# Alternative formula for the Hamiltonian

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• grzz
In summary, Fermi's equation (13) for the relativistic hamiltonian for a point charge is:0 = -\frac 1 {2m} \left( \left[ mc +\frac {W_a - ~eV} c \right ]^2 - \left[ p -\frac {eU} c \right ]^2 \right) + \frac {mc^2} 2.
grzz
In his article 'Quantum theory of radiation', Reviews of Modern Physics, Jan 1932, volume 4, Fermi gives the relativistic hamiltonian function ##W_a## for a point charge by equation (13),

## 0 = - \frac 1 {2m} \left( \left[ mc +\frac {W_a - ~eV} c \right ]^2 - \left[ p -\frac {eU} c \right ]^2 \right) + \frac {mc^2} 2. ##

Is it possible to derive the above equation from,

## \left[ {W_a - ~eV} \right ]^2 = \left[ p -\frac {eU} c \right ] ^2 c^2 + m^2c^4? ##

Thanks.

I tried the following.

## \left[ {W_a - ~eV} \right ]^2 = \left[ p -\frac {eU} c \right ] ^2 c^2 + m^2c^4 ##

## \left[ \frac{W_a - ~eV} c \right ]^2 = \left[ p -\frac {eU} c \right ] ^2 + m^2c^2 ##

## 0 = -\frac 1 {2m}\left(\left[ \frac{W_a - ~eV} c \right ]^2 - \left[ p -\frac {eU} c \right ] ^2 \right) + \frac {mc^2} 2. ##

This was as far as I could go. I am assuming that ##W_a## includes the rest mass energy ##mc^2##.
Perhaps I am missing something.

Any help?

Thanks.

Last edited:
A constant change in the total energy (e.g. adding mc2) doesn't change the physics (as long as we are not dealing with gravity).

I am still wondering from where did Fermi got his equation (13).

grzz said:
In his article 'Quantum theory of radiation', Reviews of Modern Physics, Jan 1932, volume 4, Fermi gives the relativistic hamiltonian function ##W_a## for a point charge by equation (13),

## 0 = - \frac 1 {2m} \left( \left[ mc +\frac {W_a - ~eV} c \right ]^2 - \left[ p -\frac {eU} c \right ]^2 \right) + \frac {mc^2} 2. ##

Is it possible to derive the above equation from,

## \left[ {W_a - ~eV} \right ]^2 = \left[ p -\frac {eU} c \right ] ^2 c^2 + m^2c^4? ##

Thanks.
Where did you get the second equation from?

kimbyd said:
Where did you get the second equation from?

I am using the usual

##E^2 = p^2 c^2 + {m_o}^2c^4,##

and replacing E by ##W_a##, assuming ##W_a## stands for the total energy of the particle.

I am also putting in the scalar potential V and vector potential U since the charged particle is in an electromagnetic field.

grzz said:
In his article 'Quantum theory of radiation', Reviews of Modern Physics, Jan 1932, volume 4, Fermi gives the relativistic hamiltonian function ##W_a## for a point charge by equation (13),

## 0 = - \frac 1 {2m} \left( \left[ mc +\frac {W_a - ~eV} c \right ]^2 - \left[ p -\frac {eU} c \right ]^2 \right) + \frac {mc^2} 2. ##

Is it possible to derive the above equation from,

## \left[ {W_a - ~eV} \right ]^2 = \left[ p -\frac {eU} c \right ] ^2 c^2 + m^2c^4? ##

Thanks.
It looks as if Fermi defines ##W_a## as the energy relative to the rest mass energy, so

## E_{rel} \equiv W_a-eV +mc^2 ##

Using this, his equation follows. I think he made this choice so that upon expansion, the constant term ##mc^2/2## disappears.

grzz said:
and replacing E by ##W_a##, assuming ##W_a## stands for the total energy of the particle.
That assumption seems unlikely. I believe your math above is correct, which would indicate that the relativistic Hamiltonian does not include the mass as part of the energy. So you'd have to include an extra ##mc^2## to get the total energy. I believe that resolves the discrepancy.

Does the paper define ##W_a## explicitly? That would probably be your best bet in determining whether this is correct or not.

That assumption seems unlikely.

Thank you both, 'nrqed' and 'kimbyd'!

Yes, my mistake was assuming that ##W_a## includes the rest mass energy.

I derived it again and it turned out as the one given by Fermi, as 'nrqed' said.

Last edited by a moderator:

## 1. What is the Hamiltonian in physics?

The Hamiltonian is a mathematical function that represents the total energy of a system in classical mechanics. It is used to describe the dynamics of a system over time.

## 2. Why is there a need for an alternative formula for the Hamiltonian?

The traditional formula for the Hamiltonian can be complex and difficult to use in certain situations. An alternative formula can provide a simpler and more intuitive way to calculate the energy of a system.

## 3. What are the benefits of using an alternative formula for the Hamiltonian?

An alternative formula for the Hamiltonian can provide a better understanding of the dynamics of a system and make calculations easier. It can also provide insights into new ways of approaching problems in physics.

## 4. How do alternative formulas for the Hamiltonian differ from the traditional formula?

Alternative formulas for the Hamiltonian may use different mathematical approaches or variables to represent the energy of a system. They may also take into account different factors or properties of the system.

## 5. What are some examples of alternative formulas for the Hamiltonian?

Some examples of alternative formulas for the Hamiltonian include the Lagrangian, which uses generalized coordinates to describe the energy of a system, and the Dirac Hamiltonian, which is used in quantum mechanics to describe the energy of a particle.

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