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I'm studying about quantum-spin (in a syllabus about non-relativistic quantum-mechanics though), but I have some trouble understanding everything. So I would like to ask some small questions, which may perhaps clear up the picture as a whole.

So they are trying to determine the influence of a spatial rotation, to the state in the two dimensional spin-space. So they try to determine the operator $Û_R$ responsible for this rotation (which acts on the spin-space and thus is represented by a 2x2 matrix).

To do this they first try to determine $Û_{\epsilon}$ which is the infinitesimal unitary operator for rotation about an infinitesimal angle $\epsilon$ and they state that such an operator can always be written as: $$Û_{\epsilon}= 1-i\epsilonÂ$$ with $Â$ hermitic. So $Â$ is also acting on the two dimensional spin-space and can be represented by a 2x2 matrix. So far so good.

But now they state that $Â$ has to be constructed in such a way that $<\chi\vert\chi>$ and $<\chi'\vert\chi'>$ and $<\chi'\vert\chi>$ are rotation-invariant. Where $\vert\chi>$ and $\vert\chi'>$ are states of the two dimensional spin-space and $\vert\chi'> = Û_{\epsilon}\vert\chi>$. What I don't understand is:

1) What is meant by rotation invariance? Because if we rotate $\vert\chi>$ and $\vert\chi'>$ by taking the action of a unitary rotation-operator $Û_{\theta}$ , then these inproducts are necessarily conserved, whatever $Â$ may be. So I guess they mean something else.

2) Why must $Â$ be constructed so that this is true? Perhaps this will be clear if I understand 1) though.

Furthermore they state that this can only be the case if: $$Â = \vec{1}_n\cdot\vec{Â} = \frac{1}{2}\vec{1}_n\cdot\vec{\sigma}$$ With $\vec{Â}$ and $\vec{\sigma}$ hermitical vector operators. So I wonder:

3) Why is this only the case if this is true?
4) Somewhat further they also mention that $Tr(\sigma_z) = 0$ but I don't really see here too why this has to be the case.

I hope that someone can answer one (or more) of these questions, I think it will help a lot in understanding the other claims of the chapter about quantum-spin.

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Because all the structure of quantum mechanics is encoded in the inner product, the inner products are precisely what a symmetry must preserve. However this is a question of how to implement the symmetry, whereas your other questions seem to be related to the structure of the symmetry group. First I will discuss the rotation group acting on world vectors, since this is what defines the rotation group.

A finite proper rotation acting on world vectors, i.e., a $3\times 3$ orthogonal matrix can be built from successive infinitesimal rotations: $$U_n(\theta) = (1 - \frac{\theta}{n} \hat A)^n.$$ Here proper means that $U$ preserves orientation; this implies $\det U = 1$. In the limit $n \to \infty$ this is $$U(\theta) = e^{-\theta \hat A}.$$

Now from $$1 = U^T U = e^{\theta \hat A^T}e^{\theta \hat A}$$ and taking the derivative with respect to $\theta$, $$0 = \hat A^T e^{\theta \hat A} e^{\theta\hat A} + e^{\theta\hat A^T} \hat A e^{\theta\hat A} = \hat A^T + A$$ we get that $A$ must be anti-symmetric. From the rotation matrices $$R_x(\theta)= \begin{pmatrix} \cos \theta & -\sin \theta & 0 \\ \sin\theta & \cos \theta &0 \\ 0 & 0 & 1 \end{pmatrix}$$ and similarly for $y,z$ and taking the $\theta$ derivative you find a basis for these matrices such that $$[\pi_x, \pi_y] = -\pi_z$$ and cyclically.

