How do you normalize this wave function?

[R P Stone]

TL;DR Summary

How would the wave function be normalized of an extended state for a particle moving in a delta potential in 1D?

I have a basic question in elementary quantum mechanics:

Consider the Hamiltonian $$H = -\frac{\hbar^2}{2m}\partial^2_x - V_0 \delta(x),$$ where $\delta(x)$ is the Dirac function. The eigen wave functions can have an odd or even parity under inversion. Amongst the even-parity wave functions, one represents a bound state and the rest are extended states making a continuum spectrum. The wave functions for the extended states with energy $E_k = \hbar^2 k^2/2m$ are
$$\psi_k(x) = N_k \left(e^{-ik|x|} + b_k e^{ik|x|}\right), \quad k\geq0$$ with $b_k = \frac{iV_0}{2k}+1$ and $N_k$ is the normalization factor that renders $$\int^\infty_{-\infty} dx ~ \psi^*_{k'}(x) \psi_k(x) = \delta(k-k').$$ My question is, what is $N_k$ (up to an irrelevant phase factor)?

I proceed as follows but run into a problem. First, from Schrodinger equation I could obtain that $$(E_k - E_{k'}) \int^L_{-L} dx ~ \psi^*_{k'}(x) \psi_k(x) = - \frac{\hbar^2}{2m}\int^L_{-L} dx ~ \partial_x \left(\psi^*_{k'}(x)\partial_x \psi_k(x) - \psi_k(x) \partial_x \psi^*_{k'}(x)\right).$$ Here $L$ shall be sent to infinity. Then I substitute the expression of $\psi_k$ and $\psi_{k'}$ and end up with $$\int^L_{-L} dx ~ \psi^*_{k'}(x) \psi_k(x) = 2i|N_k|^2\frac{b_kb^*_{k'} e^{i(k-k')L} - e^{i(k'-k)L}}{k-k'},$$ where I have dropped terms with $e^{\pm i(k+k')L}$, which are supposed to vanish in the limit $L$ going to infinity. I got stuck here. Would anybody show me how the wave functions are orthogonal at first? Thanks.

 

[BvU]

Hello Stone, !

Wave functions $\ e^{ikx}\$ for particles that are eigenstates of the $p$ operator can't be normalized in $x$-space (and vice versa). A property of Fourier transforms.

Still useful to study topics like scattering:
https://en.wikipedia.org/wiki/Delta_potential#Scattering_(E_>_0)

 

[R P Stone]

Hi, thanks for your responses. I know that they are not normalized to unity, but they can still be normalized to Dirac function, which is what's sought for.

 

[BvU]

Sorry I misunderstood the expression. Perhaps @vanhees71 can shed light on this ?

 

[A. Neumaier]

Hamiltonians with a potential that is this singular cannot be treated in the usual way, since as written down they cannot be interpreted as an operator densely defined on a Hilbert space. Such Hamiltonians must be renormalized to make physical sense - like the Hamiltonans appearing in relativistic quantum field theory. After renormalization, everything works out correctly.

A nice treatment of the delta potential can be found in:

• R.M. Cavalcanti, Exact Green's functions for delta-function potentials and renormalization in quantum mechanics (1998).

The renormalization of singular Hamiltonians in more general cases is discussed in my tutorial on renormalization.

 

[R P Stone]

Hamiltonians with a potential that is this singular cannot be treated in the usual way, since as written down they cannot be interpreted as an operator densely defined on a Hilbert space. Such Hamiltonians must be renormalized to make physical sense - like the Hamiltonans appearing in relativistic quantum field theory. After renormalization, everything works out correctly.

A nice treatment of the delta potential can be found in:

• R.M. Cavalcanti, Exact Green's functions for delta-function potentials and renormalization in quantum mechanics (1998).

The renormalization of singular Hamiltonians in more general cases is discussed in my tutorial on renormalization.

Thanks for your suggestion. However, the article deals with bound states and the regularization issue seems present only for D>1. Moreover, such issue does not exist if one looks at a discrete version of the Hamiltonian. Whatever, the wave functions must be orthogonal, as physically required. How to prove that?

