Acceleration in quantum mechanics

[hokhani]

TL;DR

Theoretically, acceleration seems impossible in QM

Suppose in the presence of an electric field we solve electronic Hamiltonian with eigen energies and corresponding eigenstates $|\psi\rangle$. The action of momentum operator on the stationary states $|\psi\rangle$ doesn't change by time. So, momentum-change (or acceleration) seems meaningless. Doesn't it?

 

[hokhani]

If we have the eigenstates $|\psi\rangle$ of the electronic Hamiltonian in the presence of an applied electric field, the expectation value of the location would not change by time because $\langle \psi | \vec r |\psi \rangle$ is constant. So, how we may have charge transport?

 

[PeterDonis]

the stationary states

You're assuming that there are stationary states in the presence of an electric field. That isn't always the case. For example, in a uniform electric field, say a field of constant magnitude pointing in the $x$ direction, there are no stationary states for a charged particle.

For cases where there are stationary states, for example an electron in a hydrogen atom, where the electric field comes from the Coulomb potential of the nucleus, then it is indeed the case that the electron will not accelerate. But that is just one particular case.

 

[PeterDonis]

If we have the eigenstates $|\psi\rangle$ of the electronic Hamiltonian in the presence of an applied electric field, the expectation value of the location would not change by time because $\langle \psi | \vec r |\psi \rangle$ is constant. So, how we may have charge transport?

The response to this is the same as post #3 above, which is why I moved the quoted post to this thread.

 

[Vanadium 50]

Besides all that, of course stationary states are not undergoing acceleration. They are stationary.

 

[Demystifier]

Acceleration is possible because the state can be in a superposition of energy eigenstates, which is not a stationary state.

 

[hokhani]

You're assuming that there are stationary states in the presence of an electric field. That isn't always the case. For example, in a uniform electric field, say a field of constant magnitude pointing in the $x$ direction, there are no stationary states for a charged particle.

For cases where there are stationary states, for example an electron in a hydrogen atom, where the electric field comes from the Coulomb potential of the nucleus, then it is indeed the case that the electron will not accelerate. But that is just one particular case.

Do you mean that for an electron in an electric field, the Schrödinger equation is not soluble? Then, how to solve the quantum charge transport problem?

 

[Vanadium 50]

But didn't the problem specify stationary states?
I really wish people would post their questions verbatim. You know, follow the PF Rules and all.

 

[PeterDonis]

Do you mean that for an electron in an electric field, the Schrödinger equation is not soluble?

Which Schrödinger equation?

The time-dependent Schrödinger equation is of course soluble, and gives you solutions describing electrons that are not stationary.

The time-independent Schrödinger equation, which is what you would try to solve to find stationary states, has no solutions for the case you describe, because there are no stationary states.

Note, however, that the real problem here is that you are trying to solve for an open system: you are trying to describe just the electron in the presence of a static field, but that is an incomplete description. You are ignoring the exchange of energy between the electron and the field and just looking at the electron's energy, and of course that is not going to be conserved by itself, and hence there will be no stationary states of constant energy.

A complete description would require you to include the interaction between the electron and the field, which would mean using QED instead of non-relativistic QM. That would also allow you to describe the radiation emitted by the accelerating electron, which cannot be described by the non-relativistic model you are implicitly using.

 

[PeterDonis]

Acceleration is possible because the state can be in a superposition of energy eigenstates

While this is true as a general statement, for the particular case the OP described, there are no energy eigenstates for many types of electric field (such as the field of constant magnitude and direction I described in an earlier post). The reason is, as I said in post #9 just now, that the system in such cases is an open system and trying to describe it using non-relativistic QM with the electron as the only quantum degree of freedom and the electric field described only as a potential in the Hamiltonian is not a good model because there is no conserved energy.

 

[div_grad]

TL;DR Summary: Theoretically, acceleration seems impossible in QM

@hokhani I'd say it is perfectly possible, as understood as the time-derivative of the velocity. You can work in the Heisenberg picture in which it is the operators (and not the Schrödinger wave functions) that now have explicit time-dependence. Then the Heisenberg equation for the dynamics of the momentum $\mathbf{p}$ is
$$
i\hbar\frac{\text{d} \mathbf{p}}{\text{d}t} = \left[\mathbf{p}, H\right] + i\hbar\frac{\partial \mathbf{p}}{\partial t} \rm{,}
$$
where $\left[\cdot, H\right]$ is the commutator with the Hamiltonian $H$. Note that the operator $\mathbf{p}$ corresponds to the canonical momentum, which can be different from the kinetic momentum $\mathbf{p}_\text{kin} = m\mathbf{v}$ which is the "usual" product of the particle's mass and its velocity (which has a well-defined operational meaning). In particular, the canonical momentum of the particle with electric charge $q$ placed in the electromagnetic field described by a vector potential $\mathbf{A}$ is given by $\mathbf{p} = \mathbf{p}_\text{kin} + q\mathbf{A} = m\mathbf{v} + q\mathbf{A}$. Then the kinetic energy of the particle $\frac{m\mathbf{v}^2}{2} = \frac{\mathbf{p}^2_\text{kin}}{2m}$ (which has a well defined operational meaning as it is defined with the kinetic momentum) enters into the Hamiltonian as $\frac{\mathbf{p}^2_\text{kin}}{2m} = \frac{\left(\mathbf{p}-q\mathbf{A}\right)^2}{2m}$ and it is with respect to this Hamiltonian (including also the potential-energy terms) that the commutator in the Heisenberg equation is to be calculated. So you actually can make sense of the "accelerations" that you speak about, you just need (in the presence of the electromagnetic fields) to remember the difference between $\mathbf{p}$ and $m\mathbf{v}$ when calculating your time-derivatives.

