Hamiltonian density

  • #1
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Main Question or Discussion Point

Hi. In elementary quantum mechanics the continuity equation is used to derive the electron current, i.e.
[tex]
\frac{\partial \rho(\mathbf r,t)}{\partial t}+\nabla\cdot\mathbf j(\mathbf r,t)=0
[/tex]

and one then puts [itex]\rho(\mathbf r,t)=\psi^*(r,\mathbf t)\psi(\mathbf r,t)[/itex].

Now if I want to derive an expression for the energy current, the continuity equation is
[tex]
\frac{\partial H}{\partial t}+\nabla\cdot\mathbf j_E(\mathbf r,t)=0
[/tex]
where [itex]H[/itex] is the energy density(the Hamiltonian density). But what is the Hamiltonian density?
 

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  • #2
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By the way, does anyone have an expression for the energy current, or know where I can find one? I know what the particle current is, but I have actually never seen the energy current.
 
  • #3
dextercioby
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What makes you think there is one ?
 
  • #4
tom.stoer
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One can derive a continuity equation for energy; but one needs the energy-momentum (2,0) tensor:

[tex]\partial_\mu T^{\mu\nu} = 0[/tex]

(in flat space)
 
  • #5
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I suppose the Hamiltonian you are talking about here is not the Hamiltonian operator, but its expectation value.

The expression for the expectation value involves an integral over the space, and the integrand can be considered as the "Hamiltonian density".
 
  • #6
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So if the Hamiltonian density does not exist, how does one derive the expression for the energy current? Also, does anyone know an expression for the energy current, because I cannot seem to find one.
 
  • #7
tom.stoer
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As I said:

One can derive a continuity equation for energy; but one needs the energy-momentum (2,0) tensor:

[tex]\partial_\mu T^{\mu\nu} = 0[/tex]

(in flat space)
The reason is that energy is the 0-component of a 4-vector, but energy density is the 00-component of a 4*4 tensor. In some sense you can identify T°°(x) with the Hamiltonian density.
 
  • #8
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Can you give me any reference that works out the energy current? Because as soon as you start talking about 4-vectors and tensors, I'm lost. I just find it astonishing that it is so simple to derive the electron current density, but very hard to derive the energy current density.
 
  • #9
A. Neumaier
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Can you give me any reference that works out the energy current? Because as soon as you start talking about 4-vectors and tensors, I'm lost. I just find it astonishing that it is so simple to derive the electron current density, but very hard to derive the energy current density.
Chapter 7.3 in Vol. I of the QFT treatise by Weinberg.
 
  • #10
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Chapter 7.3 in Vol. I of the QFT treatise by Weinberg.
I see him deriving some very general results, but what I am looking for is the energy current equivalent of the particle current
[tex]
j(\mathbf r)=\frac{\hbar}{2mi}\left[\Psi^\dagger (\mathbf r)\nabla\Psi(\mathbf r)-(\nabla\Psi^\dagger (\mathbf r))\Psi(\mathbf r)\right)
[/tex]

But perhaps I am looking for something which does not exist? Is there no energy current which can be written in simple terms like the particle current?
 
  • #11
A. Neumaier
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I see him deriving some very general results, but what I am looking for is the energy current equivalent of the particle current
[tex]
j(\mathbf r)=\frac{\hbar}{2mi}\left[\Psi^\dagger (\mathbf r)\nabla\Psi(\mathbf r)-(\nabla\Psi^\dagger (\mathbf r))\Psi(\mathbf r)\right)
[/tex]

But perhaps I am looking for something which does not exist? Is there no energy current which can be written in simple terms like the particle current?
The energy current _is_ a very general result. if you want to have Weinberg's formulas look like the particle current, you only need to write
[tex]j_\mu(x):=T_{0\mu}(x).[/tex]
and specialize his statements about the energy-momentum tensor to one zero component. (For the other components, you get corresponding momentum currents.)
 
  • #12
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The energy current _is_ a very general result. if you want to have Weinberg's formulas look like the particle current, you only need to write
[tex]j_\mu(x):=T_{0\mu}(x).[/tex]
and specialize his statements about the energy-momentum tensor to one zero component. (For the other components, you get corresponding momentum currents.)
Would you mind stating the expression for the energy current here in the same way as I have written the particle current? I would really appreciate it.
 
  • #13
A. Neumaier
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Would you mind stating the expression for the energy current here in the same way as I have written the particle current? I would really appreciate it.
You don't get it for free - you must work for it!

Try to derive what you want as far as you can, following Weinberg and my identification, and show your attempt here, including where you got stuck. Then I'll give you directions on how to correct or extend your reasoning.
 
  • #14
tom.stoer
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I see him deriving some very general results, but what I am looking for is the energy current equivalent of the particle current ...
From your definition of the current I think what you are looking for is not the "energy current" but the probability density

[tex]\rho\sim\psi^\ast\psi[/tex]

and the probability current density

[tex]j\sim\psi^\ast\nabla\psi + \text{c.c.}[/tex]

of the non-relativistic Schroedinger equation. Is this correct?
 
