Hamiltonian in Quantum vs Classical

[Electric to be]

The Hamiltonian in classical mechanics is not always equal to the total energy of the system. I believe this is only true if there is only a potential field and no vector potential. However, in quantum mechanics for a particle in an EM field, even if a vector potential is used the energy operator E -> H. Is this an inconsistency?

While H is not always equal to E in classical mechanics, H is still always regarded as the operator for E when there are not exclusively scalar potentials in QM?

Thanks.

 

[Electric to be]

Bumping it with a little addition. If there isn't a scalar potential energy term at all, (so no conservative field) how is total energy even defined in quantum mechanics (at least non-relativistic QM) ?

Is it simply the Eigenvalue of the Hamiltonian? (if it isn't K + U? ) Is it simply just the relative state levels between which photons can be absorbed? Obviously energy is usually a relative concept defined up to a constant, but here it seems that energy is simply defined as being the eigenvalue of the Hamiltonian if it isn't equal to K+U. Just like in classical mechanics where the Hamiltonian isn't always equal to the total energy K+U.

 

[PeterDonis]

The Hamiltonian in classical mechanics is not always equal to the total energy of the system. I believe this is only true if there is only a potential field and no vector potential.

Can you give a reference? You might be confusing energy with momentum.

If there isn't a scalar potential energy term at all, (so no conservative field) how is total energy even defined in quantum mechanics (at least non-relativistic QM) ?

Energy is an observable, so in QM it's a self-adjoint operator, like every observable. The particular operator for the energy is the Hamiltonian operator.

 

[vanhees71]

Do you have a concrete example, where the Hamiltonian is not the total energy of the system?

In #1 you claimed that the Hamiltonian for a particle in an electromagnetic field is not the total energy, but that's obviously not true. For simplicity let's consider the non-relativistic approximation for a single particle in an electromagnetic field. We start from the Lagrangian
$$L=\frac{m}{2} \dot{\vec{x}^2}-q \phi(t,\vec{x}) + \frac{q}{c} \dot{\vec{x}} \cdot \vec{A}(t,\vec{x}),$$
where $\phi$ is the scalar and $\vec{A}$ the vector potential of the electromagnetic field.

The canonical momenta are
$$\vec{p}_{\text{can}}=\frac{\partial L}{\partial \dot{\vec{x}}}=m \dot{\vec{x}}+\frac{q}{c} \vec{A}=\vec{p}_{\text{mech}} + \frac{q}{c} \vec{A}.$$
So in this case the canonical momentum is not the mechanical momentum. Maybe that's the cause of your confusion.

The Hamiltonian is
$$H=\dot{\vec{x}} \cdot \vec{p}_{\text{can}}-L=\frac{m}{2} \dot{\vec{x}}^2+q \phi(t,\vec{x}),$$
which is the total energy of the system.

Of course for the Hamilton formalism you have to express $\dot{\vec{x}}$ in terms of the canonical (!) momentum, leading to
$$H=\frac{1}{2m} \left [\vec{p}_{\text{can}}-\frac{q}{c} \vec{A}(t,\vec{x}) \right]^2+q \phi(t,\vec{x}).$$
Since in the square brackets you have $\vec{p}_{\text{mech}}$, it's still the standard expression for the total energy.

That's also clear from the equations of motion, following from the Lagrange or Hamilton equations:
$$m \ddot{\vec{x}}=q \left (\vec{E} + \frac{1}{c} \dot{\vec{x}} \times \vec{B} \right ),$$
where
$$\vec{E}=-\vec{\nabla} \phi-\frac{1}{c} \partial_t \vec{A}, \quad \vec{B}=\vec{\nabla} \times \vec{A}.$$
When using the heuristics of canonical quantization in the wave-mechanics (position-space) representation, it's $p_{\text{can}}$ which is substituted with $$\hat{\vec{p}}_{\text{can}}=-\mathrm{i} \hbar \vec{\nabla},$$
and thus you get the Schrödinger equation in the form
$$\mathrm{i} \partial_t \psi(t,\vec{x})=\hat{H} \psi(t,\vec{x})=-\frac{1}{2m} \left [\vec{\nabla}-\frac{q}{c} \vec{A}(t,x) \right ]^2 \psi + q \phi(t,\vec{x}) \psi(t,\vec{x}).$$
The Hamiltonian is still the operator representing the total energy of the particle.

