Finding the magnetic field based on vector potential

[jm16180339887]

Homework Statement

If you have |$\psi$>=cos($\theta$/2)e^{i*l*$\varphi$/2}+sin($\theta$/2)e^{-i*l*$\varphi$/2}e^i$\varphi$
and A=<$\psi$|$\partial$_$\varphi$|$\psi$>$\hat{r}$
find B in polar coordinates

Homework Equations

B=$\nabla$xA

The Attempt at a Solution

So far I got B=$\frac{1}{r}$[-ie^i$\varphi$($\frac{l(l-1)}{4}$e^(l-1)-$\frac{(1-l)(1-\frac{l}{2})}{2}$e^(1-l))$\hat{\theta}$-($\frac{l}{2}$cos($\theta$/2)sin($\theta$/2)+$\frac{l}{4}$sin²($\theta$/2)e^{i$\varphi$(l-1)}-$\frac{l}{4}$cos²($\theta$/2)e^{i$\varphi$(l-1)}+($\frac{1}{2}$cos²($\theta$/2)-$\frac{1}{2}$sin²($\theta$/2)-$\frac{l}{4}$cos²($\theta$/2)+$\frac{l}{4}$sin²($\theta$/2)e^{i$\varphi$(1-l)}-cos($\theta$/2)sin($\theta$/2)+$\frac{l}{2}$sin($\theta$/2)cos($\theta$/2))$\hat{\varphi}$]

but I was told that the answer is l+1. How do I simplify this equation?

 

[mighty2000]

First, let's rewrite the given wave function as:

|\psi>=cos(\theta/2)e^{i(l+1)\varphi/2}+sin(\theta/2)e^{-i(l-1)\varphi/2}

Next, we can calculate the gradient of A using the product rule:

\nabla A = \frac{\partial}{\partial r}(\frac{1}{r}A)\hat{r}+\frac{1}{r}\frac{\partial A}{\partial \theta}\hat{\theta}+\frac{1}{r\sin{\theta}}\frac{\partial A}{\partial \varphi}\hat{\varphi}

Now, let's calculate each component separately:

\frac{\partial}{\partial r}(\frac{1}{r}A) = \frac{1}{r^2}A

\frac{\partial A}{\partial \theta} = <\psi|\frac{\partial}{\partial \theta}|\psi> = -\sin{\theta}cos{\theta}e^{i(l+1)\varphi/2}+\sin{\theta}cos{\theta}e^{-i(l-1)\varphi/2} = 0

\frac{\partial A}{\partial \varphi} = <\psi|\frac{\partial}{\partial \varphi}|\psi> = -\frac{i(l+1)}{2}e^{i(l+1)\varphi/2}+\frac{i(l-1)}{2}e^{-i(l-1)\varphi/2}

Substituting these components back into the gradient equation, we get:

\nabla A = \frac{1}{r^2}A\hat{r}+\frac{i(l+1)}{2r}e^{i(l+1)\varphi/2}\hat{\varphi}-\frac{i(l-1)}{2r}e^{-i(l-1)\varphi/2}\hat{\varphi}

Finally, we can use the definition of B to simplify this expression:

B = \nabla \times A = \frac{1}{r^2}A\hat{\varphi}+\frac{i(l+1)}{2r}e^{i(l+1)\varphi/2}\hat{r}-\frac{i(l-1)}{2r}e