Is the Inverse Momentum Operator an Essential Tool in Quantum Mechanics?

[hilbert2]

In QM, the inverse distance operator $\hat{r}^{-1}$ appears often because of the association to Coulomb potential. The operator of inverse momentum, $\frac{1}{\hat{p}}$ is a lot more rare.

In the book "Exploring Quantum Mechanics: A Collection of 700+ Solved Problems for Students, Lecturers and Researchers", readable at Google books, there's an exercise where it's shown that in the position representation the inverse momentum operation is

$\hat{p}^{-1}\psi(x) = \frac{i}{\hbar}\int_{-\infty}^{x}\psi(x')dx'$,

which is easy to confirm by operating with $\hat{p} = -i\hbar\frac{d}{dx}$ and using the fundamental theorem of calculus.

Another way to write the inverse momentum operator would seem to be by expressing the wave function $\psi(x)$ as a linear combination of eigenstates of $\hat{p}$:

$\psi(x) = \int_{-\infty}^{\infty}\phi(p)e^{ipx}dp$,

or something like that, and then replacing this with

$\int_{-\infty}^{\infty}\frac{\phi(p)e^{ipx}}{p}dp$

where every component has been divided with the corresponding momentum.

But the problem is, the component with zero momentum will lead to division by zero. How to get around this problem?

Edit: A strange thing about this operator is that the modified time dependent Schrödinger equation $i\hbar \frac{\partial \Psi}{\partial t} = -\frac{\hbar^2}{2m}\frac{\partial^2 \Psi}{\partial x^2} + \frac{1}{2}kx^2 \Psi + \alpha Im\left( \int_{-\infty}^{x}\Psi (x’,t)dx’\right)$, which I mentioned in an Insights article some time ago and which makes a wavepacket behave like a damped harmonic oscillator, also contains an integral that is equivalent to that in the inverse momentum operator.

 

[strangerep]

But the problem is, the component with zero momentum will lead to division by zero. How to get around this problem?

Cauchy Principal Value + contour integration?

 

[hilbert2]

Cauchy Principal Value + contour integration?

Thanks for the reply. The division by zero probably isn't a problem after all if the momentum space wavefunction $\phi(p)$ approaches zero fast enough when $p\rightarrow 0$ (I wasn't really thinking this carefully enough when posting it).

As for the damped Schrödinger equation, many of the equations of that type that are found in the literature seem to contain a term where the composite operator of the inverse momentum with some nonlinear operator is acting on the wavefunction, such as the $\int_{0}^{x}|\psi(x')|^2 dx'$ in this article. The imaginary part operator $\hat{A}\psi(x) = I am (\psi(x))$ in the equation at the end of my post is also nonlinear, because it only satisfies $\hat{A}(\psi_1 (x) + \psi_2 (x))$ and not $\hat{A}(c\psi(x)) = c\hat{A}(\psi(x))$ if $c$ can be a complex number.

When I made a simulation using Crank-Nicolson method calculating the evolution of a Gaussian wavepacket in a system with Hamiltonian

$\hat{H} = \frac{\hat{p}^2}{2m} + \frac{1}{2}k\hat{x}^2 + C\hat{p}^{-1}$,

the norm of the wavepacket seemed to decay exponentially as if $\hat{p}^{-1}$ wasn't a hermitian operator, which is a bit strange... Probably an error in my code, though.

 

[DeathbyGreen]

How did you get from $\psi(x) = \int_{\infty}^{\infty}\phi(p)e^{ipx}dp$ to $\int_{\infty}^{\infty}\frac{\phi(p)e^{ipx}}{p}dp$? I'm not sure just dividing by $\hat{p}$ is correct; taking the inverse of an operator can change the domain that the operator is applicable in. Inverse operators can be tricky. Was this rigorously derived or eye-balled? Could you post explicitly how you did this? Also, simply by looking at the result, assuming it is correct, it's essentially a delta function. It blows up as you approach the origin from either side, so the previous suggestion of the principle value might be a way to approach it. As an approximation, you could say

$\int_{\infty}^{\infty}\frac{\phi(p)e^{ipx}}{p}dp \approx \int_{\infty}^{\infty}\delta(p)\phi(p)e^{ipx}dp$

 

[hilbert2]

If the $\phi (p)$ were a Dirac comb function $\phi(p) = \sum\limits_{k=0}^{\infty}\delta(p - p_k )$ then it would obviously be correct. That's basically my reasoning for that expression for $\hat{p}^{-1}$.