 

[Electric to be]

Do you have a concrete example, where the Hamiltonian is not the total energy of the system?

In #1 you claimed that the Hamiltonian for a particle in an electromagnetic field is not the total energy, but that's obviously not true. For simplicity let's consider the non-relativistic approximation for a single particle in an electromagnetic field. We start from the Lagrangian
$$L=\frac{m}{2} \dot{\vec{x}^2}-q \phi(t,\vec{x}) + \frac{q}{c} \dot{\vec{x}} \cdot \vec{A}(t,\vec{x}),$$
where $\phi$ is the scalar and $\vec{A}$ the vector potential of the electromagnetic field.

The canonical momenta are
$$\vec{p}_{\text{can}}=\frac{\partial L}{\partial \dot{\vec{x}}}=m \dot{\vec{x}}+\frac{q}{c} \vec{A}=\vec{p}_{\text{mech}} + \frac{q}{c} \vec{A}.$$
So in this case the canonical momentum is not the mechanical momentum. Maybe that's the cause of your confusion.

The Hamiltonian is
$$H=\dot{\vec{x}} \cdot \vec{p}_{\text{can}}-L=\frac{m}{2} \dot{\vec{x}}^2+q \phi(t,\vec{x}),$$
which is the total energy of the system.

Of course for the Hamilton formalism you have to express $\dot{\vec{x}}$ in terms of the canonical (!) momentum, leading to
$$H=\frac{1}{2m} \left [\vec{p}_{\text{can}}-\frac{q}{c} \vec{A}(t,\vec{x}) \right]^2+q \phi(t,\vec{x}).$$
Since in the square brackets you have $\vec{p}_{\text{mech}}$, it's still the standard expression for the total energy.

That's also clear from the equations of motion, following from the Lagrange or Hamilton equations:
$$m \ddot{\vec{x}}=q \left (\vec{E} + \frac{1}{c} \dot{\vec{x}} \times \vec{B} \right ),$$
where
$$\vec{E}=-\vec{\nabla} \phi-\frac{1}{c} \partial_t \vec{A}, \quad \vec{B}=\vec{\nabla} \times \vec{A}.$$
When using the heuristics of canonical quantization in the wave-mechanics (position-space) representation, it's $p_{\text{can}}$ which is substituted with $$\hat{\vec{p}}_{\text{can}}=-\mathrm{i} \hbar \vec{\nabla},$$
and thus you get the Schrödinger equation in the form
$$\mathrm{i} \partial_t \psi(t,\vec{x})=\hat{H} \psi(t,\vec{x})=-\frac{1}{2m} \left [\vec{\nabla}-\frac{q}{c} \vec{A}(t,x) \right ]^2 \psi + q \phi(t,\vec{x}) \psi(t,\vec{x}).$$
The Hamiltonian is still the operator representing the total energy of the particle.

I will look for an example and update when I get the chance to do so. I just have read this statement on other parts of the Internet, but I attempt to provide a source to answer this question.However, in terms of what you said. I am not very familiar with canonical momentum, but does that classically capture the total energy of the system? In a classical syste the total energy of a charge in an EM field should be the energy density of the E and B field integrated over all space + kinetic energy of the particle. (And also rest mass if you want to go relativistic)

Since the canonical momentum isn't simply the momentum of the particle, does it somehow bring in the energy in the magnetic and electric field?