 

[A. Neumaier]

Thanks for your suggestion. However, the article deals with bound states and the regularization issue seems present only for D>1.

Indeed, the 1D case is still finite; I forgot that you had posed the 1D version. The Green's function
$$G(E;x,y)=\langle x|(E-H)^{-1}|y\rangle=\int dE'\frac{\psi(E',x)\psi(E',y)}{E-E'}$$
encodes all eigenvector information $\psi(E',x)$ of the continuous spectrum without running into normalization problems.

Moreover, such issue does not exist if one looks at a discrete version of the Hamiltonian.

In a discrete version of the problem, there is no delta function. This is the situation discussed in detail in my tutorial. Renormalization is still needed in dimension >1 if infinitely many modes are used.
Or do you mean the continuos version on a bounded interval? Then you have square integrable solutions and just divide by the norm to normalize.

Whatever, the wave functions must be orthogonal, as physically required. How to prove that?

Only when the eigenvalues are distinct. One proves it by evaluating $\psi_jH\psi_k$ in two ways, using that $\psi_j$ or $\psi_j$ (2 cases) are eigenvectors. For the continuous spectrum, this argument becomes mathematically dubious, and one needs to use spectral projections to eigenspaces on disjoint energy intervals.

 

[R P Stone]

Only when the eigenvalues are distinct. One proves it by evaluating $\psi_jH\psi_k$ in two ways, using that $\psi_j$ or $\psi_j$ (2 cases) are eigenvectors. For the continuous spectrum, this argument becomes mathematically dubious, and one needs to use spectral projections to eigenspaces on disjoint energy intervals.

You mean they can not be normalised even to the delta function? Thanks.

 

[A. Neumaier]

You mean they can not be normalised even to the delta function? Thanks.

This is not what one would call normalized. In the continuous spectrum one works with the density of states, not with a pseudo normalization.

 

[Isaac0427]

Disclaimer: I've never actually solved for the scattering state wavefunctions as opposed to just the T and R coefficients. So I am assuming you have the correct wavefunction.

My tip for the integral is this: you are integrating an even function over a symmetric interval. Exploit this, and that should get rid of the absolute values. If you know the integral of a complex exponential from zero to infinity or negative infinity to zero (which does have something do do with the delta function), you should arrive at your answer.

This is not what one would call normalized. In the continuous spectrum one works with the density of states, not with a pseudo normalization.

While the terminology is not technically correct, Griffiths makes use of it in his textbook. He refers to the concept that for scattering states
$$\int_{-\infty}^{\infty}\psi_{k'}(x)^*\psi_k(x)dx=\delta(k-k')$$
as "'orthonormality'" (in quotes). I believe he uses the term Dirac orthonormality somewhere, which I prefer. While it may not be an entirely accurate use of terminology, I think its the best way to get the point across to students.

 

[vanhees71]

Though the $\delta$-distribution potential is a bit tricky and quite artificial, as @A. Neumaier stated above, here's the pragmatic physicist's approach.

My suggestion is to use another regularization than the "cut-off" method you used. The idea is that you have integrals of the form
$$\int_{0}^{\infty} \exp(\mathrm{i} a x), \quad a \in \mathbb{R}.$$
This you can regularize by introducing a regulator $\epsilon>0$ such as to make the integral convergent:
$$\int_{0}^{\infty} \exp[(-\epsilon + \mathrm{i} a) x]=\frac{1}{\epsilon-\mathrm{i} a}=\frac{\mathrm{i}}{a-\mathrm{i} \epsilon} \stackrel{\epsilon \rightarrow 0^+}{=}\mathrm{i} [\mathcal{P} \frac{1}{a} + \mathrm{i} \pi \delta(a)].$$
Here $\mathcal{P}$ stands for "principle value" and $\delta$ is the Dirac $\delta$ distribution.

 

[R P Stone]

@vanhees71

thanks for your suggestion. Actually I tried it but it did not work: I could not get rid of the principal parts.

