Relativistic derivation of E=1/2MV^2 from QFT or Diriac or other

[PBTR3]

It is easy to derive E=1/2mv^2 from the Schroedinger equation for the nonrelativistic one dimensional case where e^ipx-iEt/\hbar is the free traveling wave function:
i\hbar x -iE/\hbar x e^ipx-iEt/\hbar = - - \hbar^2/2m x p^2/2m x e^ipx-iEt/\hbar
which reduces to E=1/2mv^2

Where should I start to do the same thing for the relativistic, free one dimensional case? I would guess that it should reduce to E=mvc in this case if we are taking about photons.

 

[PeterDonis]

@PBTR3 please use the PF LaTeX feature when posting math, it makes it much easier to read. You can find help on it here:

https://www.physicsforums.com/help/latexhelp/

It is easy to derive E=mv^2

I think you mean $E = \frac{1}{2} m v^2$, correct?

 

[PBTR3]

@PBTR3 please use the PF LaTeX feature when posting math, it makes it much easier to read. You can find help on it here:

https://www.physicsforums.com/help/latexhelp/I think you mean $E = \frac{1}{2} m v^2$, correct?

Yes

 

[PBTR3]

c

It is easy to derive E=1/2mv^2 from the Schroedinger equation for the nonrelativistic one dimensional case where e^ipx-iEt/\hbar is the free traveling wave function:
i\hbar x -iE/\hbar x e^ipx-iEt/\hbar = - - \hbar^2/2m x p^2/2m x e^ipx-iEt/\hbar
which reduces to E=1/2mv^2

Where should I start to do the same thing for the relativistic, free one dimensional case? I would guess that it should reduce to E=mvc in this case if we are taking about photons.

 

[Nugatory]

You're not going to find a relativistic derivation of $E=mv^2/2$ because that is a non-relativistic formula; it doesn't hold when relativistic effects are significant. The relationship you're looking for is $E^2=(m_0c^2)^2+(pc)^2$ - and that's not derived from relativistic quantum mechanics; it's the other way around.

 

[PBTR3]

I agee but I should be able to show E=pc for a photon. I can use Lagrange's diff equation to recover F=ma for a nonrelativistic free particle. I can use Schoedinger's diff equation to recover E=1/2mv^2 for a nonrelativistic free particle. There should be a diff equation that recovers E=pc for a relativistic free photon or other massless particle?

 

[kith]

Light is already described as a wave in classical electromagnetism so the equation you are looking for is simply the classical electromagnetic wave equation.

However, there are a couple of issues: in relativistic quantum theory, single particle equations lead to problems like negative energies or even negative probabilities. Also the photon doesn't have a position operator which makes it difficult to introduce a wavefunction for it. These problems are solved by promoting the equations to field operator equations in quantum field theory.

 

[PBTR3]

Thanks. The photon should have an energy operator(Hamiltonian?) I need to learn how to use LaTex. It does not seem to work with Android or Linux. When I get that worked out I will resume this thread.

 

[kith]

Thanks. The photon should have an energy operator(Hamiltonian?)

Yes, one can write down such an expression (see http://www.cft.edu.pl/~birula/publ/CQO7.pdf by Bialynicki-Birula). But it is rarely used because as I said, single particle equations are problematic in the relativistic domain.

 

[king vitamin]

Start with the Klein-Gordon Hamiltonian,

$$
H = \int d^d x \, \left[ \frac{1}{2} \Pi(x,t)^2 + \left( \nabla \phi(x,t) \right)^2 + \frac{1}{2}m^2 \phi(x,t)^2 \right],
$$

where $[\phi(x,t),\Pi(x',t)] = i \delta^d(x - x')$. Then expand the fields as

$$
\phi(x,t) = \int \frac{d^d p}{(2 \pi)^d} \frac{1}{\sqrt{2 \sqrt{p^2 + m^2}}} \left( a(\mathbf{p},t) e^{i \mathbf{p} \cdot \mathbf{x}} + a^{\dagger}(\mathbf{p},t) e^{-i \mathbf{p} \cdot \mathbf{x}} \right),
$$

$$
\Pi(x,t) = - i\int \frac{d^d p}{(2 \pi)^d} \sqrt{\frac{\sqrt{p^2 + m^2}}{2}} \left( a(\mathbf{p},t) e^{i \mathbf{p} \cdot \mathbf{x}} - a^{\dagger}(\mathbf{p},t) e^{-i \mathbf{p} \cdot \mathbf{x}} \right),
$$

where the operators in the expansion must now satisfy $[a(\mathbf{p},t),a^{\dagger}(\mathbf{p}',t)] = (2 \pi)^d \delta^d(\mathbf{p} - \mathbf{p}')$. Plugging this into the Hamiltonian, you can find that (up to a constant)
$$
H = \int \frac{d^d p}{(2 \pi)^d} \sqrt{p^2 + m^2} a^{\dagger}(\mathbf{p}) a(\mathbf{p}).
$$
One can show that the spectrum of this theory is that of particles with momentum $\mathbf{p}$ and energy $E(\mathbf{p}) = \sqrt{p^2 + m^2}$. In the limit $m \gg p$, we have $E(\mathbf{p}) \approx m + \frac{p^2}{2m}$, which is basically what you're asking for. (I'm using units where the speed of light is 1.)

 

[PBTR3]

I

Great. Now I will spend some time (maybe weeks)going through your math (at 4AM?) in detail. This also implies that if m approaches zero and light is c that E=pc for a free, massless, relativistic particle, which is what I am trying to prove.