Phase factors and Modulus Square?

[kq6up]

If I have something like:

$$\lvert \langle M \lvert \hat{L}_x+i\hat{L}_y \rvert M-1 \rangle \rvert ^2=c$$.

where $c$ is any old real number. If I undid the modulus square to find:

$$ \langle M \lvert \hat{L}_x+i\hat{L}_y \rvert M-1 \rangle=\pm \sqrt{c} $$ Would I not have to consider $c$ as being negative, positive as well as imaginary negative positive? So:

$$ \langle M \lvert \hat{L}_x+i\hat{L}_y \rvert M-1 \rangle=\pm i \sqrt{c} $$ as well.

I am trying to get to:

$$ \langle M \lvert \hat{L}_x \rvert M-1 \rangle=\frac{1}{2} \sqrt{c} $$

and:

$$ \langle M \lvert \hat{L}_y \rvert M-1 \rangle=-\frac{i}{2} \sqrt{c} $$

From Landau and Lifshitz QM 3ed page 89, and I am not following. Any tips will be appreciated.

Thanks,
KQ6UP

 

[Charles Link]

You need the raising and lowering operators: $L_+=L_x+iL_y$ and $L_-=L_x-iL_y$. You can solve for $L_x$ and $L_y$ and substitute. The computations are straightforward using the formulas for the result of $L_+$ and $L_-$ operating on the state $| M >$ (or $| M-1>$, etc.).

 

[kq6up]

You need the raising and lowering operators: $L_+=L_x+iL_y$ and $L_-=L_x-iL_y$. You can solve for $L_x$ and $L_y$ and substitute. The computations are straightforward using the formulas for the result of $L_+$ and $L_-$ operating on the state $| M >$ (or $| M-1>$, etc.).

That is where I started before $\langle M \lvert \hat{L}_+ \rvert M-1 \rangle=\sqrt{(L+M)(L-M+1)}$, and I am trying to get to:

$\langle M \lvert \hat{L}_x \rvert M-1 \rangle=\frac{1}{2}\sqrt{(L+M)(L-M+1)}$

Thanks,
KQ6UP

 

[Charles Link]

That is where I started before $\langle M \lvert \hat{L}_+ \rvert M-1 \rangle=\sqrt{(L+M)(L-M+1)}$, and I am trying to get to:

$\langle M \lvert \hat{L}_x \rvert M-1 \rangle=\frac{1}{2}\sqrt{(L+M)(L-M+1)}$

Thanks,
KQ6UP

You can readily compute $L_x=(L_++L_-)/2$ . $\\$ $L_- |M-1>$ will give some number times $|M-2>$. The matrix element $<M|M-2> =0$ and $<M|M> =1$. Also note $L_+|M-1>=\sqrt{(L+M)(L-M+1)}|M>$.

 

[kq6up]

You can readily compute $L_x=(L_++L_-)/2$ . $\\$ $L_- |M-1>$ will give some number times $|M-2>$. The matrix element $<M|M-2> =0$ and $<M|M> =1$. Also note $L_+|M-1>=\sqrt{(L+M)(L-M+1)}|M>$.

This is true. However, I am still wondering about my original question. When you "undo" the square of a modulus, it seems reasonable to me that you would have a family of solutions as detailed above. Is this true?

Thanks,
KQ6UP

 

[Charles Link]

This is true. However, I am still wondering about my original question. When you "undo" the square of a modulus, it seems reasonable to me that you would have a family of solutions as detailed above. Is this true?

Thanks,
KQ6UP

Your "c" above is always positive. I think one question you are asking is might these matrix elements always contain an arbitrary phase factor of the form $exp(i \phi)$, and I think the answer to this is yes.

 

[blue_leaf77]

Just confirming what Charles has said above, one can always replace $c$ with $ce^{i\phi}$ as this will only impose a constant phase factor on the state. It's chosen as real positive just a matter of convention, pretty much the same case as the constants $\sqrt{n+1}$ and $\sqrt{n}$ in the raising and lowering operators in harmonic oscillator.