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Normalizing a wave function  how the integration is done? 
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#1
Feb1813, 01:31 PM

P: 205

I have been searching for an anwser everywhere, but i can't seem to understand something. In this topic (you don't need to read it) i managed to find out that "we can calculate normalisation factor ##\Psi_0## of a wavefunction ##\Psi## if we integrate probability ##\Psi^2## over some volume and equate it to 1". Hence:
[itex] \int\limits_{V} \Psi^2 \, \textrm{d}V= 1 [/itex] Now how exactly do we integrate this? Please be specific, because in the post i linked to i got an anwser that the result of integration is [itex] \int\limits_{V} \Psi^2 \, \textrm{d}V = \psi_0^2 V [/itex] and i don't know how is this possible. Maybee my interpretation of this is wrong and this is why below i am supplying you with my interpretation. My interpretation: For the sake of clarity i will just choose some wave function for example ##\Psi = \Psi_0 \sin(\omega t  kx)##. I chose this as it is similar to an already known wave function of a sinusoidal wave ##A = A_0 \sin(\omega t  kx)## which i have been using allover wave physics. I don't know if i am allowed to choose the ##\Psi## like that because for now i don't know enough to know what i am alowed/not allowed to do in QM. If i understand this ##\Psi_0## in a vave function ##\Psi = \Psi_0 \sin(\omega t  kx)## is the normalisation factor i am seeking? (Please confirm this). So now i take an integral of the wavefunction and equate it to 1: [itex] \begin{split} \int \limits^{}_{V} \left\Psi \right^2 \, \textrm{d} V &= 1\\ \int \limits^{}_{V} \big\Psi_0 \sin (\omega t  kx) \big^2 \, \textrm{d} V &= 1\\ &\dots \end{split} [/itex] I get lost at the spot where i wrote down "##\dots##". I really don't know how to get ##\psi_0^2 V## as a result of integration. 


#2
Feb1813, 01:55 PM

Sci Advisor
Thanks
P: 4,160

Of course the answer you get depends on what wavefunction you start with. If, as in the thread you quoted, you choose ψ = ψ_{0} exp^{i(kx  ωt)}, this is a traveling plane wave. Its probability density is ψ^{2} = ψ*ψ = ψ_{0}^{2} = const, so the integral gives you ψ_{0}^{2} V.
If, on the other hand you choose ψ = ψ_{0} sin(kx  ωt), this is a standing wave. Its probability density is ψ_{0}^{2} sin^{2}(kx  ωt) which is not constant, and you'll get a different integral. 


#3
Feb1813, 02:21 PM

P: 205

[itex] \Psi^2 = \left \Psi_0 e^{i(\omega t  kx)} \right ^2 = \overline{\Psi} \Psi = \underbrace{\Psi_0 e^{i(\omega t  kx)}}_{conjugate} \Psi_0 e^{i(\omega t  kx)} = {\Psi_0}^2 \frac{\Psi_0 e^{i(\omega t  kx)}}{\Psi_0 e^{i(\omega t  kx)}} = \Psi_0^2 [/itex] Is my calculation legit? Please confirm. And please tell me how do i know that ##\Psi_0^2## is a constant and i should therefore integrate it as such? 


#4
Feb1813, 05:28 PM

Sci Advisor
Thanks
P: 4,160

Normalizing a wave function  how the integration is done?
Yes, that's correct. And a plane wave will have a constant amplitude, so ψ_{0} will be constant.



#5
Feb1913, 12:10 AM

P: 205

But how can i calculate integral for wave function ##\Psi = \Psi_0 \sin(\omega t  kx)##. Could i simplify this by stating that the wave is travelling in ##x## direction and only integrate over ##x## or should i use a triple ##\iiint## and integrate over ##x##, ##y## and ##z##? I need some advice on how to calculate this integral:
[itex] \int\limits_V \left \Psi_0 \sin(\omega t  kx) \right^2 \, \textrm{d} V [/itex] 


#6
Feb1913, 12:18 AM

Mentor
P: 11,777

A plane wave that extends to infinity in all directions cannot be normalized in the usual sense, i.e.
$$\int_{all space} {\Psi^* \Psi dx dy dz} = 1$$ (in three dimensions) Such waves are not physically realistic. The amplitude of a realworld wave function has to drop off to zero as we go "far enough" away from the center of the system. This leads to the concept of wave packets. Nevertheless, we often talk about plane waves as convenient idealizations or approximations over small regions of space. 


#7
Feb1913, 02:02 AM

P: 205




#8
Feb2013, 04:10 AM

P: 362

I think it would be best to go back and look at the fundamentals of QM, once you have, the answers to your questions will become obvious.
The answer to your question is complicated and i don't know where to start explaining. 


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