Canonical transformation in Hamiltonian

[jackychenp]

Hamiltonian H=\frac{1}{2m}(P+\frac{e}{c}A)^{2} - e\phi and H^{'}=\frac{1}{2m}(P+\frac{e}{c}A^{'})^{2} - e\phi^{'}

With gauge: A^{'}=A+\nabla\chi and \phi^{'}=\phi-\frac{1}{c}\dot{\chi}

Why H^{'}-\frac{e}{c}\dot{\chi}=e^{-\frac{ie\chi}{\hbar c}}He^{\frac{ie\chi}{\hbar c}}? Thanks.

 

[A. Neumaier]

Hamiltonian H=\frac{1}{2m}(P+\frac{e}{c}A)^{2} - e\phi and H^{'}=\frac{1}{2m}(P+\frac{e}{c}A^{'})^{2} - e\phi^{'}

With gauge: A^{'}=A+\nabla\chi and \phi^{'}=\phi-\frac{1}{c}\dot{\chi}

Why H^{'}-\frac{e}{c}\dot{\chi}=e^{-\frac{ie\chi}{\hbar c}}He^{\frac{ie\chi}{\hbar c}}? Thanks.

replace chi by s*chi and differentiate both sides. Then find a differential equation in s satisfied by both sides of the (modified) equation you are looking for. Since for s=0 the two sides are equal and the initial value problem has a unique solution, the two sides agree for any s, and in particular for s=1.

 

[jackychenp]

replace chi by s*chi and differentiate both sides. Then find a differential equation in s satisfied by both sides of the (modified) equation you are looking for. Since for s=0 the two sides are equal and the initial value problem has a unique solution, the two sides agree for any s, and in particular for s=1.

Hi Neumaier,

Thanks for your reply! In your method, shall I differentiate the euqation H^{single-quote}-\frac{e}{c}\dot{s\chi}=e^{-\frac{ies\chi}{\hbar c}}He^{\frac{ies\chi}{\hbar c}} with t? I didn't find a differential equation in s satisfied by both sides of the (modified) equation .
I was thinking to use taylor expansion to expand e^{-\frac{ies\chi}{\hbar c}} and prove both sides are equal, but it is a little complicate.

 

[A. Neumaier]

Hi Neumaier,

Thanks for your reply! In your method, shall I differentiate the euqation H^{single-quote}-\frac{e}{c}\dot{s\chi}=e^{-\frac{ies\chi}{\hbar c}}He^{\frac{ies\chi}{\hbar c}} with t? I didn't find a differential equation in s satisfied by both sides of the (modified) equation .

Differentiate with respect to s. You should get a differential equation of the form Hdot(s)-Chidot=[H(s),Chi], with constants and signs adapted to your actual formula.

 

[jackychenp]

Differentiate with respect to s. You should get a differential equation of the form Hdot(s)-Chidot=[H(s),Chi], with constants and signs adapted to your actual formula.

Thanks. I got \frac{\partial H^{'}(s)}{\partial s}-\frac{e}{c}\dot{\chi}=\frac{ie}{\hbar c} [e^{\frac{-ies\chi}{\hbar c}}He^{\frac{ies\chi}{\hbar c}} ,\chi].
By setting s=0 and expand H'(s), the result becomes \frac{1}{m} (P+\frac{e}{c}A)\bigtriangledown\chi=\frac{ie}{\hbar c}[H,\chi]. It looks both sides are not equal.

 

[A. Neumaier]

Thanks. I got \frac{\partial H^{'}(s)}{\partial s}-\frac{e}{c}\dot{\chi}=\frac{ie}{\hbar c} [e^{\frac{-ies\chi}{\hbar c}}He^{\frac{ies\chi}{\hbar c}} ,\chi].
By setting s=0 and expand H'(s), the result becomes \frac{1}{m} (P+\frac{e}{c}A)\bigtriangledown\chi=\frac{ie}{\hbar c}[H,\chi]. It looks both sides are not equal.

You forgot to evaluate the commutator [H,Chi], using the definitions and the CCR.

 

[jackychenp]

Thanks a lot!
I have one more question :
We can set F(s)= H^{'}-\frac{e}{c}\dot{s\chi} - e^{-\frac{ies\chi}{\hbar c}}He^{\frac{ies\chi}{\hbar c}}
So \frac{\partial F(s))}{\partial s}|_{s=0} = 0 means the F(s) has a min or max value at s=0. Why F(s) has unique solution and agree for any s?
One simple example is F(s,x)=s^2x^2, \frac{\partial F(s,x)}{\partial s} = 0|_{s=0}, but F(s,x)=0 only agrees for s=0.

 

[A. Neumaier]

Thanks a lot!
I have one more question :
We can set F(s)= H^{'}-\frac{e}{c}\dot{s\chi} - e^{-\frac{ies\chi}{\hbar c}}He^{\frac{ies\chi}{\hbar c}}
So \frac{\partial F(s))}{\partial s}|_{s=0} = 0 means the F(s) has a min or max value at s=0. Why F(s) has unique solution and agree for any s?
One simple example is F(s,x)=s^2x^2, \frac{\partial F(s,x)}{\partial s} = 0|_{s=0}, but F(s,x)=0 only agrees for s=0.

To argue that some function F(s) vanishes for all s you need to show that F(0)=0 _and_ that F(s) satisfies a differential equation Fdot(s)=R(F(s),s) with some R _not_ depending on s and R(0,s)=0 for all s.

Since (under appropriate conditions usually taken fro granted in physics) the solution of an initial value problem for ordinary differential equations is unique and the zero function satisfies the equation and the initial condition, you know that F must be identically zero.

 

[jackychenp]

Neumaier, thanks for your patience. I appreciate your help!