Alternative formula for the Hamiltonian

  • I
  • Thread starter grzz
  • Start date
  • #1
993
13
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.
 

Answers and Replies

  • #2
993
13
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:
  • #3
35,268
11,534
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).
 
  • #4
993
13
I am still wondering from where did Fermi got his equation (13).
 
  • #5
kimbyd
Science Advisor
Gold Member
1,181
630
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?
 
  • #6
993
13
Where did you get the second equation from?
Thanks 'Kimbyd' for your interest.

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.
 
  • #7
nrqed
Science Advisor
Homework Helper
Gold Member
3,737
279
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.
 
  • #8
kimbyd
Science Advisor
Gold Member
1,181
630
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.
 
  • #9
993
13
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:

Related Threads on Alternative formula for the Hamiltonian

  • Last Post
Replies
7
Views
3K
Replies
8
Views
2K
  • Last Post
Replies
3
Views
690
Replies
1
Views
294
Replies
7
Views
5K
Replies
5
Views
667
  • Last Post
Replies
6
Views
1K
Replies
0
Views
1K
  • Last Post
Replies
4
Views
2K
Replies
4
Views
2K
Top