The propagator of eigenstates of the Total Angular Momentum

[mathsisu97]

Homework Statement

What is the explicit expression of the propagator for a particle in a magnetic field along with z-axis

Relevant Equations

$H = \omega J_z$
$J^2 |jm> = \hbar^2 j(j+1) |jm>$
$J_z |jm> = \hbar m |jm >$

To show that when $[J^2, H]=0$ the propagator vanishes unless $j_1 = j_2$ , I did ($\hbar =1$)
$$ K(j_1, m_1, j_2 m_2; t) = [jm, e^{-iHt}]= e^{iHt} (e^{iHt} jm e^{-iHt}) - e^{-iHt} jm $$
$$ = e^{iHt}[jm_H - jm] $$
So we have
$$ \langle j_1 m_1 | [jm, e^{-iHt} ] | j_2 m_2 \rangle $$
$$ = (j_1 m_1 - j_2 m_2) \langle j_1 m_1 | e^{-iHt} | j_2 m_2 \rangle $$
$$ = \langle j_1 m_1 | e^{-iHt} [ jm_H -jm] | j_2 m_2 \rangle $$
$$ \frac{d jm_H}{dt} = \frac{i}{\hbar} [ H, jm_H] $$
Thus
$$ jm_H = jm $$
And
$$ (j_1 m_1 - j_2 m_2) \langle j_1 m_1 | e^{-iHt} | j_2 m_2 \rangle = \langle j_1 m_1 | e^{-iHt} | j_2 m_2 \rangle $$
$$ (j_1 m_1 - j_2 m_2) K(j_1 m_1, j_2 m_2; t) =0 $$
So the propagator vanishes unless $j_1 = j_2$
b) Splitting the time evolution
$$ \langle j m_1 | e^{-iHt} | j m_2 \rangle = \langle | e^{-iH(t-t_1)} e^{-iHt_1} | j m_2 \rangle $$
In the continuous (usual) case we have
$$ K(x,t,x', 0) = \int dx'' K(x,t, x'', t_1) K(x'', t_1, x', 0) $$
In the discrete case this is simply just a sum
$$ K(x,t,x', 0) = \sum K(x,t, x'', t_1) K(x'', t_1, x', 0) $$
$$ = K (t-t_1) K(t_1) $$
I am stuck on part c (see attached file) where I need to show that when
$$ H= \omega J_z $$
for a particle in a magnetic field, what is the propagator?

 

[vanhees71]

I can't make sense of what you wrote as solution for (a).

Hint: Assuming $[\hat{H},\hat{J}^2]=0$, what is
$$\hat{J}^2 \exp(-\mathrm{i} \hat{H} t)|j,m \rangle?$$

For (b) it's ok.

For (c) just check, what $\hat{J}_z$ does when applied to $|jm \rangle$ and what does this mean for an arbitrary function $f(\hat{J}_z)$ when applied to this angular-momentum eigenvector.

 

[mathsisu97]

For part (a) I see what you are hinting at. I still do not understand part (c) or what you have said. Applying $J_z$ to $|jm>$ gives the eigenstates. I am trying to draw parallel with the free particle propagator but not getting anywhere.

 

[vanhees71]

It should be clear that $\hat{J}_z |j m \rangle=m |j m \rangle$. Now think what this implies for $f(\hat{J}_z) |j m \rangle$ with $f$ an arbitrary function of $\hat{J}_z$!

 

[mathsisu97]

Ahh yes sorry! I already had that $\vec{J}^2 |jm \rangle = \hbar^2j(j+1) |jm \rangle$ and $\vec{J}_z |jm \rangle = \hbar m|jm \rangle$ . I know that for the orbital angular momentum quantum numbers $L$ and $L_z$ we get the Spherical Harmonics from which we can derive the Legendre functions, does the same apply here?

 

[vanhees71]

It's much simpler! You don't need the explicit calulation of the $|jm \rangle$ states in position space (which are of course the spherical harmonics, if you deal with orbital angular momentum. You can calculate the propagator very simply here. Think again, what's $f(\hat{J}_z)|jm \rangle$ and then set $$f(\hat{J}_z)=\exp(-\mathrm{i} \omega \hat{J}_z)$$.

Note that I assume everywhere that we use units with $\hbar=1$ (just being lazy writing all the $\hbar$ factors ;-)).

 

[mathsisu97]

A function of the $J_z$ is a rotation? Sorry for not getting the point! I am currently looking back on my notes on Total Angular momentum and not see what youre are getting at with the arbitrary function application. Thanks for all the help so far

 

[vanhees71]

It's very simple! You know that $\hat{J}_z |jm \rangle=m |j m \rangle$. Now think about $f(\hat{J}_z) |j m \rangle$. You can start by thinking about $f(\hat{J}_z)=\hat{J}_z^k$ (with $k \in \mathbb{N}$) and then about a operator-valued Taylor expansion $f(\hat{J}_z)=\sum_{k=0}^{\infty} \frac{1}{k!} \hat{J}_z^k$. At the end apply the result to $f(\hat{J}_z)=\exp(-\mathrm{i} \omega t \hat{J}_z)$.

 

[mathsisu97]

So we get
$$ f(\vec{J}_z) | jm \rangle = \sum_{k=0}^{\infty} \frac{1}{k!} \vec{J}_z^k |jm \rangle $$
$$ = \sum_{k=0}^{\infty} \frac{1}{k!} m^k |jm \rangle $$
So the propagator is
$$ = \sum_{k=0}^{\infty} \frac{1}{k!} m^k |jm \rangle = \exp(-i \omega t \vec{J}_z) $$
Doesn't quite seem right or at least the full answer.
Edit:
The LHS is simply just the exponential function? Therefore
$$ = \exp(m) = \exp(-i \omega t \vec{J}_z) $$
$$ m = -i \omega t \vec{J}_z? $$

 

[nrqed]

So we get
$$ f(\vec{J}_z) | jm \rangle = \sum_{k=0}^{\infty} \frac{1}{k!} \vec{J}_z^k |jm \rangle $$

Watch out, your exponent is not just $J_z$ (by the way, there is no vector symbol here), right? It is
$-i \omega t J_z$, so you need to correct slightly what you wrote here.

$$ = \sum_{k=0}^{\infty} \frac{1}{k!} m^k |jm \rangle $$

The next step that vanhees71 was pointing to is this. Let's take what you wrote just above (again, it has to be changed because of the factor $-i \omega t$ so that you have
$$ = \sum_{k=0}^{\infty} \frac{1}{k!} (-i \omega t m)^k |jm \rangle .$$
Now, you just have an ordinary function of time (not an operator) multiplying a ket, so you can just think of this as
$$ \Biggl( \sum_{k=0}^{\infty} \frac{1}{k!} (-i \omega t m)^k \Biggr) |jm \rangle .$$
Now you can do the sum inside the parenthesis, right? Next, you will apply on this the appropriate bra to finally get the propagator. Note that the expression in the parenthesis is not an operator, so when you apply a bra, it will directly get applied on the ket.