 

[hilbert2]

I guess it's possible to add a term like $H'(t) = Ae^{Bt}$, where $A$ and $B$ are constants and $t$ is the time variable, in the Hamiltonian function to get an explicit time dependence that doesn't affect any actual dynamics. Then it wouldn't really make sense to say that the Hamiltonian is equal to any total energy that is defined in a practical manner.

 

[Electric to be]

I guess it's possible to add a term like $H'(t) = Ae^{Bt}$, where $A$ and $B$ are constants and $t$ is the time variable, in the Hamiltonian function to get an explicit time dependence that doesn't affect any actual dynamics. Then it wouldn't really make sense to say that the Hamiltonian is equal to any total energy that is defined in a practical manner.

Here is a thread that talks about this issue. https://physics.stackexchange.com/q...ian-of-a-system-not-equal-to-its-total-energy

I do understand your derivation now, thought would seem to work for a static, conservative electric field & static magnetic field. If the EM field included some non-conservative electric field then it would break down and the Hamiltonian wouldn't equal to the total energy.

So - it seems that while there are times in classical mechanics where the Hamiltonian isn't equal to the total energy, we still always take the eigenvalue of the quantum Hamiltonian to be defined as the energy simply as a postulate. Obviously there is no requirement for the QM to satisfy this classical analogy, but it would have been nice :)

In this case, I guess it doesn't really make sense to talk about energy as having any individual components in a classical sense like the sum of kinetic and potential energy, but the energies of the quantum Hamiltonian are simply "some quantity" DEFINED as the eigenvalue of this problem, in which the differences between photon frequency jumps can be calculated from. And in cases where the classical hamiltonian IS equal to the total classical energy, it just so happens that these two "energies" match.

At least that's what I seem to have gathered.

 

[PeterDonis]

@Electric to be please do not quote the entire post you're responding to. Just quote the specific parts you're responding to. If you highlight a particular part of a post, and use the "reply" popup, it will appear quoted in the edit box.

 

[vanhees71]

I will look for an example and update when I get the chance to do so. I just have read this statement on other parts of the Internet, but I attempt to provide a source to answer this question.However, in terms of what you said. I am not very familiar with canonical momentum, but does that classically capture the total energy of the system? In a classical syste the total energy of a charge in an EM field should be the energy density of the E and B field integrated over all space + kinetic energy of the particle. (And also rest mass if you want to go relativistic)

Since the canonical momentum isn't simply the momentum of the particle, does it somehow bring in the energy in the magnetic and electric field?

No, in the static case, i.e., for time-indpendent $\vec{E}$ and $\vec{B}$ the total energy of the particle in the external field is $\mathcal{E}=T+V=m \vec{v}^2/2+q \phi$.

This you get unambiguously from the equations of motion
$$m \ddot{\vec{x}}=q \left (\vec{E}+\frac{\dot{\vec{x}}}{c} \times \vec{B} \right)$$
Multiplying with $\dot{\vec{x}}$ gives
$$\frac{\mathrm{d}}{\mathrm{d} t} \left (\frac{m}{2} \dot{\vec{x}}^2 \right)=-q \frac{\mathrm{d} \phi}{\mathrm{d} t},$$
if we have static fields, i.e., $\vec{E}=-\vec{\nabla} \phi$, and then the Hamiltonian is the total energy.

For time dependent fields, you are right that $H$ is not the total energy of the system, and because then $\vec{E}=-\vec{\nabla} \phi-\partial_t \vec{A}$ is no longer a potential field anymore, you cannot derive the force from a potential, and it's not clear how to define total energy. You can only define the work along the trajectory of the particle, as usual for non-conservative forces! So I was inaccurate in claiming that also here $H=E$! In the most general case it's even difficult to define what total energy means.

