Boundary conditions for the momentum space SE

[andresB]

Usually, energy eigenvalue problems in QM are defined by boundary conditions given in terms of the particle position. For example, we have the particle in an infinite square well
$$
-\frac{\hbar^{2}}{2m}\frac{d^{2}\psi_{n}(x)}{dx^{2}} =E_{n}\psi_{n}(x),$$
$$\psi_{n}(0) =\psi_{n}(L)=0,$$

or the bouncing neutron problem
$$-\frac{\hbar^{2}}{2m}\frac{d^{2}\psi_{n}(x)}{dx^{2}} +mgx=E_{n}\psi_{n}(x),$$
$$\psi_{n}(0) =0.$$My question is: How to solve, from the very beginning, the above problems (and related ones) in momentum space? The equations to solve are clear to me. For example
$$\frac{p^{2}}{2m}\phi_{n}(p)=E_{n}\phi_{n}(p)$$
and

$$\frac{p^{2}}{2m}\phi_{n}(p) +i\hbar mg\frac{d}{dx}\phi_{n}(p)=E_{n}\phi_{n}(p).$$

But how to apply boundary conditions so I get the same energy levels as in configuration space?

 

[PeterDonis]

how to apply boundary conditions

What do the boundary conditions you gave in position space translate to in momentum space?

 

[vanhees71]

Caveat: You can only use eigenstates of self-adjoint operators. For the infinite interval there's no self-adjoint momentum operator.

For the bouncing-neutron problem you rather first solve the problem without the reflecting mirror. Then you get continuous energies and you can Fourier-transform to position space. The result is an Airy function. Then you get the solutions with the reflecting mirror by fulfilling the boundary condition in momentum space, $\psi_E(x=0)=0$, which can only be fulfilled for discrete energies determined by the zeroes of the Airy function.

 

[andresB]

What do the boundary conditions you gave in position space translate to in momentum space?

Ok, I get it. Mathematically, the problems were way easier than I thought initially.

However, I'm not entirely satisfied. I can't make physical sense of the boundary conditions in momentum space. For example, for the square well we have
$$\int_{-\infty}^{\infty}\phi_{n}(p)dp=\int_{-\infty}^{\infty}\phi_{n}(p)e^{ipL/\hbar}dp=0.$$
What do those tell us about $\phi_{n}(p)$ as an amplitude of probability?

 

[PeterDonis]

Ok, I get it.

I'm not sure you do. You didn't answer the question I asked. For example, for the square well the position space boundary condition is $\psi_n(0) = \psi_n(L) = 0$. What does that translate to in momentum space?

for the square well we have
$$\int_{-\infty}^{\infty}\phi_{n}(p)dp=\int_{-\infty}^{\infty}\phi_{n}(p)e^{ipL/\hbar}dp=0.$$

Where did the exponential factor come from? As you write this, it doesn't make sense. For one thing, you should have $\int_{-\infty}^{\infty} \phi_n(p) dp = 1$ for correct normalization of the probability.

 

[andresB]

Where did the exponential factor come from? As you write this, it doesn't make sense. For one thing, you should have $\int_{-\infty}^{\infty} \phi_n(p) dp = 1$ for correct normalization of the probability.

Normalization is for the square of the wavefunction.

I'll write the solutions to both problems tomorrow.

 

[PeterDonis]

Normalization is for the square of the wavefunction.

Ah, yes, you're right. Sorry for the mixup on my part.

However, your equation does assume that the momentum space wave function is odd. Where does that assumption come from?

 

[vanhees71]

Ok, I get it. Mathematically, the problems were way easier than I thought initially.

However, I'm not entirely satisfied. I can't make physical sense of the boundary conditions in momentum space. For example, for the square well we have
$$\int_{-\infty}^{\infty}\phi_{n}(p)dp=\int_{-\infty}^{\infty}\phi_{n}(p)e^{ipL/\hbar}dp=0.$$
What do those tell us about $\phi_{n}(p)$ as an amplitude of probability?

