Electron in free space and Schrodinger's equation

Question

Topic: Quantum Theory of Solids

I have come across the following text:

Electron in Free Space:

Consider the motion of an electron in free space. If there is no force acting on the particle, then potential function $V(x)$ will be constant and we must have $E > V(x)$.

$E=$ Total Energy of the Particle

The reasoning that I have applied is:

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Since Force $(F) =0$ => Electric Field $(W)=0 $

and the electric field is the negative gradient of the potential function

$W=-\frac{dV(x)}{dx}=0$

Which implies that $V(x)=$ constant

But can I apply this reasoning?

This result has been obtained from a classical phenomenon(Electrostatics).

Can this apply to quantum mechanical models? Or is there some other explanation?

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Secondly why in this case $E > V(x)$. I know that electron in free space has the maximum energy of 0 eV. An electron bound to a nucleus has negative energy. Does this have to do anything with it?

Answer 1

The key point here is that in quantum mechanics, all forces are conservative - that is, they all preserve the law of conservation of energy exactly and absolutely. You should expect this to make sense, because all forces that in our every-day experience appear to be "non-conservative", like friction, are really just the manifestations of conservative forces acting on a microscopic scale. Quantum mechanics is the physics of that microscopic scale, and its mathematical formulation bears witness to this fact with exactly what you have described.

In classical mechanics, a conservative force is defined as a force field which conserves energy - meaning you can take a particle in and responsive to that force field, move it around a closed circuit in space, and you will end up with no more and no less energy when you finished as when you started. Mathematically, that means

$$\mbox{Work done (energy released)} = \oint_{\gamma} \mathbf{F} \cdot d\mathbf{l} = 0$$

where $\gamma$ is some closed loop, and $d\mathbf{l}$ is just a tiny vectorial increment of length therealong representing a little piece of motion. Using some theorems of vector calculus, you get that it is then possible to describe the force field $\mathbf{F}$ as the gradient of a scalar function (i.e. one returning only a number), $U$:

$$\mathbf{F} = -\nabla U$$

and this function $U(\mathbf{r})$ has the dimensions of an energy, and is what we call the potential energy. This $U$ is what is typically (oddly) denoted $V$ in many quantum mechanics texts. Because all forces are conservative like this on the scale in which we use quantum theory, we can effectively ditch the force $\mathbf{F}$ specifically as redundant and work only with the potential energy function $U$, and when one goes deeper still, one finds this makes more sense because really, energy is the more fundamental physical quantity than force.

If the force $\mathbf{F}$ is zero, then you must have $\nabla U = 0$, and it is easy to see that means $U$ is constant. In other words, yes, you can apply the reasoning you give, though with the caveat that the force in question need not be electric. An electron being sucked into at least a Newtonian gravity well is entirely treatable the same way - just assign $U(\mathbf{r}) := -\frac{GMm_e}{r}$.

Now if your suspicion about classical reasoning is that we still have a classical-looking potential energy function and that the intuition behind it is still based on the classical notion of force, you are right! In fact, to fully treat force fields quantum-mechanically, we need quantum field theory - which constitutes the basis of the very best theories we have for understanding the basic constituents of our Universe.

Answer 2

But can I apply this reasoning? This result has been obtained from a classical phenomenon(Electrostatics).

Any time you take a classical problem and quantize it, you are making an educated guess as to what the Hamiltonian will look like. In general, there are many different quantum mechanical systems which have the same classical limit, so there's no unique way of translating a classical problem into the "correct" underlying quantum mechanical problem.

Your reasoning makes sense, and in accordance with Ehrenfest's theorem, it respects the correspondence principle. I don't think you can say (or ask for) any more validation than that.

Secondly why in this case $E>V(x)$

You could answer this a few different ways. From a physical standpoint, one could argue that the free particle Hamiltonian $H_{free}=-\frac{d^2}{dx^2}$ is just the kinetic energy operator, and for a that state with definite kinetic energy, that kinetic energy should be positive.

You could also observe that the eigenstates corresponding to negative energies take the form $e^{\pm\alpha x}$ with $\alpha >0$, and "blow up" either at large positive or large negative values.

Of course, you could also argue that neither of these arguments is satisfactory. The first, for example, assumes something about kinetic energy (whatever that is, in the context of quantum mechanics) - maybe in quantum mechanical systems, kinetic energy can be negative. The second argument is also problematic because while it's true that functions of the form $e^{\pm i\alpha x}$ don't diverge to infinity, they are also not normalizable, so by a similar argument we should throw them away as well.