 

[muscaria]

and because then ⃗E=−⃗∇ϕ−∂t⃗A\vec{E}=-\vec{\nabla} \phi-\partial_t \vec{A} is no longer a potential field anymore, you cannot derive the force from a potential, and it's not clear how to define total energy

.
You can define the total energy as $K+V$ if you include the field energy term $\textbf{A}\cdot\textbf{v}$ in your definition of the potential energy. In terms of the work function $U(q_i,\dot{q}_i)$, the Lorentz force emerges as a result of
$$F_k=\frac{\partial U}{\partial q_k}-\frac{d}{dt}\frac{\partial U}{\partial\dot{q}_k}$$.
If you define $$V=\sum_i\frac{\partial U}{\partial\dot{q}_i}\dot{q}_i -U$$ then the total energy is potential plus kinetic.

 

[atyy]

For time dependent fields, you are right that $H$ is not the total energy of the system, and because then $\vec{E}=-\vec{\nabla} \phi-\partial_t \vec{A}$ is no longer a potential field anymore, you cannot derive the force from a potential, and it's not clear how to define total energy.

Could one define the energy as the conserved quantity due to time reversal symmetry in Noether's theorem?

Then the question is whether H = E even in the presence of a vector potential. I found an interesting discussion in http://users.physics.harvard.edu/~morii/phys151/lectures/Lecture18.pdf. It seems that there is indeed a problem with potentials. However it turns out the problem goes away, unless there is explicit time dependence in the Hamiltonian.

 

[muscaria]

Could one define the energy as the conserved quantity due to time reversal symmetry in Noether's theorem?

When you make time a dynamical variable in phase space and treat it as a variational variable, the canonical momentum of the time coordinate is the Hamiltonian. Does this answer your question?

EDIT: See Variational principles by "Lanczos"

 

[Electric to be]

In the most general case it's even difficult to define what total energy means.

For a classical system - a charge interacting with an electric field will always have the total energy conserved. However, instead of attributing the potential energy to saw energy associated with the charge, the energy is store in the field. (Energy density is proportional to B^2 and E^2)

I believe there are lagrangian formalisms that cover the evolution of the electromagnetic field with a certain amount of charges. However, you're right when a single particle Hamiltonian is introduced with the fields essentially as a set parameter that aren't influenced by the motions of the particle it isn't exacty possible to define total energy. As I stated earlier, this case was my source of confusion in quantizing the Hamiltonian and calling it the operator for energy. Like I said above, it isn't necessary for this to hold true though. Energy is simply defined as the eigenvalues of the Hamiltonian, not as K+U.

 

[vanhees71]

Could one define the energy as the conserved quantity due to time reversal symmetry in Noether's theorem?

Then the question is whether H = E even in the presence of a vector potential. I found an interesting discussion in http://users.physics.harvard.edu/~morii/phys151/lectures/Lecture18.pdf. It seems that there is indeed a problem with potentials. However it turns out the problem goes away, unless there is explicit time dependence in the Hamiltonian.

Energy is the conserved quantity due to time-translation symmetry. If, however, the external fields are explicitly time-dependent that's no longer a symmet since for this the condition is $(\partial_t L)_{\text{expl}}=0$.

 

[vanhees71]

For a classical system - a charge interacting with an electric field will always have the total energy conserved. However, instead of attributing the potential energy to saw energy associated with the charge, the energy is store in the field. (Energy density is proportional to B^2 and E^2)

I believe there are lagrangian formalisms that cover the evolution of the electromagnetic field with a certain amount of charges. However, you're right when a single particle Hamiltonian is introduced with the fields essentially as a set parameter that aren't influenced by the motions of the particle it isn't exacty possible to define total energy. As I stated earlier, this case was my source of confusion in quantizing the Hamiltonian and calling it the operator for energy. Like I said above, it isn't necessary for this to hold true though. Energy is simply defined as the eigenvalues of the Hamiltonian, not as K+U.

That's a very tricky issue. A fully self-consistent classical Hamiltonian for classical charged point particles + field, including radiation reaction, doesn't exist. That's the limit, where the classically finally breaks down, and the correct description is QED, which is at least well defined order by order in perturbation theory.

