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In section 4.3 of Griffths' "Introduction to Quantum Mechanics", just below Figure 4.6, the sentence begins

Let $\hbar \ell$ be the eigenvalue of $L_z$ at this top rung...

Why is this valid? In the previous pages, there is no derivation of this fact. It's not surprising that this eigenvalue has $\hbar$ in it, but I don't see why I should expect it to be an integer multiple of $\hbar$.

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Why does this shock you? You expect to get $\hbar$, but are surprised that it's an integer multiple of it? I'm not sure what's bugging you. –  CHM May 6 '12 at 22:58
    
If it helps, you could just ignore my last sentence. In that case, I'm just wondering what allows us to claim that $\hbar \ell$ is an eigenvalue of $L_z$. In the text, this is stated and not justified (as far as I can tell). –  unit3000-21 May 6 '12 at 23:06

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When you initially set the eigenvalue at the top rung to $\hbar l$, you don't need to assume that $l$ is an integer, you can think of it as any multiplicative constant. Clearly there is no loss of generality there. The beautiful aspect of the ladder operator approach is that you can use it to prove that $l$ must be a non-negative integer or half-integer.

This argument is presented clearly in Griffiths, at least in the second edition (perhaps you are using the first edition?). Using the ladder operators $L_+$ and $L_-$, and the conditions that there must be a top rung and a bottom rung for the ladder of eigevnalues, you automatically find that

the eigenvalues of $L_z$ are $m \hbar$, where $m$ ... goes from $-l$ to $+l$ in $N$ integer steps. In particular, it follows that $l = -l + N$, and hence $l = N/2$, so $l$ must be an integer or a half-integer.

So, the nature of $l$ is discovered as a conclusion - there is no initial assumption.

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Ah, you're absolutely right. This is my fault: I was looking for justification in the pages before, not the pages after. The very next page describes exactly what you said. Thanks. –  unit3000-21 May 7 '12 at 0:09

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