# Space Quantization of Quantum Angular Momentum

I am trying to understand what my book is trying to convey.

Quantum angular momentum is $L_z = m_l \hbar$

"Choosing arbitrarily a z axis and using an appropriate experimental technique, we measure the z component of $\vec{L}$. We find $L_z = -\hbar,0, \hbar$. Choosing another z axis and repeating the measurement we again find $L_z = -\hbar,0, \hbar$. This behaviour is completely different from classical vectors."

The Idea of a classical vector is that it exists independent of the coordinate system used to examine it. Why is this not the case with this quantum system? how does the vector "Know" where the z axis is?

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## 2 Answers

Perhaps this will be illuminating. Consider the operator $\vec n\cdot \vec L$ corresponding to measuring angular momentum in an arbitrary direction specified by a unit vector $\vec n$, then I make the following:

Claim. For any two unit vectors $\vec n_1$ and $\vec n_2$, the spectra (sets of eigenvalues) of $\vec n_1\cdot\vec L$ and $\vec n_2\cdot \vec L$ are the same.

I'll prove this claim below, but before I do, let's think about what this means. If we were to measure the component of angular momentum pointing in any direction we choose, this result says that we always get the same, discrete sets of possible measured values. In particular, if we choose to call a certain direction in space our $z$-axis, or if we rotate our coordinate system and call another direction in space our $z$-axis, then the set of possible measured values we will get for the $z$-component of the angular momentum will be the same in both cases.

Proof of claim.

Since $\vec n_1$ and $\vec n_2$ are unit vectors, there is a rotation (special orthogonal transformation) $R$ for which $$\vec n_1 = R\vec n_2$$ It follows that \begin{align} \vec n_2\cdot \vec L &= (R^t\vec n_1)\cdot \vec L \\ &= \vec n_1\cdot (R\vec L) \\ &= (n_1)_iR^i_jL^j \tag{$\star$} \end{align} Now, recall that $\vec L$ is a vector operator; this means that there exists a unitary representation $U$ of the rotation group acting on the Hilbert space such that $$U(R)^{-1} L^iU(R) = R^i_jL^j$$ so that for any unit vector $\vec n$, one has $$U(R)^{-1} n_iL^iU(R) = n_iR^i_jL^j$$ which in vector notation can be written as $$U(R)^{-1} (\vec n\cdot\vec L) U(R) = \vec n\cdot(R\vec L) \qquad \tag{\star\star}$$ Upon combining $(\star)$ and $(\star\star)$, we find that $$\vec n_2\cdot\vec L = U(r)^{-1} (\vec n_1\cdot\vec L) U(R)$$ This shows that the two operators $\vec n_1\cdot\vec L$ and $\vec n_2\cdot\vec L$ are related by a unitary similarity transformation, so they have the same spectra.

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By the way, you can use \tag{x} to tag an equation with x so you can reference it later. Though I'm not sure if it allows \star as an argument, I've only ever done it with numbers, e.g. \tag{1}. –  David Z Mar 13 '13 at 1:08
@DavidZaslavsky Oh, awesome thanks! I'm used to the align environment auto-tagging in TeX so I was unsure of how to do it... –  joshphysics Mar 13 '13 at 1:13

This is the case with this system. As your book says, we get the same result ($-\hbar,0,\hbar$) no matter which z axis we choose. The "This behaviour is completely different from classical vectors" part refers to the fact that the length of $\vec{L}$ is quantized, it is not supposed to mean that we have somehow broken rotational invariance.

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