The boundary conditions are given in position space. It's not so easy to translate them to conditions in momentum space. So you have to calculate the energy eigenfunctions in position space. For your problem of the "free falling" neutrons above a "mirror" you can get the ansatz function most easily in momentum space, though. Then you transform back to the position representation and impose the boundary condition, which selects the energy eigenvalues.

For the infinite-square-well problem of course your particle is restricted to the interval $[0,L]$. You have the rigid boundary conditions $\psi(0)=\psi(L)=0$. Then you solve the energy-eigenvalue equation in position space directly, which is very easy in this case. Then you impose the boundary conditions, which select the energy eigenvalues. Note that here there's no momentum operator, because $-\mathrm{i} \partial_x$ has no domain on this space as a self-adjoint operator. The scalar product in this case is
$$\langle \psi_1|\psi_2 \rangle=\int_0^L \mathrm{d} x \psi_1^*(x) \psi_2(x).$$
It doesn't make sense to define the wave function on entire $\mathbb{R}$.

 

[vanhees71]

Ah, yes, you're right. Sorry for the mixup on my part.

However, your equation does assume that the momentum space wave function is odd. Where does that assumption come from?

For the infinite square well there's no self-adjoint momentum operator, i.e., you cannot define momentum as an observable for a particle confined to a finite spatial interval. Nevertheless the Hamiltonian, $\hat{H} \propto -\partial_x^2$ can be defined as a self-adjoint operator and thus energy is a self-adjoint operator.

To get used to the important question of domains and co-domains of self-adjoint operators, I recommend to think about the wave function $\psi(x)=N x(1-x)$ and think about calculating $\langle E \rangle$ and $\langle E^2 \rangle$ and then think about, what you get if you do this in the "naive way" for $\Delta E^2 =\langle E^2 \rangle -\langle E \rangle^2$ and then think further, how to define these moments of the energy-probabilities in a correct way.

The infinite-square-well problem is NOT an easy problem, if discussed in a complete way!

 

[andresB]

Infinite square well
we want to solve the equation
$$\frac{p^{2}}{2m}\phi_{n}(p)=E_{n}\phi_{n}(p)\quad\qquad(1)$$
with the boundary conditions
$$\psi(x=0)=\psi(x=L/2)=0.$$
Remember that the relation between position and momentum wavefunctions is given by the Fourier transform
$$\psi(x)=\frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty}\phi(p)e^{ipx/\hbar}dp.$$
Hence, in momentum space the boundary condition read
$$\int_{-\infty}^{\infty}\phi_{n}(p)dp=\int_{-\infty}^{\infty}\phi_{n}(p)e^{ipL/\hbar}dp=0.$$
Equation (1) has solutions of the form
$$\phi_{n}(p)=A\delta(p+\sqrt{2mE_{n}})+B\delta(p-\sqrt{2mE_{n}}).$$
Using the boundary condition leads to
$$B =-A,$$
and
$$\exp[-\frac{iL}{\hbar}\sqrt{2mE_{n}}]=\exp[\frac{iL}{\hbar}\sqrt{2mE_{n}}].$$

This last equation can only be true if
$$\frac{L}{\hbar}\sqrt{2mE_{n}}=n\pi\;(n\in\mathbb{Z})\rightarrow E_{n}=\frac{n^{2}\pi^{2}\hbar^{2}}{2L},$$
hence
$$\phi_{n}(p)\sim\left[\delta(p+\frac{n\pi\hbar}{L})-\delta(p-\frac{n\pi\hbar}{L})\right],$$
where we are ignoring any normalization constant. Notice that the Fourier transform of the above leads to the well-kwon solution in position space
$$\psi(x)\sim\int_{-\infty}^{\infty}\left[\delta(p+\frac{n\pi\hbar}{L})-\delta(p-\frac{n\pi\hbar}{L})\right]e^{ipx/\hbar}dp=e^{i\frac{n\pi\hbar}{L}x}-e^{-i\frac{n\pi\hbar}{L}x}\sim\sin(\frac{n\pi}{L}x).$$

