# Alternative formula for the Hamiltonian

• I
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.

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mfb
Mentor
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).

kimbyd
Gold Member
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?

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.

nrqed
Homework Helper
Gold Member
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.

kimbyd
Gold Member
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.

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