 

[Electric to be]

That's a very tricky issue. A fully self-consistent classical Hamiltonian for classical charged point particles + field, including radiation reaction, doesn't exist. That's the limit, where the classically finally breaks down, and the correct description is QED, which is at least well defined order by order in perturbation theory.

Well I think there is some form of "Hamiltonian field theory" for classical fields that can describe evolution of fields and particles within it. Though if I had to guess this would probably only work for relatively simple configurations, and would otherwise have to be numerically solvable? I'm not an expert in Lagrangian mechanics. http://www.phys.lsu.edu/~jarrell/COURSES/ELECTRODYNAMICS/Chap12/chap12.pdf

This wouldn't be used for Schrodinger equation like QM like you said, but would be used (and is) in QED.

 

[atyy]

For a classical system - a charge interacting with an electric field will always have the total energy conserved. However, instead of attributing the potential energy to saw energy associated with the charge, the energy is store in the field. (Energy density is proportional to B^2 and E^2)

I believe there are lagrangian formalisms that cover the evolution of the electromagnetic field with a certain amount of charges. However, you're right when a single particle Hamiltonian is introduced with the fields essentially as a set parameter that aren't influenced by the motions of the particle it isn't exacty possible to define total energy. As I stated earlier, this case was my source of confusion in quantizing the Hamiltonian and calling it the operator for energy. Like I said above, it isn't necessary for this to hold true though. Energy is simply defined as the eigenvalues of the Hamiltonian, not as K+U.

Have you seen http://users.physics.harvard.edu/~morii/phys151/lectures/Lecture18.pdf ? It discusses that H = E even if there is a vector potential.

 

[Electric to be]

Have you seen http://users.physics.harvard.edu/~morii/phys151/lectures/Lecture18.pdf ? It discusses that H = E even if there is a vector potential.

Notice however, they do assume that a scalar electric potential can at least be written (even if it happens to be zero). However, if you have a non-conservative e-field (so time varying EM field) then this concept breaks down. That link also talks about that (must be time independent).

 

[vanhees71]

Have you seen http://users.physics.harvard.edu/~morii/phys151/lectures/Lecture18.pdf ? It discusses that H = E even if there is a vector potential.

It depends on, how you define "Energy". For a particle in an explicity time-dependent electromagnetic field $H$ is not the expression occurring in the work-energy theorem. Take my calculation from #9 and apply it to this more general case of non-static fields. Then the work-energy theorem reads, using $T=m \dot{\vec{x}}^2/2$
$$\dot{T}=q \dot{\vec{x}} \cdot \vec{E}=q \dot{\vec{x}} (-\vec{\nabla} \phi -\partial_t \vec{A})= -q (\dot{\phi}-\partial_t \phi) -q \dot{\vec{x}} \cdot \partial_t \vec{A}.$$
Or
$$\frac{\mathrm{d}}{\mathrm{d} t} (T+q \phi)=q (\partial_t \phi - \dot{\vec{x}} \cdot \partial_t \vec{A}).$$
Of course you can derive this also from
$$\frac{\mathrm{d} H}{\mathrm{d} t}=\frac{\partial H}{\partial t}=-\frac{\partial L}{\partial t} = q \partial_t \phi -q \dot{\vec{x}} \cdot \partial_t \vec{A}.$$
Note that all this holds only for the trajectory of the particle, i.e., the solution of the equations of motion!

 

[atyy]

It depends on, how you define "Energy". For a particle in an explicity time-dependent electromagnetic field $H$ is not the expression occurring in the work-energy theorem. Take my calculation from #9 and apply it to this more general case of non-static fields. Then the work-energy theorem reads, using $T=m \dot{\vec{x}}^2/2$
$$\dot{T}=q \dot{\vec{x}} \cdot \vec{E}=q \dot{\vec{x}} (-\vec{\nabla} \phi -\partial_t \vec{A})= -q (\dot{\phi}-\partial_t \phi) -q \dot{\vec{x}} \cdot \partial_t \vec{A}.$$
Or
$$\frac{\mathrm{d}}{\mathrm{d} t} (T+q \phi)=q (\partial_t \phi - \dot{\vec{x}} \cdot \partial_t \vec{A}).$$
Of course you can derive this also from
$$\frac{\mathrm{d} H}{\mathrm{d} t}=\frac{\partial H}{\partial t}=-\frac{\partial L}{\partial t} = q \partial_t \phi -q \dot{\vec{x}} \cdot \partial_t \vec{A}.$$
Note that all this holds only for the trajectory of the particle, i.e., the solution of the equations of motion!