Now this is all acting on world vectors and not quantum states. To implement this on quantum states any set of matrices $\sigma_x, \sigma_y, \sigma_z$ such that these commutation relations hold will do, not necessarily $3\times 3$ and possibly complex. This gives you the representations of infinitesimal rotations. Do get finite rotations, you use the exponential map. Now since the symmetry should preserve inner products, we must have (by arguments in the other answers) $$U^\dagger U =e^{\theta \sigma_x^\dagger}e^{\theta \sigma_x} = 1.$$ Repeating the argument above shows that $\sigma_x$ must be anti-Hermitian. However, if $\sigma_x$ is anti-Hermitian, $\tilde\sigma_x = -i\sigma_x$ is Hermitian and the commutator is $$[\tilde \sigma_x, \tilde \sigma_y] = \tilde\sigma_z$$ and this is the more familiar physicist's version.

From the commutation relations it follows that the eigenvalues of the $\tilde\sigma$ matrices come in integer steps and are symmetric around $0$. Since the trace is the sum of the eigenvalues, the matrices must be traceless.

Thus in conclusion that $$U = e^{-i\theta\tilde{\sigma}}$$ is unitary requires only that $\tilde{\sigma}$ is Hermitian. However from the structure of the rotation group, it follows that $\tilde{\sigma}$ is also traceless.

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1) I think that they mean a rotation with a unitary matrix. But $\hat U$ will be unitary not for all $\hat A$ : you need that $\hat A$ is hermitian (this is a constraint on $\hat A$).

2) If $\hat A$ is not such that these products are invariant, then the matrix $\hat U$ is not unitary, and this is a problem. All that gives constraints on $\hat A$.

3) All two-by-two unitary matrix can be written as a sum $a \mathbf{1}+b_1 \sigma_x+b_2 \sigma_y+b_3 \sigma_z$ (that is $\mathbf{1}$ plus the Pauli matrices form a basis in the space of 2-by-2 unitary matrices). Then, because $Tr(\hat A)=0$ (see answer to the fourth question), $a=0$.

4) Because $\hat U$ is unitary, $\det \hat U=1$, which implies that $Tr \hat A=0$.

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    $\begingroup$ Are you sure that unitarity automatically implies $\mathrm{det}(U)=1$? If so what is the difference between $\mathrm{SU(N)}$ and $\mathrm{U(N)}$? I thought that one imposes $\mathrm{det}(U)=1$, because it implies that we are then preserving the handedness of the system. $\endgroup$ – Hunter Dec 18 '13 at 19:53
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    $\begingroup$ @Hunter: You're right, unitary only implies $|\det U|=1$. But here $U$ is assumed to be close to the identity, which has a determinant $1$, so we are effectively studying unitary matrix with determinant $1$, that is SU(2). Even if it wasn't the case, a global phase does not play a role in QM, so we could always absorb it in the definition of the states. $\endgroup$ – Adam Dec 18 '13 at 21:06
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1 & 2) I think that with rotational invariant they mean transformations that do not change the inner product: \begin{equation} \langle \chi|\chi\rangle \rightarrow \langle \chi'|\chi'\rangle = \langle \chi|U^\dagger U |\chi\rangle \end{equation} which is clearly preserved if $U$ is unitary. This should also answer question 2, because a true symmetry transformation does not change the inner product (so that probabilities do not change before and after the symmetry transformation). However, $\langle \chi'|\chi\rangle$ is not preserved and I am wondering if this may be a typo?

As a side note: mathematically, $U$ could also be anti-unitary: \begin{equation} \langle \chi_1|\chi_2\rangle \rightarrow \langle \chi'_1|\chi'_2\rangle = \langle \chi_1|U^\dagger U |\chi_2\rangle = \langle \chi_2|\chi_1\rangle \end{equation} but this is pretty unusual (and I have never formally studied it, so do not have much info on it).

3) Any $2 \times 2$ traceless and Hermitian matrix can be written as: \begin{equation} x=x_i \sigma_i = \begin{pmatrix} x_3 & x_1 - ix_2 \\ x_1 + ix_2 & -x_3 \end{pmatrix} \end{equation} where $\sigma_i$ denote the Pauli matrices.

4) Rember that $\sigma_z$ denotes the $z$ Pauli matrix.

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