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Normalization of step potential eigenfunction 
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#1
Dec612, 07:22 AM

P: 7

Hi,
I am stuck with a problem which effectively boils down to this: Given the eigenstates of a Hamiltonian with a step potential in the x direction [itex]H=\hbar^2/2m \nabla^2 + V_0 \Theta(x)[/itex] [itex]\psi(q)_{in}=cos(qx)\frac{\sqrt{K_{V_0}^2q^2}}{q}sin(qz) \qquad x<0[/itex] [itex]\psi(q)_{out}=e^{\sqrt{K_{V_0}^2q^2}x} \qquad x>0[/itex] where [itex]K_\xi[/itex] refers to [itex]\sqrt{2m \xi /\hbar^2}[/itex] and q is the momentum quantum number labelling the eigenstate with eigenenergy [itex]\frac{\hbar^2q^2}{2m}[/itex]. How do I normalize this? It seems to me that I should be able to get normalized states of the form [itex]\int_{\infty}^{0}\psi(q)_{in}\psi(q')_{in}^* dx+\int_{0}^{∞}\psi(q)_{out}\psi(q')_{out}^* dx=\delta(qq')[/itex] but I cannot get anything resembling a delta function out of the left hand side when I plug the eigenfunctions into this equation. Where am I going wrong? Shouldn't it be possible to get orthonormal states for this scenario? Thanks... 


#2
Dec612, 08:02 AM

PF Gold
P: 1,148

It seems you are trying to get [itex]\delta(qq')[/itex] by integrating over [itex] q[/itex]. That does not seem to be a good idea.
Usual approach is this: in case you have two functions [itex]f_E(q),f_{E'}(q)[/itex] that are eigenfunctions of the Hamiltonian for energies [itex]E, E'[/itex], it should be possible to normalize them in such way that their scalar product is delta function in energy: [tex] \int_{\infty}^{\infty} f^*_E(q) f_{E'}(q) dq = \delta(EE') [/tex] 


#3
Dec612, 01:37 PM

P: 7

Oh... no the integration is over space... i should have mentioned that... i'll edit the post right away...



#4
Dec712, 09:46 AM

PF Gold
P: 1,148

Normalization of step potential eigenfunction
Ah I see, [itex]q[/itex] is a wave number. Then it should be possible to get
[tex] \int_{\infty}^{\infty} \psi_q(x) \psi_{q'}(x) dx = \delta(qq') [/tex] The way to verify this is so is to integrate the lefthand side (I'll denote it by [itex]F(q,q')[/itex] ) with some continuous function of [itex]q[/itex]: [tex] \int_{\infty}^{\infty} F(q,q') f(q) dq [/tex] If the result is [itex]f(q')[/itex] multiplied by some constant, then one can normalize the original functions [itex]\psi_q(x)[/itex] so as to make [itex]F[/itex] exactly into delta distribution. Concerning the normalization, you may find this useful: http://www.physicsforums.com/showthr...ighlight=delta 


#5
Dec712, 10:00 AM

Mentor
P: 11,589

If x is not bounded in some way, you won't get normalizable eigenstates. Any prefactor (!=0) leads to an undefined integral for ##\psi(q)_{in}##.



#6
Dec912, 10:28 AM

P: 7

@Jano L.: Well F(q,q') unfortunately doesnt resemble a delta function, hence the problem.
@mfb: Thanks... I found something relevant in R Shankar's QM text. He deals with a similar problem of normalization by setting the constant such that the incoming wave is normalized in the usual delta function sense. His rationale in doing so is that this reproduces a wave pulse located initially very far from the step will be properly reproduced. I dont really buy his argument because it seems to me he's making two conflicting assumptions, one that the wave packet is strongly peaked in momentum space and two, that its far enough from the step that only the incoming wave is relevant at the initial time. 


#7
Dec912, 11:27 AM

P: 356

The right handed portion is not a physical wave function anyways, I don't see the point of this exercise.



#8
Dec1012, 04:13 AM

P: 7

and by the right handed portion you mean?



#9
Dec1012, 11:44 AM

P: 356

the wave function for x<0 is not a physical wave function



#10
Dec1012, 01:22 PM

Sci Advisor
P: 1,190

It's a little easier to verify the orthogonality of two of these wave functions with different values of q. On the left side, you need to insert a convergence factor like exp[epsilon x], and take epsilon to zero after doing the integral. Then adding the contribution of the right side, you should get zero.
If the two values of q are the same, then you're integrating the square of a function, and it's pretty clear that the value is infinite. So, this tells you that the result is some qdependent number times a delta function. If you want the qdependent number, you have to be more careful, retain epsilon, and then compute the integral over one of the q's, and then take epsilon to zero. EDIT: OK, I actually checked the orthogonality (doing the math with Mathematica), and it worked. With the convergence factor, the cos*cos and sin*sin terms on the left both integrate to zero (in the limit epsilon > 0); the sin*cos and cos*sin terms do not, and are exactly canceled by the contribution of the rightside integral (after some algebra). 


#11
Dec1012, 11:23 PM

P: 7

@avodyne: Thanks... thats a good idea... and it gives me some confidence that the normalization CAN be done... it would be great if you could give me the reference to where you first came across this trick, if you happen to be remember...
however the orthogonality here strictly holds only for q and q' are unrelated not unequal. For instance if you put q'=q you can check that in the limit the integral again diverges and the said terms do not cancel each other. What i'm concerned about is that infinity at q=q' doesnt give me δ(qq') it could also be δ(q^2q'^2) or δ(any function with roots at q' and q'). Thanks again for the convergence factor idea... i'm working on it some more... 


#12
Dec1112, 11:03 AM

P: 7

ok thanks, i managed to get it normalized using the convergence factor and the definition of a delta function as a limiting lorentzian...
the normalization constant i get is √(q/∏V_0) i'll edit this post to show the steps if anybody faces the same problem later... 


#13
Dec1112, 01:05 PM

Sci Advisor
P: 1,190




#14
Dec1512, 07:55 AM

P: 7

@Avodyne:Yes, that is true... my bad...
A clarification on the earlier post: The integral evaluates to [itex]\frac{\pi m V_0}{\hbar^2 q^2} \delta(qq')[/itex] which is equivalent to [itex]\frac{\pi V_0}{q} \delta(E_qE_{q'})[/itex] so the normalization constant would be [itex]\sqrt{\frac{\pi m V_0}{\hbar^2 q^2}}[/itex] if we're integrating over quantum numbers in our subsequent calculations or [itex]\sqrt{\frac{\pi V_0}{q}}[/itex] if we're integrating over energy. 


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