In the case of explicit time-dependence, does the Noether's theorem energy still exist? My intuition is that explicit time-dependence breaks time translation, so the Noether theorem would fail, and energy would not be defined.

 

[Orodruin]

In the case of explicit time-dependence, does the Noether's theorem energy still exist? My intuition is that explicit time-dependence breaks time translation, so the Noether theorem would fail, and energy would not be defined.

Does momentum exist when a system is not translationally invariant?

 

[vanhees71]

In the case of explicit time-dependence, does the Noether's theorem energy still exist? My intuition is that explicit time-dependence breaks time translation, so the Noether theorem would fail, and energy would not be defined.

Noether's theorem tells you that if the Hamiltonian is explicitly time dependent, then it is no longer transverse. With Poisson brackets it's very easy to see:
$$\frac{\mathrm{d}}{\mathrm{d} t} H=\{H,H\}+\partial_t H=\partial_t H.$$
So $H$ is conserved if and only if it is not explicitly time dependent. Nevertheless by definition it still generates time translations, and as such you can define it to be the energy of the system, even if it's no longer a conserved quantity if the Hamiltonian is explicitly time dependent.

 

[vanhees71]

Does momentum exist when a system is not translationally invariant?

You can still define it as the quantity that generates spatial translations, but it won't be conserved anymore.

 

[Orodruin]

You can still define it as the quantity that generates spatial translations, but it won't be conserved anymore.

Obviously. It was a leading question for atyy.

 

[atyy]

Noether's theorem tells you that if the Hamiltonian is explicitly time dependent, then it is no longer transverse. With Poisson brackets it's very easy to see:
$$\frac{\mathrm{d}}{\mathrm{d} t} H=\{H,H\}+\partial_t H=\partial_t H.$$
So $H$ is conserved if and only if it is not explicitly time dependent. Nevertheless by definition it still generates time translations, and as such you can define it to be the energy of the system, even if it's no longer a conserved quantity if the Hamiltonian is explicitly time dependent.

But then H=E by definition, so it doesn't make sense to discuss conditions for H=E.

 

[Electric to be]

So my question still kind of remains. If there is a non-conservative field, say an E field with Curl. The Hamiltonian will still be defined as the operator for energy in QM, but how is energy in general defined then? (since there isn't any scalar potential energy). Is it simply some quantity that defines the differences between photon absorption levels?

 

[muscaria]

The Hamiltonian in classical mechanics is not always equal to the total energy of the system. I believe this is only true if there is only a potential field and no vector potential.

So my question still kind of remains. If there is a non-conservative field, say an E field with Curl. The Hamiltonian will still be defined as the operator for energy in QM, but how is energy in general defined then? (since there isn't any scalar potential energy).

