Understanding the Relationship between the Hamiltonian and Magnetic Field

[ziyad]

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

 

[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

 

[ziyad]

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

 

[dextercioby]

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

 

[clive]

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

 

[dextercioby]

E sarcina elementara "e".

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

 

[Eye_in_the_Sky]

... 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 ...

No, this is not the way.

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

What you have written down comes from

e^iR∙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:

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

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

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

 

[ziyad]

clive 'e' is the charge

ok i wrote it wrong. here i scanned the equations
http://www.public.asu.edu/~ziyads/hamiltonian.gif

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.

 

[dextercioby]

Use the Latex code.

Daniel.

 

[Eye_in_the_Sky]

e^iR∙a/h_bar P e^{-iR∙a/h_bar} = P - a .

... The general expression goes like this:

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

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

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.

[R_k, R_l] = [P_k, P_l] = 0 ,

[R_k, P_l] = ih_bar δ_kl .

Alternatively, they are equivalent to:

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

or

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

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.
_____________

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

You have made an error here. The right-hand side should read H1, as before.
_____________

here i scanned the equations
http://www.public.asu.edu/~ziyads/hamiltonian.gif

... A is a constant vector.

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):

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

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.
_____________

how r u able to write equations in this forum.

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.

 

[ziyad]

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

 

[Eye_in_the_Sky]

We have

e^{-ieR∙A/h_bar} H_o e^{ieR∙A/h_bar} = H₁ ,

where A is a constant vector, and

H_o = P²/2m + V(R) ,

and

H₁ = (P + eA)²/2m + V(R) .

Now what exactly do you want to show?

 

[ziyad]

That these both r same


e^{-ieR∙A/h_bar} H_o e^{ieR∙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

 

[Eye_in_the_Sky]

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

 

[ziyad]

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

 

[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)H_oS^†(R) .

Therefore, a ket |ψ> is such that

H_o|ψ> = E|ψ> ,

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

H₁|φ> = E|φ> .

This shows that H_o 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 H_o (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.)

 

[ziyad]

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

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


YAY!

 

[Eye_in_the_Sky]

So glad to have been of help!

(Good luck on the exam! )