- #1
antibrane
- 38
- 0
I am attempting to find the probability, after time [itex]t[/itex], of a hydrogen atom in a magnetic field [itex]\vec{\mathbf{B}}=B_0\hat{\mathbf{z}}[/itex] to go from
[tex]
\left|n,l,s,j,m_j\right\rangle \longrightarrow \left|n',l',s,j',m_j'\right\rangle
[/tex]
where [itex]j=l+\frac{1}{2}[/itex] and [itex]j'=l'+\frac{1}{2}[/itex] or [itex]j'=l'-\frac{1}{2}[/itex]. Since it is hydrogen then [itex]s=s'=\frac{1}{2}[/itex].
What I thought I should do was use the Hamiltonian
[tex]
\hat{H}=-\vec{\boldsymbol{\mu}}\cdot\vec{\mathbf{B}}=\frac{e}{2m}(\vec{\mathbf{L}}+2\vec{\mathbf{S}})
[/tex]
and then I get the matrix element for the transition,
[tex]
\left\langle n',l',s',j',m_j'\right|\hat{H}\left|n,l,s,j,m_j \right\rangle=\frac{\hbar e}{2m} B_0g_Jm_j\langle n',l',s',j',m_j'|n,l,s,j,m_j\rangle
[/tex]
where [itex]g_F[/itex] is the Lande g-factor. I would have used this in time-dependent perturbation theory to get the probability after time [itex]t[/itex], however, while this does give me a non-zero element for [itex]j'=l'+\frac{1}{2}[/itex] (where we would have [itex]l'=l[/itex]), it is zero due to orthogonality for the other case ([itex]j'=l'-\frac{1}{2}[/itex]). Is there something wrong with the way I am going about this?
I really appreciate any comments.
[tex]
\left|n,l,s,j,m_j\right\rangle \longrightarrow \left|n',l',s,j',m_j'\right\rangle
[/tex]
where [itex]j=l+\frac{1}{2}[/itex] and [itex]j'=l'+\frac{1}{2}[/itex] or [itex]j'=l'-\frac{1}{2}[/itex]. Since it is hydrogen then [itex]s=s'=\frac{1}{2}[/itex].
What I thought I should do was use the Hamiltonian
[tex]
\hat{H}=-\vec{\boldsymbol{\mu}}\cdot\vec{\mathbf{B}}=\frac{e}{2m}(\vec{\mathbf{L}}+2\vec{\mathbf{S}})
[/tex]
and then I get the matrix element for the transition,
[tex]
\left\langle n',l',s',j',m_j'\right|\hat{H}\left|n,l,s,j,m_j \right\rangle=\frac{\hbar e}{2m} B_0g_Jm_j\langle n',l',s',j',m_j'|n,l,s,j,m_j\rangle
[/tex]
where [itex]g_F[/itex] is the Lande g-factor. I would have used this in time-dependent perturbation theory to get the probability after time [itex]t[/itex], however, while this does give me a non-zero element for [itex]j'=l'+\frac{1}{2}[/itex] (where we would have [itex]l'=l[/itex]), it is zero due to orthogonality for the other case ([itex]j'=l'-\frac{1}{2}[/itex]). Is there something wrong with the way I am going about this?
I really appreciate any comments.