Energy is defined in general as follows for the case I think you have in mind. We can split the Lagrangian $L(q,\dot{q})=T(q,\dot{q})-U(q,\dot{q})$ into a quadratic part (kinetic) and whatever remains, as
$$L=\frac{1}{2}\sum_{ij}c_{ij}\dot{q}_i\dot{q}_j-U(q,\dot{q}).$$
$U$ is the generalised potential, which will contain both scalar and vector potentials. The Hamiltonian is defined in terms of the Legendre transform $$H=\sum_i \frac{\partial L}{\partial\dot{q}_i}\dot{q}_i-L=\sum_i p_i\dot{q}_i-L.$$ In terms of the above expression for $L$, the canonical momenta read $$p_k=\sum_{ki}c_{ki}\dot{q}_i-\frac{\partial U}{\partial \dot{q}_i},$$
where we use the fact that $c_{ij}=c_{ji}$. In the case of electromagnetism, the 2nd term on the right hand side of the above expression is $-A_i$, i.e. the negative of the vector potential.
Inserting the canonical momenta into the above Legendre tranform, we find that $$H=\sum_{ij}c_{ij}\dot{q}_i\dot{q}_j-\sum_i\frac{\partial U}{\partial\dot{q}_i}\dot{q}_i-L=T+U-\sum_i\frac{\partial U}{\partial\dot{q}_i}\dot{q}_i$$.
This is a general expression for the total energy of a system with fixed energy (may need to be a Galilean invariant system though) when the generalised potential is velocity dependent (Is this the general expression for the energy you are looking for?). The point is that it isn't the scalar potential energy $V$ which is the fundamental quantity but the generalised scalar potential $U$. The vector potential can be derived from $U$ through partial differentiation, the components of $A$ are the slopes of $U$ with respect to the velocities. Any curl of an electric field etc.. will be encoded in this.
However you can still define the total energy as being "kinetic" + "potential" energies: $H=T+V$, so long as you define $$V=U-\sum_i\frac{\partial U}{\partial\dot{q}_i}\dot{q}_i.$$
In the case of EM: $$U=-\sum_i A_i\dot{q}_i+\phi$$ and so the potential energy $V$ is $\phi$ and the Hamiltonian is $$H=\frac{(\mathbf{p}-\mathbf{A})^2}{2m}+\phi=T+V.$$
You can also see that this gives the correct expression for the Lorentz force, as follows.
The Euler-Lagrange equations $$\frac{\partial L}{\partial q_k}-\frac{d}{dt}\left(\frac{\partial L}{\partial\dot{q}_k}\right)=0,$$ splits off into a generalised force (The F in Newton's equation, here associated with the generalised potential): $$F_k=-\frac{\partial U}{\partial q_k}+\frac{d}{dt}\left(\frac{\partial U}{\partial\dot{q}_k}\right),$$
and a force of inertia (The $ma$ in Newton's equation, here absorbed in the kinetic energy function): $$-\frac{\partial T}{\partial q_k}+\frac{d}{dt}\left(\frac{\partial T}{\partial\dot{q}_k}\right)=F_k.$$ Substituting the earlier general definition of $V$ in terms of $U$ into the above expression for $F_k$, you find that the generalised force takes the Lorentz form $$F_k=-\frac{\partial V}{\partial q_k}-\frac{\partial A_k}{\partial t}+\sum_j\dot{q}_j\left(\frac{\partial A_j}{\partial q_k}-\frac{\partial A_k}{\partial q_j}\right),$$
where $A_i=\frac{\partial U}{\partial\dot{q}_i}$ and we use the fact that the total time derivative $$\frac{dA_k}{dt}=\frac{\partial A_k}{\partial t}+\sum_i\dot{q}_i\frac{\partial A_k}{\partial q_i}.$$

Does this help at all?

 

[Electric to be]

Does this help at all?

I am unfortunately not super familiar with Lagrangian mechanics, and mainly have just studied its' connection to QM. From what I saw here, the quantity known as energy was defined in some sense using generalized potentials. But I don't believe this is the same as the traditional energy which should only include real quantities of energy, such as kinetic energy, and energy stored in fields. In the case of a single particle, the energy stored in fields can be assigned to a single particle through the use of a scalar potential (for the case of a static E-field), but the energy is still in reality stored in the field.

So my question. Say we have a single particle? In non-relativistic QM there is no notion of assigning energy to the field since the field isn't quantized. However, say there is a changing EM field that will have curls etc. and not be static. Some energy can still be found as the eigenvalue of the Hamiltonian, but this certainly isn't equal to the kinetic energy + (some energy stored in the field). It seems like your definition of energy is simply defining energy as something, but it doesn't the components of energy present in classical mechanics?

