Phase factors and Modulus Square?

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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
 

Answers and Replies

  • #2
Charles Link
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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.).
 
  • #3
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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
 
  • #4
Charles Link
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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> ##.
 
  • #5
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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
 
  • #6
Charles Link
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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.
 
  • #7
blue_leaf77
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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.
 
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