 

[blue_leaf77]

Taking analogy with the position operator counterpart, $1/r$, the first thing about this operator is that its spectrum lies in the positive half of the axis. Your reciprocal $p$ operator does not seem to have this property, it might help defining it with an absolute sign $|p|$. Second, the operator $1/r$ AFAIK only appears in 3D in which the integration element is of the form $r^2\sin\theta\ dr\ d\theta\ d\phi$, the presence of $r^2$ saves us from the trouble with the singular nature of $1/r$. There are cases in which a reciprocal operator is used to model a Coulomb attraction in 1D toy system but this particular operator is equipped with what they call a smoothing factor $\alpha$ (usually some small number),
$$
O(x) = \frac{1}{\sqrt{x^2+\alpha^2}}
$$
which again eliminate the possibility of division by zero.

You may have a similar reciprocal magnitude momentum operator in 3D or 2D since the integration elements are $p^2\sin\theta\ dp\ d\theta\ d\phi$ and $p\ dp\ d\phi$ respectively, but I doubt that the 1D version will be a sound-defined operator.

In the book "Exploring Quantum Mechanics: A Collection of 700+ Solved Problems for Students, Lecturers and Researchers", readable at Google books, there's an exercise where it's shown that in the position representation the inverse momentum operation is

^p−1ψ(x)=iℏ∫x−∞ψ(x′)dx′\hat{p}^{-1}\psi(x) = \frac{i}{\hbar}\int_{-\infty}^{x}\psi(x')dx',

which is easy to confirm by operating with ^p=−iℏddx\hat{p} = -i\hbar\frac{d}{dx} and using the fundamental theorem of calculus.

Does the book also shows the 3D version of this operator, i.e. one proportional to the gradient operator. As far as I know, inverting a gradient requires the gradient to be a conservative vector field, that is given $\nabla \ \psi$, one can find $\psi$ if $\nabla \times \nabla \ \psi = 0$ for if this condition is fulfilled, one can for instance follow what section 4 in this paper describes. But the aforementioned condition is not general.

 

[hilbert2]

The description of this is on page 24 of the book, and it's about the inverse of the cartesian 1D momentum operator $\hat{p}_x$. I maybe should have talked about the position operator $\hat{x}$ instead of the distance operator $\hat{r}$ in the beginning of the post, as the "radial momentum operator" $-i\hbar\frac{d}{dr}$ is known to be not hermitian and therefore is not a valid observable. The 3D version is not mentioned in there.

About the nonlinear Schrödinger equation with an integral term in it... There seems to be a typo in the formula above, and the original version of the equation was

$i\hbar \frac{\partial \Psi}{\partial t} = -\frac{\hbar^2}{2m}\frac{\partial^2 \Psi}{\partial x^2} + \left[\frac{1}{2}kx^2+ \alpha Im\left( \int_{-\infty}^{x}\Psi (x’,t)dx’\right)\right] \Psi$,

where the nonlinear term is a part of the potential energy and not an independent term in the equation. This equation can be solved by Crank-Nicolson method by linearizing it in a way where the linear system of equations to be solved does not have equal matrices on each time step.

However, if I apply the C-N method on the equation

$i\hbar \frac{\partial \Psi}{\partial t} = -\frac{\hbar^2}{2m}\frac{\partial^2 \Psi}{\partial x^2} + \frac{1}{2}kx^2 \Psi + C\int_{-\infty}^{x}\Psi (x’,t)dx’$,

which now contains something proportional to the operator $\hat{p}^{-1}$ in the last term, I get a triangular matrix system to be solved on each timestep and it still seems to make the position expectation value of the wavepacket behave like the position of a damped classical oscillator. Whether it keeps the norm of the wave function constant depends on whether $C$ is real or imaginary.