 

[vanhees71]

The Hamiltonian is defined as the operator of time evolution. If it's explicitly time dependent it's not representing a conserved quantity. It's up to you, whether you call it then still "energy" or not. It's just semantics.

 

[Electric to be]

The Hamiltonian is defined as the operator of time evolution. If it's explicitly time dependent it's not representing a conserved quantity. It's up to you, whether you call it then still "energy" or not. It's just semantics.

But the operator for the "energy" observable is defined as the Hamiltonian.

 

[PeterDonis]

The Hamiltonian is defined as the operator of time evolution.

This is not quite correct. The time evolution operator is $e^{i H t}$, where $H$ is the Hamiltonian.

the operator for the "energy" observable is defined as the Hamiltonian.

Yes, that's correct. But @vanhees71 is correct that, if the Hamiltonian is explicitly time dependent, then "energy" is not a conserved quantity--which violates many people's intuition about how "energy" should work. In other words, the issue here is not whether the operator for "energy" is the Hamiltonian or something else; the issue is that you are expecting "energy" to be conserved in situations where it isn't.

 

[vanhees71]

If $\hat{H}$ is time dependent the unitary time-evolution operator (for states in the Schrödinger picture) is
$$\hat{U}(t,t_0) = \mathcal{T} \exp \left [-\mathrm{i} \int_{t_0}^t \mathrm{d} t' \hat{H}(t') \right].$$
So strictly speaking the Hamiltonian is the generator for time evolution.

 

[Orodruin]

This is not quite correct. The time evolution operator is $e^{i H t}$, where $H$ is the Hamiltonian.

In other words, the Hamiltonian is the generator of time translations. This is true clasically as well as in QM.

 

[Electric to be]

the issue is that you are expecting "energy" to be conserved in situations where it isn't.

Well no I'm not expecting energy to be conserved. I'm perfectly fine with the energy having some time dependence - what I am concerned with is how "energy" is defined in these situations. It seems that the basic definition of energy in QM is simply the eigenvalue of the Hamiltonian. While in situations like the hydrogen atom this makes sense classically (sum of kinetic and potential energies & when E>0 we have free particles), but in other situations it doesn't and seems to be simply defined as the Eigenvalue of the Hamiltonian without a real physical meaning like other physical observables like position and momentum.

Classically, the energy of a system that has work done on it by an external system (and therefore has the total energy increased) still has a very clear definition for it's total energy. It's the sum of kinetic energies, energy stored in fields, etc. However, here I just don't see how the energy is actually defined in the case of a changing EM field since in QM we aren't assigning any energy to the field itself, and we are treating the particle as "holding" all the energy through things like scalar potentials (when they can be defined).

One possible explanation (?) is that in certain cases the overall "energy" of the particle resembles the classical situation: kinetic + potential. This applies for static fields. Otherwise for a general EM field the "energy" simply represents possible states between which photons can be absorbed, and not the sum of it's parts (kinetic + potential).

 

[PeterDonis]

It seems that the basic definition of energy in QM is simply the eigenvalue of the Hamiltonian.

Only if the system is in an eigenstate of the Hamiltonian. If the system is not in an eigenstate of the Hamiltonian, then it doesn't have a definite energy; all you have is probabilities of different energies being measured if you measure the energy.

in QM we aren't assigning any energy to the field itself

What makes you think that? The field is part of the total system, and the Hamiltonian operator acts on the total system.

What you can't usually do in QM is split up the energy of the total system into "energy of the particle" and "energy of the field". You can only do this in particular cases when the system's state happens to be separable, i.e., it's just a single product of a particle state and a field state. But in most cases the state of the total system is entangled, so the "particle" and "field" don't have well-defined states by themselves, which means they don't have well-defined energies by themselves either.

Basically it seems to me that you are trying to apply classical thinking to the quantum case where it doesn't work.