 

[R P Stone]

@Isaac0427

thanks for your help. As you might see in my question, I’ve already tried what you suggested, but went nowhere.

 

[Isaac0427]

@Isaac0427

thanks for your help. As you might see in my question, I’ve already tried what you suggested, but went nowhere.

I looked a little more into it, and I don't think you have the proper wavefuntion. In that case, the integral may not work out. I believe the wavefunction for scattering from the left should be
$$
\psi_k(x)=\left\{\begin{matrix}
N_k\left(e^{ikx}+\sqrt{R_k}e^{-ikx}\right) & \textrm{when }x<0\\
N_k\sqrt{1-R_k}e^{ikx} & \textrm{when }x>0
\end{matrix}\right.
$$
where R_k is the (entirely real and positive) reflection coefficient. You can express it in terms of m, k, and V₀ (although I would caution you in using V₀ as it does not have the dimensions of energy, which may be misleading), but the exact expression for R shouldn't matter for the calculation you are trying to do.

The integrals appeared to work out, and I got an answer for $N_k$ that I will have in a spoiler. But I would highly recommend trying to do the integral yourself. There was nothing too advanced, but a few tricks you have to use.

Spoiler: My answer for N_k

I got plain old $N_k=\frac{1}{\sqrt{2\pi}}$ which is the same as it is for the free particle. I'm not 100% sure this is correct, but someone can let me know if they got a different answer.

 

[R P Stone]

@Isaac0427

really appreciate your effort. While the wave function you gave is certainly proper, the one I gave is also legitimate. yours can be obtained by combining an odd and an even parity wave functions mentioned in my question. i would like to have wave functions with definite parity.

 

[A. Neumaier]

@Isaac0427

really appreciate your effort. While the wave function you gave is certainly proper, the one I gave is also legitimate. yours can be obtained by combining an odd and an even parity wave functions mentioned in my question. i would like to have wave functions with definite parity.

You can take pairs of his solutions with the same energy, and apply a 2D unitary transformation that sends them into parity eigenstates.

 

[R P Stone]

@A. Neumaier
Thanks. That’s what I did, but it did not work out.

 

[A. Neumaier]

@A. Neumaier
Thanks. That’s what I did, but it did not work out.

What went wrong?

 

[R P Stone]

@A. Neumaier

I could not get @Isaac0427 ‘s result. I ran into the same problem as described in the question.

 

[Isaac0427]

@A. Neumaier

I could not get @Isaac0427 ‘s result. I ran into the same problem as described in the question.

You're just looking for N_k right? So you set R_k=R_k’ and remember that the integral from negative infinity to infinity of exp[i(k-k’)x]=2πδ(k-k’).

 

[R P Stone]

@Isaac0427 @A. Neumaier :

Taking Isaac's wave function,

$$I = \int^\infty_{-\infty} dx ~ \psi^*_{k'}(x) \psi_k(x) = I_> + I_<,$$ where
$$I_> = \int^\infty_{0} dx ~ \psi^*_{k'}(x) \psi_k(x), \quad I_< = \int^0_{-\infty} dx \psi^*_{k'}(x) \psi_k(x) .$$ Substituting the expressions,
$$I_> = N^2_k |1-R_k| \int^\infty_0 dx ~ e^{i(k-k')x}$$
and
$$I_< = N^2_k \int^0_{-\infty} dx ~ \left(e^{i(k-k')x} + |R_k| e^{i(k'-k)x} + \left[\sqrt{R_k}^* e^{i(k+k')x} + c.c. \right]\right).$$ Rearranging,
$$I = N^2_k \left[\int^\infty_{0} dx ~ \left(e^{i(k'-k)x} + (|R_k| + |1-R_k|) e^{i(k-k')x}\right) + \int^\infty_0 dx ~ \left(\sqrt{R_k}^*e^{-i(k+k')x} + \sqrt{R_k} e^{i(k+k')x}\right)\right].$$ How could you go further from here? Thanks.