EDIT: Also, if I construct a classical mechanical system of an oscillator with a Lagrangian function that has a reciprocal mometum term in it,

$L = \frac{1}{2}m(\dot{x})^2 - \frac{1}{2}kx^2 + \frac{C}{m\dot{x}}$,

the equation of motion becomes nonlinear and probably impossible to solve exactly, and solving it numerically will just lead to $x(t)$ blowing up when the velocity gets too close to zero. The quantum version of this seems to be possible to solve numerically without blowup and behaves somewhat like a damped oscillator, which is quite strange...

 

[Demystifier]

Cauchy Principal Value + contour integration?

What do you mean by "+"? Is it a logical "and" or a logical "or"? I think it should be "or". Either you take the principal value or deform the contour of integration.

 

[hilbert2]

If I represent a wave function $\psi (x)$ with a discrete object $\psi_l$, with $l$ a position index, the matrix representing the operator $\hat{p}^{-1}$ would seem to become a lower triangular matrix with all nonzero elements having value $\frac{i\Delta x}{\hbar}$ ($\Delta x$ is the step size in the discretization). This would then approximate the integration from $-\infty$ to $x$ with a Riemann sum. Obviously this isn't hermitian because the elements on one side of the diagonal are zero and the elements on the other size are nonzero. So it's probably not surprising that putting a term proportional to $\hat{p}^{-1}$ to a Hamiltonian operator will lead to non-unitary dynamics.

The problem of the state with $p_{x}=0$ becoming infinite when acted on by $\hat{p}_x^{-1}$ isn't a problem for any bound state where the particle in confined to some limited space, because in the $p_x =0$ state the wave function is a nonzero constant on the whole $x$-axis and that's why the integral from $-\infty$ to $x$ becomes infinite.

 

[vanhees71]

Well, one has to be careful with the domains and co-domains, i.e., $1/\hat{p}$ is only defined on such Hilbert-space vectors $|\psi \rangle$, for which $1/\hat{p} |\psi \rangle$ not only exists but is also in the Hilbert space. In other words in the momentum representation with $\psi(p)=\langle p|\psi \rangle$ also $\psi(p)/p$ must be square integrable. With this caveat there shouldn't be too much of a problem to define the inverse momentum operator.

 

[strangerep]

What do you mean by "+"? Is it a logical "and" or a logical "or"? I think it should be "or". Either you take the principal value or deform the contour of integration.

Well, not quite. You can interpret the original integral as a CPV, and then, (if you wish), use contour integration to evaluate it.

 

[DrDu]

I think you should clarify whether you are really interested in the inverse momentum operator or a Schrödinger operator with an inverse momentum contribution. These are two completely different situations. Specifically, I would consider your Hamiltonian with a quadratic potential in momentum representation. then $\hat{p}=p$ and $\hat{x}=i\hbar \partial_p$. Then changing the names $x \leftrightarrow p$, the Hamiltonian can be seen to be identical to that of a harmonic oscillator with an additional 1/x term which has been studied intensively.

 

[hilbert2]

I think you should clarify whether you are really interested in the inverse momentum operator or a Schrödinger operator with an inverse momentum contribution. These are two completely different situations. Specifically, I would consider your Hamiltonian with a quadratic potential in momentum representation. then $\hat{p}=p$ and $\hat{x}=i\hbar \partial_p$. Then changing the names $x \leftrightarrow p$, the Hamiltonian can be seen to be identical to that of a harmonic oscillator with an additional 1/x term which has been studied intensively.

The most interesting thing I saw in this subject was that if I add an inverse momentum term in a 1D pos. rep. Schrödinger equation, the Crank-Nicolson numerical scheme has a triangular linear system instead of the usual tridiagonal. Solving this in the momentum space will probably solve the problem of the operator being nonhermitian for bound states, thanks for the idea.