  • #15
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Can you give me any reference that works out the energy current? Because as soon as you start talking about 4-vectors and tensors, I'm lost. I just find it astonishing that it is so simple to derive the electron current density, but very hard to derive the energy current density.
For a closed electromagnetic system, there is a similar relationship between the energy density and the energy flux density which you can think as the energy current density:
[tex]\dfrac{{\partial u}}{{\partial t}} + \nabla \cdot{\boldsymbol{S}} = 0[/tex]
 
  • #16
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From your definition of the current I think what you are looking for is not the "energy current" but the probability density

[tex]\rho\sim\psi^\ast\psi[/tex]

and the probability current density

[tex]j\sim\psi^\ast\nabla\psi + \text{c.c.}[/tex]

of the non-relativistic Schroedinger equation. Is this correct?
Hi Tom. No I am in fact interested in the energy current. I am just mentioning the probability current in order to explain that I am looking for the energy equivalent of the probability current. But it is true that what I am doing is completely non-relativistic. I need this energy current for calculating the thermal conductivity of a substance.

By the way, I have a paper which discusses this topic a little, but it expresses the energy current as an integral, which is not what I am looking for. See the attachment.
 

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  • #17
A. Neumaier
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Hi Tom. No I am in fact interested in the energy current. I am just mentioning the probability current in order to explain that I am looking for the energy equivalent of the probability current. But it is true that what I am doing is completely non-relativistic. I need this energy current for calculating the thermal conductivity of a substance.

By the way, I have a paper which discusses this topic a little, but it expresses the energy current as an integral, which is not what I am looking for. See the attachment.
The formula in the attachment is not correct since the j given there has no longer an r dependence. You get a correct formula by applying the generalities of http://en.wikipedia.org/wiki/Noether_theorem#Field-theoretic_derivation to the time translation symmetry of your system. (Weinberg is not appropriate in your case since your system is nonrelativistic.)
 
  • #18
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The formula in the attachment is not correct since the j given there has no longer an r dependence. You get a correct formula by applying the generalities of http://en.wikipedia.org/wiki/Noether_theorem#Field-theoretic_derivation to the time translation symmetry of your system. (Weinberg is not appropriate in your case since your system is nonrelativistic.)
If you want to you can see the full article in this attachment. Also in formula (2) he defines a current which has no r-dependence. I think the article looks pretty legitimate, so it would be strange if it had large errors.
 

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  • #19
dextercioby
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That current is just a number, because the fields under the integral are solutions of the field equations. It loses any r or t dependence, since the space dependence is all integrated wrt, while the t dependence is lost because j is a conserved current. jt_1)=j(t_2).
 
  • #20
A. Neumaier
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That current is just a number, because the fields under the integral are solutions of the field equations. It loses any r or t dependence, since the space dependence is all integrated wrt, while the t dependence is lost because j is a conserved current. jt_1)=j(t_2).
How can such a current satisfy the continuity equation, which characterises the fact that the total energy is conserved?
 
  • #21
tom.stoer
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It is by no means clear why they call it a 'current' - it's integrated over x. For a conserved current one would expect a local conservation law, a continuity equation for charge and current density.
 
  • #22
A. Neumaier
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If you want to you can see the full article in this attachment. Also in formula (2) he defines a current which has no r-dependence. I think the article looks pretty legitimate, so it would be strange if it had large errors.
It is dangerous to rely on formulas written elsewhere without being able to check them yourself. Being published does not imply being correct.

The notation in the paper is _very_ sloppy. You can check this already by looking at his formula (2) for the charge current, specialize to the case of zero external vector potential, and comparing it with the correct expression that you wrote down in #10 (apart perhaps from the factor).

So you need to learn the tools of the trade that enable you to apply Noether's theorem yourself!
 
  • #23
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It is dangerous to rely on formulas written elsewhere without being able to check them yourself. Being published does not imply being correct.

The notation in the paper is _very_ sloppy. You can check this already by looking at his formula (2) for the charge current, specialize to the case of zero external vector potential, and comparing it with the correct expression that you wrote down in #10 (apart perhaps from the factor).

So you need to learn the tools of the trade that enable you to apply Noether's theorem yourself!
I completely agree with you. I have to learn those things, I just don't have the time right now. Maybe later in the week. I was just hoping I could derive the heat current from the continuity equation and avoid Lagrangian densities for now. By the way, I think the article uses a different definition of the current than what we are used to. If you look at pages 30-31 of G.D. Mahan(attached), he has a current which is defined as the volume integral of the current density, when he is discussing tight binding. (on top page 31 he puts the current equal to dP/dt, which on bottom page 30 is defined as the volume integral of the current density)
 

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  • #24
A. Neumaier
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I completely agree with you. I have to learn those things, I just don't have the time right now. Maybe later in the week. I was just hoping I could derive the heat current from the continuity equation and avoid Lagrangian densities for now. By the way, I think the article uses a different definition of the current than what we are used to. If you look at pages 30-31 of G.D. Mahan(attached), he has a current which is defined as the volume integral of the current density, when he is discussing tight binding. (on top page 31 he puts the current equal to dP/dt, which on bottom page 30 is defined as the volume integral of the current density)
(1.2.18) has a q-dependent current density, as it should be.
 
  • #25
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Hey. I did a calculation now(in second quantization, nonrelativistic), and I found an expression for the energy current.
 

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