The correct answer to this question is a bit technical. Roughly speaking, operators with continuous spectra (such as the momentum and energy operators in this case) do not actually have eigenvectors and eigenvalues, because if $\psi$ obeys the equation

$$P \psi = \lambda \psi$$

then it can be shown that $\psi \notin L^2(\mathbb R)$. In physics, we still hold on to those states, but simply note that the are not physical (i.e. in real life, you cannot truly have a free-particle state of definite momentum or energy). In such cases, we replace the requirement that such states be square-integrable with the requirement that their inner products with physical states be well-defined. In other words, if $\psi_P$ is an unphysical state of definite momentum and $\phi$ is an actually physical state, then we demand that $$\langle \psi_P , \phi\rangle < \infty$$

This is what separates functions like $e^{i\alpha x}$ from functions like $e^{\alpha x}$. The former satisfy the above requirement, while the latter do not. Therefore, even though neither are proper, normalizable states, we keep the former (and call them "generalized eigenstates") but must throw away the latter.

From this, it follows that the only allowed states of definite energy take the from $e^{\pm i\alpha x}$, which correspond to strictly positive kinetic energy terms when you plug them into $H_{free}$.

Answer 3

The equation that rules the motion of a free electron in Quantum Mechanics is the Schroedinger equation

$$ i\hbar\frac{\partial}{\partial t}\psi(x,t)=H\psi(x,t)$$

which in its time independent version says ($\psi(x,t)=\psi_{x}(x)\cdot \psi_t(t)$)

$$-\frac{\hbar^2}{2m}\nabla^2\psi_{x}(x)+V(x)\psi_{ x}(x)=E\psi_{x}(x)$$

As the electron is free, the first part of your reasoning bring you correctly to the conclusion that $V(\vec x)$ must be constant, so the time independent Schroedinger equation reduces to the eigenvalue problem

$$-\frac{\hbar^2}{2m}\nabla^2\psi_{x}(x)=(E-V)\psi_{x}(\vec x)\equiv \tilde E\psi_{x}(x)$$

If we solve formally, without making any assumption on $\tilde E$, we get

$$\phi(x)=C_0\exp\left[\pm i\sqrt{\frac{2m\tilde E}{\hbar^2}}\right]x$$

Now, mathematically both $\tilde E>0$ and $\tilde E<0$ would be acceptable, but physically we would expect something from a free electron, for example that it conserves its momentum during motion.

Indeed, the Hamiltonian also commutes with momentum operator $\hat{P}$, so it's physically reasonable to search for solutions which diagonalize at the same time $\hat H$ and $\hat{P}$.

As $\hat{P}$ is Hermitian, in order for $\phi(x) $ to be an eigenstate of $\hat{P}\stackrel{\cdot}{=}-i\hbar\frac{\partial}{\partial x}\delta (x-x')$ yielding a real eigenvalue, we must have $i\sqrt{\frac{2m\tilde E}{\hbar^2}}$ imaginary, i.e. $\tilde E>0$.

So, for the free electron, we consider only the case $E>V$.

Answer 4

Okay, I'm posting this because it's too big for a comment, and it contains kind of answer contents.

Firstly, I must address that you must be very careful with notation. What is $V$? What is $W$? Indeed, you'd better change this notation as soon as possible, because it's the perfect recipe for a mess.

Note that, in electromagnetism, the people who haven't gone crazy yet still use $V$ for "potential". For example, electrostatic potential, which is volts, symbol $V$.

Yet some others prefer greek letters for scalars, yes, the same people who use $E$ for energy and $W$ for work haha. Absolute lack of coherence. Just kidding, but there's truth about that.

Okay, so, if you use $V$ for potential (so do I), use another letter for potential energy. Remember that sChrödinger's equation is

$$-\frac{\hbar^2}{2m} \nabla^2\psi + E_p(x)\cdot\psi = E\cdot\psi$$

Where $E_p$ is the potential energy. I'm the only freak who uses that notation, and I'm proud of it. For me, chosing $V$ both for potential and also potential energy is not just a bad notation, it is an evil notation. So be careful.

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And now, I'll add one possible "explanation". It's not very formal, not mathematical, but it's intuitive.

Yes, you can apply the same reasoning.

But one way to understand is that, if the particle is in free space, all points must be equivalent.

Classically, you'd say that you can choose the $gauge$ wherever, so you can say $V=0$ in free space, or $V=1Volt$, or any other. $0$ is an easy value, pick it.

Okay but, what if there's some kind of "vacuum energy" or something like that? No one would expect that in CM, but in QM, there can be such weird things.

So, at one point, we can have $V\neq 0$. The thing is that, if you are actually in free space, if it is really free, away from any perturbation, then all points are equivalent. So it makes no sense that the point next to you has a different potential. If so, the point would be distinguishable, and so you could choose that different point as a reference. You'd have a measurable perturbation, and it would be no longer free space.

So free space must imply constant potential.

Then it turns out that you can also choose the gauge wherever, so it is the same as CM, but if it weren't, it'd be constant as well.

Hope this helped.

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One more thing, the particle doesn't have to satisfy $E>E_p(x)$ neccesarily. Especially because you can choose the reference for $V(x)$. It can be $E>E_p$ or not, depending on the origin.