Bouncing neutrons

The problem here is to solve the equation
$$\frac{p^{2}}{2m}\phi_{n}(p)+img\hbar\frac{\partial}{\partial p}\phi_{n}(p)=E_{n}\phi_{n}(p);\quad\psi_{n}(0)=0.$$

Remember that the Airy functions are defined by
$$Ai(x)\sim\int_{-\infty}^{\infty}\exp\left[i\left(xp+\frac{p^{3}}{3}\right)\right]d\text{p}\sim-i\int_{-i\infty}^{i\infty}\exp\left(xp+\frac{p^{3}}{3}\right)d\text{p}$$
(Vallée and Soares, Airy Functions and Applications to Physics, 2nd Edition, 2010, page 6)
First, we make a change of variables to work with dimensionless quantities
$$q_{0} =\left(\frac{\hbar^{2}}{2m^{2}g}\right)^{1/3},\qquad\qquad\mathfrak{q}=q/q_{0},$$

$$p_{0} =\left(2\hbar m^{2}g\right)^{1/3},\qquad\qquad\mathfrak{p}=p/p_{0},$$

$$E_{0}=\left(\frac{\hbar^{2}mg^{2}}{2}\right)^{1/3},\qquad\qquad a_{n}=-E_{n}/E_{0},$$
leading to the simplified equation
$$\mathfrak{p}^{2}\phi_{n}(\mathfrak{p})+i\frac{\partial}{\partial\mathfrak{p}}\phi_{n}(\mathfrak{p})=-a_{n}\phi_{n}(\mathfrak{p}),$$
with solutions given by
$$\phi_{n}(\mathfrak{p})\sim\exp\left[i\left(a_{n}\mathfrak{p}+\frac{\mathfrak{p}^{3}}{3}\right)\right].$$

Notice that our boundary conditions lead to
$$\psi(0)=0\rightarrow Ai(x=a_{n})=0,$$
hence, $a_{n}$ is a zero of the Airy function, and the energies are related to these zeroes by $E_{n}=-a_{n}\left(\frac{\hbar^{2}mg^{2}}{2}\right)^{1/3}.$

Finally, the full (not normalized) wavefunctions are

$$\psi(\mathfrak{q})\sim\int_{-\infty}^{\infty}\exp\left[i\left(a_{n}\mathfrak{p}+\frac{\mathfrak{p}^{3}}{3}\right)\right]e^{i\mathfrak{p\mathfrak{q}}}d\mathfrak{p}\sim\int_{-\infty}^{\infty}\exp\left[i\left((\mathfrak{q}+a_{n})\mathfrak{p}+\frac{\mathfrak{p}^{3}}{3}\right)\right]d\mathfrak{p}=Ai(\mathfrak{q}+a_{n}),$$

agreeing with literature (Spectra of neutron wave functions in Earth's gravitational fields, by Suda et al.).

 

[andresB]

@vanhees71

I understand what you say about self-adjoint operators, but it seems that no problem arises when solving these system in momentum space.

 

[andresB]

So, the mathematics in momentum space seems to work. Yet, I'm still puzzled by the boundary conditions in momentum space. I don't see any logic, from the probabilistic point of view, that allows me to get those boundary conditions without using a Fourier transform.

 

[vanhees71]

@vanhees71

I understand what you say about self-adjoint operators, but it seems that no problem arises when solving these system in momentum space.

For sure for the infinite square well, there's no self-adjoint momentum operator. Since momentum is the generator of time [EDIT: corrected in view of #16] space translations it must be of the form $-\mathrm{i} \partial_x$ (in postion space representation and with natural units, $\hbar=1$). Assuming it were self-adjoint, then due to the commutation relation $[\hat{x},\hat{p}]=\mathrm{i}$, it should have a continuous real spectrum. The eigenfunctions, however, must be $\sim \exp(-\mathrm{i} p x)$, which doesn't fulfill the boundary conditions $u_p(0)=u_p(L)=0$.