 

[Isaac0427]

@Isaac0427 @A. Neumaier :

Taking Isaac's wave function,

$$I = \int^\infty_{-\infty} dx ~ \psi^*_{k'}(x) \psi_k(x) = I_> + I_<,$$ where
$$I_> = \int^\infty_{0} dx ~ \psi^*_{k'}(x) \psi_k(x), \quad I_< = \int^0_{-\infty} dx \psi^*_{k'}(x) \psi_k(x) .$$ Substituting the expressions,
$$I_> = N^2_k |1-R_k| \int^\infty_0 dx ~ e^{i(k-k')x}$$
and
$$I_< = N^2_k \int^0_{-\infty} dx ~ \left(e^{i(k-k')x} + |R_k| e^{i(k'-k)x} + \left[\sqrt{R_k}^* e^{i(k+k')x} + c.c. \right]\right).$$ Rearranging,
$$I = N^2_k \left[\int^\infty_{0} dx ~ \left(e^{i(k'-k)x} + (|R_k| + |1-R_k|) e^{i(k-k')x}\right) + \int^\infty_0 dx ~ \left(\sqrt{R_k}^*e^{-i(k+k')x} + \sqrt{R_k} e^{i(k+k')x}\right)\right].$$ How could you go further from here? Thanks.

Here is my color-coded work. Just to expand on a few things: First, you should keep in mind that R is real and positive. Second, where I do the u-substitution, the negative sign can be used to swap the bounds of integration. Finally, when k'=-k, you get an interesting result as well. I think if you use your even-parity case, that problem will be sorted out. It shouldn't be surprising that scattering from the left is not orthogonal to scattering from the right with the same energy (although it would be if there is no reflection. This should make sense intuitively). The work I did here is still valid for k'=-k, as $R_{-k}=R_k$.

 

[R P Stone]

Here is my color-coded work. Just to expand on a few things: First, you should keep in mind that R is real and positive. Second, where I do the u-substitution, the negative sign can be used to swap the bounds of integration. Finally, when k'=-k, you get an interesting result as well. I think if you use your even-parity case, that problem will be sorted out. It shouldn't be surprising that scattering from the left is not orthogonal to scattering from the right with the same energy (although it would be if there is no reflection. This should make sense intuitively). The work I did here is still valid for k'=-k, as $R_{-k}=R_k$.
View attachment 266751

Many thanks! But you can’t assume that $R_k$ is real. It is in general complex.

 

[Isaac0427]

Many thanks! But you can’t assume that $R_k$ is real. It is in general complex.

In fact, it is necessarily real and positive. The reflection coefficient is a probability. Theoretically, one can add a complex phase out in front of the square root, but that is arbitrary and should not affect the calculations at all.

 

[R P Stone]

In fact, it is necessarily real and positive. The reflection coefficient is a probability. Theoretically, one can add a complex phase out in front of the square root, but that is arbitrary and should not affect the calculations at all.

The wave function has to satisfy
$$\psi'(0_+) - \psi'(0_-) = - 2mV_0\psi(0)/\hbar^2.$$ Using your expression,
$$\psi'(0_+) = N_k \sqrt{1-R_k} ik, \quad \psi'(0_-) = N_k(1-\sqrt{R_k}) ik, \quad \psi(0) = N_k \sqrt{1-R_k}.$$ Substituting this in the above condition leads to
$$1 - \frac{1 - \sqrt{R_k}}{\sqrt{1-R_k}} = i\frac{2mV_0}{k\hbar^2}.$$ This cannot be satisfied by a real $R_k$.

 

[Isaac0427]

The wave function has to satisfy
$$\psi'(0_+) - \psi'(0_-) = - 2mV_0\psi(0)/\hbar^2.$$ Using your expression,
$$\psi'(0_+) = N_k \sqrt{1-R_k} ik, \quad \psi'(0_-) = N_k(1-\sqrt{R_k}) ik, \quad \psi(0) = N_k \sqrt{1-R_k}.$$ Substituting this in the above condition leads to
$$1 - \frac{1 - \sqrt{R_k}}{\sqrt{1-R_k}} = i\frac{2mV_0}{k\hbar^2}.$$ This cannot be satisfied by a real $R_k$.

