# Hamiltonian In magnetic field

1. Mar 20, 2005

I have a question regarding the Hamiltonian in a magnetic field.
First Hamiltonian with potential V is given by
Ho = (1/2m)*p^2 + V

but if a vector potential A is also present then
H1 = (1/2m)*(p+eA)^2 + V

there is a way to write H1 interm of Ho
H1 = exp(-ier.A) Ho exp(ier.A)
where r is position
and r.A is dot product of position and vector potential

Any one knows Why? are these both forms equal and how to convert one into another

2. Mar 20, 2005

### clive

The expressions you wrote are two forms of the same Hamiltonian. The difference consists of p (momentum). In H_0 p is the classical momentum (mv) whereas in H_1 p is the canonical momentum (mv+eA).

BTW. In H_1 you must have (p-eA)

clive

3. Mar 20, 2005

Hey clive, yeah it has to be p-eA for it to be equal,
but how is the third equation equal to the second.

Last edited: Mar 21, 2005
4. Mar 21, 2005

### dextercioby

That $\hat{U}$ is the potential energy operator.Is there any connection between this operator and the magnetic potential vector operator...($\hat{A}$)?

5. Mar 21, 2005

### clive

What do you mean by "e" (in the third equation) ziyad?

6. Mar 21, 2005

### dextercioby

E sarcina elementara "e".

It's the elementary charge "e",i'm sure of it.

7. Mar 21, 2005

### Eye_in_the_Sky

No, this is not the way.

------------------------

What you have written down comes from

eiR∙a/h_bar P e-iR∙a/h_bar = P - a .

In this relation, however, the vector a is fixed. It has no dependence on space (although, it could depend on time). If it were the case that a had some kind of functional dependence on space – and hence, we would have "±iR∙a(R)/h_bar" in the exponentials – then it would no longer be the case that the left-hand side equals the right-hand side.

------------------------

The general expression goes like this:

eiG(R,t)/h_bar P e-iG(R,t)/h_bar = P – (grad G)(R,t) .

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There is definitely something more to be said here (though, for the moment, I'm not quite sure what it is).

8. Mar 21, 2005

clive 'e' is the charge

ok i wrote it wrong. here i scanned the equations

eye in the sky, yes ur correct. A is a constant vector.

what is that equality called. and where can i read more about it.

looking at the equation u mentioned. by changes signs
exp(-ier.A/hbar) Ho exp(ier.A/hbar) = Ho+eA

Now if i expand the square in the original H1 i get

H1 = Ho + (1/2m)(ep.A + eA.p+ (eA)^2)

How do i make this second term equal to just eA so i can use the equality u mentioned.

eye in the sky, how r u able to write equations in this forum.

Last edited by a moderator: May 1, 2017
9. Mar 21, 2005

### dextercioby

Use the Latex code.

Daniel.

10. Mar 22, 2005

### Eye_in_the_Sky

I don't know of any special names for these equalities. The first relation (in which the vector a is a constant) is directly related to the notion of the operator mR as the generator of "boosts". A similar equality, in which the roles of P and R are reversed, is directly related to the notion of the operator P as the generator of "(spatial) translations". All of these equalities are equivalent to the basic canonical commutation relations for R and P; i.e.

[Rk, Rl] = [Pk, Pl] = 0 ,

[Rk, Pl] = ihbar δkl .

Alternatively, they are equivalent to:

(i) the action of P in the {|r>} representation is given by -ihbargrad ;

or

(ii) the action of R in the {|p>} representation is given by ihbargradp ;

As for where to read more about these matters, I'm sorry but I can't think of any good elementary references. Perhaps someone else can point you in the right direction.
_____________
You have made an error here. The right-hand side should read H1, as before.
_____________
If A is a constant vector, then there is no reason to include it in the Hamiltonian. More generally, if curl A = 0, then there is no reason to include A in the Hamiltonian at all. In that case, the equations of motion for the system can be derived from a "scalar potential" alone without the need for the introduction of a "vector potential". (This fact has a deep connection with the equality quoted above in which the expression P – (grad G)(R,t) appears on the right-hand side.)

Having said that, I am beginning to see a little better what I was talking about when I said (back in post #7):
Yes, yes. There is something here about being able to 'remove' the A(R,t) term by means of a certain kind of unitary transformation if, and only if, curl A = 0 (i.e. A(r,t) = grad G(r,t) , for some function G). ... But it's all still a bit too 'fuzzy' for me to put my finger on.
_____________
If you click on the "quote" tab on any given post, you will be able to see precisely how the equations have been produced. Whenever I need special symbols (e.g. Greek letters), I copy and paste them from a "storehouse" of symbols in a WORD document I keep. Those symbols have been gathered by clicking on INSERT → SYMBOL in the upper menu of Microsoft WORD.

Perhaps using the Latex code can be easier. I haven't tried.

Last edited by a moderator: May 1, 2017
11. Mar 22, 2005

This is getting really complicated

a friend told me to look at H1*Phi(r) for arbitrary phi(r)
where phi(r) is teh wavefunction

and both forms will give the same result.
although i'm not sure how exactly

12. Mar 22, 2005

### Eye_in_the_Sky

We have

e-ieR∙A/h_bar Ho eieR∙A/h_bar = H1 ,

where A is a constant vector, and

Ho = P2/2m + V(R) ,

and

H1 = (P + eA)2/2m + V(R) .

Now what exactly do you want to show?

13. Mar 22, 2005

That these both r same

e-ieR∙A/h_bar Ho eieR∙A/h_bar = H1 ,
H1 = (P + eA)2/2m + V(R) .
one can be derived from another, or H1*phi(r) for arbitrary Phi(r) should give same result using either formulas.

Sorry dude. I'm bugging u so much

14. Mar 22, 2005

### Eye_in_the_Sky

It sounds like you want to say that Ho and H1 have the same spectrum of eigenvalues. Is that what you want to say?

15. Mar 22, 2005

actually both forms of H1 should have the same spectrum of eigenvalues.

16. Mar 22, 2005

### Eye_in_the_Sky

Okay, now we are "rolling".

First of all note that the operator

S(R) ≡ e-ieR∙A/h_bar

is a unitary operator. That is,

S-1(R) = S(R) .

So,

H1 = S(R)HoS(R) .

Therefore, a ket |ψ> is such that

Ho|ψ> = E|ψ> ,

if, and only if, the ket |φ> ≡ S(R)|ψ> is such that

H1|φ> = E|φ> .

This shows that Ho and H1 have same spectrum of eigenvalues.

Next, observe that

φ(r) ≡ <r|φ>

= <r|S(R)|ψ>

= <r|e-ieR∙A/h_bar|ψ>

= e-ier∙A/h_bar <r|ψ>

= e-ier∙A/h_bar ψ(r) .

This shows that corresponding eigenvectors of H1 and Ho (i.e. the ones which have the same eigenvalue) differ by a mere phase factor (given by
e-ier∙A/h_bar ).

... Is there anything else?

Note that to be completely rigorous, however, we would need to say a few words also about how the above conclusions are still valid in case of degeneracies. (This requires further thought.)

17. Mar 22, 2005

That wasn't so bad. It looks so simple now.

Eye in the sky. thanks for ur help dude. I think thats the solution, i don't think degeneracy is needed. this is perfect as it stands.

YAY!!!!

18. Mar 22, 2005

### Eye_in_the_Sky

So glad to have been of help!

(Good luck on the exam!!! )