 

[hokhani]

@hokhani I'd say it is perfectly possible, as understood as the time-derivative of the velocity. You can work in the Heisenberg picture in which it is the operators (and not the Schrödinger wave functions) that now have explicit time-dependence. Then the Heisenberg equation for the dynamics of the momentum $\mathbf{p}$ is
$$
i\hbar\frac{\text{d} \mathbf{p}}{\text{d}t} = \left[\mathbf{p}, H\right] + i\hbar\frac{\partial \mathbf{p}}{\partial t} \rm{,}
$$
where $\left[\cdot, H\right]$ is the commutator with the Hamiltonian $H$. Note that the operator $\mathbf{p}$ corresponds to the canonical momentum, which can be different from the kinetic momentum $\mathbf{p}_\text{kin} = m\mathbf{v}$ which is the "usual" product of the particle's mass and its velocity (which has a well-defined operational meaning). In particular, the canonical momentum of the particle with electric charge $q$ placed in the electromagnetic field described by a vector potential $\mathbf{A}$ is given by $\mathbf{p} = \mathbf{p}_\text{kin} + q\mathbf{A} = m\mathbf{v} + q\mathbf{A}$. Then the kinetic energy of the particle $\frac{m\mathbf{v}^2}{2} = \frac{\mathbf{p}^2_\text{kin}}{2m}$ (which has a well defined operational meaning as it is defined with the kinetic momentum) enters into the Hamiltonian as $\frac{\mathbf{p}^2_\text{kin}}{2m} = \frac{\left(\mathbf{p}-q\mathbf{A}\right)^2}{2m}$ and it is with respect to this Hamiltonian (including also the potential-energy terms) that the commutator in the Heisenberg equation is to be calculated. So you actually can make sense of the "accelerations" that you speak about, you just need (in the presence of the electromagnetic fields) to remember the difference between $\mathbf{p}$ and $m\mathbf{v}$ when calculating your time-derivatives.

It was very interesting point. Using Heisenberg representation for our definite problem we have time-dependent momentum operator as $\vec p(t)=e\vec E t+\vec p(0)$ which is more understandable without the quantum filed considerations. Furthermore, the position operator is time dependent as $\vec x(t) =\int_0^t \vec p /m dt +\vec x(0)$ which explicitly explains the charge transport.

 

[hokhani]

Which Schrödinger equation?

The time-dependent Schrödinger equation is of course soluble, and gives you solutions describing electrons that are not stationary.

The time-independent Schrödinger equation, which is what you would try to solve to find stationary states, has no solutions for the case you describe, because there are no stationary states.

Note, however, that the real problem here is that you are trying to solve for an open system: you are trying to describe just the electron in the presence of a static field, but that is an incomplete description. You are ignoring the exchange of energy between the electron and the field and just looking at the electron's energy, and of course that is not going to be conserved by itself, and hence there will be no stationary states of constant energy.

A complete description would require you to include the interaction between the electron and the field, which would mean using QED instead of non-relativistic QM. That would also allow you to describe the radiation emitted by the accelerating electron, which cannot be described by the non-relativistic model you are implicitly using.

It was a great explanation. So, it seems that in semiclassical approach to solve quantum transport, like Boltzmann equation, the interaction with electric field is ignored. Isn't it?

 

[PeterDonis]

in semiclassical approach to solve quantum transport

Isn't this a contradiction in terms?

 

[PeterDonis]

like Boltzmann equation

How is this "semiclassical" as opposed to just classical?

 

[hokhani]

Isn't this a contradiction in terms?

Right, sorry for using the word "quantum".

 

[hokhani]

How is this "semiclassical" as opposed to just classical?

In the semiclassical approach the energy states are treated quantum mechanically and the effect of external forces is included in the distribution function, obtained by Boltzmann equation.

 

[PeterDonis]

In the semiclassical approach the energy states are treated quantum mechanically and the effect of external forces is included in the distribution function, obtained by Boltzmann equation.

Do you have a reference illustrating this approach?

 

[hokhani]

Do you have a reference illustrating this approach?

Please see:
https://www.sciencedirect.com/science/article/abs/pii/B9780123044600500116