This is my mistake. I forgot that the phase does matter on the coefficients. If you do
$$\sqrt{R_k}\rightarrow \frac{i\beta_k}{1-i\beta_k}$$
and
$$\sqrt{1-R_k}\rightarrow \frac{1}{1-i\beta_k}$$
where $\beta_k$ is real, the integral should work as well. In fact, everything but the cross terms will work out exactly the same way, and the cross terms work out pretty easily (my mistake was assuming |A|²=R means A=√R). I'm sorry about that, but again the integration is almost exactly the same. You'll notice that if $\left|\frac{i\beta_k}{1-i\beta_k}\right|^2=R_k$ then $\left|\frac{1}{1-i\beta_k}\right|^2=1-R_k$. I believe the answer for $N_k$ is the same as well, ignoring the cross terms that have to do with the delta of k+k'.

Note: I edited one of the equations.

 

[R P Stone]

This is my mistake. I forgot that the phase does matter on the coefficients. If you do
$$\sqrt{R_k}\rightarrow \frac{i\beta_k}{1-i\beta_k}$$
and
$$\sqrt{1-R_k}\rightarrow \frac{1}{1-i\beta_k}$$
where $\beta_k$ is real, the integral should work as well. In fact, everything but the cross terms will work out exactly the same way, and the cross terms work out pretty easily (my mistake was assuming |A|²=R means A=√R). I'm sorry about that, but again the integration is almost exactly the same. You'll notice that if $\left|\frac{i\beta_k}{1-i\beta_k}\right|^2=R_k$ then $\left|\frac{1}{1-i\beta_k}\right|^2=1-R_k$. I believe the answer for $N_k$ is the same as well, ignoring the cross terms that have to do with the delta of k+k'.

Note: I edited one of the equations.

Unfortunately, it won't work. You'll run into the same kind of problems described in in the original question. In fact, it has to do the principal parts @ vanhees71.

 

[Isaac0427]

Unfortunately, it won't work. You'll run into the same kind of problems described in in the original question. In fact, it has to do the principal parts @ vanhees71.

What did you do for $\int_0^{\infty}\sin(k+k')dx$?

If you see Griffith's Introduction to Quantum Mechanics, problem 2.25 or 2.26 depending on the edition you have, he suggests a way of how to handle an integral like that.

 

[R P Stone]

What did you do for $\int_0^{\infty}\sin(k+k')dx$?

If you see Griffith's Introduction to Quantum Mechanics, problem 2.25 or 2.26 depending on the edition you have, he suggests a way of how to handle an integral like that.

Thanks! But I don't have the book at hand. Would you sketch how to do it?

 

[Isaac0427]

Thanks! But I don't have the book at hand. Would you sketch how to do it?

Griffiths states for an integral like that (specifically with showing that the integral of exp(ikx) from negative infinity to infinity is zero), although it does not converge, you can replace infinity with L (which I see you have done), and take the average value of the integral as L goes to infinity. This takes you to zero. Thus (based on Griffith’s text— and he acknowledges that it would drive a mathematician insane) I would conclude that the previously mentioned sine integral would be equal to zero when k=k’ (which, of course, is the result we want).

 

[R P Stone]

Griffiths states for an integral like that (specifically with showing that the integral of exp(ikx) from negative infinity to infinity is zero), although it does not converge, you can replace infinity with L (which I see you have done), and take the average value of the integral as L goes to infinity. This takes you to zero. Thus (based on Griffith’s text— and he acknowledges that it would drive a mathematician insane) I would conclude that the previously mentioned sine integral would be equal to zero when k=k’ (which, of course, is the result we want).

Really many thanks. But I don’t think that strategy works here. That average would not yield zero for the integral.

 

[R P Stone]

@BvU @Isaac0427 @vanhees71 @A. Neumaier

Any new ideas? Please help! Thanks!