The same holds for the problem with the ideally reflecting "ground", i.e., the boundary condition, $\psi(0)=0$, cannot be fulfilled for any momentum eigensolution.

In both cases, however, $\hat{p}^2=-\partial_x^2$ can be defined as a self-adjoint operator. Indeed its eigenfunctions are of the form
$u_{p^2}(x)=A \cos(p x) + B \sin(p x),$
and you can fulfill the boundary conditions $u_p(0)=0$, which simply leads to $A=0$, i.e.,
$$u_{p^2}(x)=N \sin(p x).$$
For the infinite square well you have the additional boundary condition $u_{p^2}=(L)=0$, which leads to the discrete eigenvalues $p_j^2=j \pi L$ with $j \in \mathbb{N}=\{1,2,\ldots \}$.

For the ideally reflecting ground you have $p^2 \in \mathbb{R}_+$ as eigenvalues, and you can solve the neutrons in the gravitational potential $U(x)=-mg x$ in terms of these eigenfunctions of $p^2$ (I guess, I haven't checked it).

It's, however simpler, to first solve the problem with $U(x)=-mgx$ in unrestricted space. The corresponding energy eigenfunctions are Airy functions and the spectrum is $\mathbb{R}$. However, you can impose the reflecting-mirror boundary condition $u_E(0)=0$ also simply to these solutions, which leads to a discrete energy spectrum as it must be for neutrons bound in the potential + "mirror", i.e., in this case you can first work in the usual momentum representation for the energy eigenproblem without the mirror and select the corresponding solutions with the boundary condition for the case with mirror afterwards.

 

[andresB]

For sure for the infinite square well, there's no self-adjoint momentum operator. Since momentum is the generator of time translations it must be of the form $-\mathrm{i} \partial_x$ (in postion space representation and with natural units, $\hbar=1$). Assuming it were self-adjoint, then due to the commutation relation $[\hat{x},\hat{p}]=\mathrm{i}$, it should have a continuous real spectrum. The eigenfunctions, however, must be $\sim \exp(-\mathrm{i} p x)$, which doesn't fulfill the boundary conditions $u_p(0)=u_p(L)=0$.

I understand what you say, but it seems that doesn't create any problem in momentum space since not only we can get the correct energy levels but we can also get the stationary states in configuration space by a Fuorier transform.

 

[vanhees71]

What you get are the complete set of energy eigenfunctions. For the square well you have
$$u_j(x)=\sqrt{2/L} \sin(k_j x), \quad k_j=\pi j/L, \quad j \in \mathbb{N}.$$
These are a complete orthonormal set of eigenfunctions of $\hat{H}=\hat{p}^2/(2m)$ with eigenvalues $E_j=k_j^2/(2m)$ on the space $\mathrm{L}^2([0,L])$ with wave functions obeying the boundary conditions $\psi(0)=\psi(L)=0$. You can expand all such functions in terms of these eigenfunctions,
$$\psi(t,x)=\sum_{j=1}^{\infty} \psi_j \exp(-\mathrm{i} E_j t) u_j(x).$$
That's an expansion in energy eigenfunctions. It's NOT an expansion in momentum eigenfunctions, because you cannot define a momentum for a particle confined rigidly to a finite interval, as demonstrated in my previous posting. As I always say: The infinite square-well problem is only a simple problem, if you don't think enough about the self-adjointness of the involved operators.

Another quite revealing task is:

Consider the wave function
$$\psi(x)=\sqrt{\frac{30}{L^5}} x(L-x).$$
This is clearly in the Hilbert space under consideration.

Now calculate $\langle E \rangle$ and $\langle E^2 \rangle$ as well as $\Delta E^2$ for this wave function. Where does the trouble come from and how to solve it?

 

[PeterDonis]

momentum is the generator of time translations

I think you mean space translations.

 

[vanhees71]

Of course. I've corrected the typo.

 

[andresB]

What you get are the complete set of energy eigenfunctions.

Well, yes. I'm interested in the energy eigenvalue problem with its its boundary conditions using the momentum SE.