we aren't square rooting a negative I don't see where the i comes in. Also, I know the magnitude of a complex number is real, so are we saying that A itself could be complex? I don't want to make any assumptions about this. Thanks!
 
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You have thought it right, 'A' can be complex indeed. In fact, A=\sqrt\frac 2 a\,e^{i\phi} satisfies the the normalization constraint for any real \phi. Mind that the phase of the wave function is arbitrary.
 
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csopi said:
You have thought it right, 'A' can be complex indeed. In fact, A=\sqrt\frac 2 a\,e^{i\phi} satisfies the the normalization constraint for any real \phi. Mind that the phase of the wave function is arbitrary.

Cool, thanks!
 
I was passing here by coincidence, and after reading the post I had few questions indeed:
If A is just the magnitude of the normalization constant, then will it differ if we took the complex complete part ? I mean, why do we always take the positive part?

Also why does A only represent the magnitude? Where does this appear from?
I always solve the particle in a box case taking the solution of A as granted without noticing that it is just a magnitude !
 
I was passing here by coincidence, and after reading the post I had few questions indeed:
If A is just the magnitude of the normalization constant, then will it differ if we took the complex complete part ? I mean, why do we always take the positive part?

Also why does A only represent the magnitude? Where does this appear from?
I always solve the particle in a box case taking the solution of A as granted without noticing that it is just a magnitude !
A is a constant that we have added after integration Sorry I do not mean added in the sense of + but more in the sense of incoperated in our equation!
 
Here is the equation:

We start of with:

1=∫ abs(psi(x))^2 dx

substitute psi(a)= A sin((npix)/a))

1=abs(A)^2∫ abs( sin((npix)/a)))^2 dx

so A=(2/a)^1/2

where I have simply replaced the k of in the top most post with sinnpi/a
 
Ibrahim Hany said:
If A is just the magnitude of the normalization constant, then will it differ if we took the complex complete part ? I mean, why do we always take the positive part?

Strictly speaking, the normalization constant can be as csopi noted, letting ##\sqrt{2/a}## be real and using any value for φ. However, when you use the wave function to calculate anything that is physically observable (e.g. a probability or an expectation value), φ always cancels out. Therefore, for simplicity, we usually choose φ = 0.
 

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