# matrix of rotation for quantum states [duplicate]

For the two-level quantum system, we have the bloch sphere representation, and for a rotation we have the exponential operator: $$\text{exp}(\frac{-i \sigma \cdot \hat{n} \phi}{2})$$ where $\sigma = (\sigma_x, \sigma_y, \sigma_z)$ has Pauli matrices as components. Hence if we start with spin up $\ \begin{bmatrix} 1 \\ 0 \end{bmatrix}$ and we apply a rotation around the y-axis then we are left with $$\text{exp}(\frac{-i \sigma_y \phi}{2})\ \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \ \begin{bmatrix} \cos(\frac{\phi}{2}) &- \sin(\frac{\phi}{2}) \\ \sin(\frac{\phi}{2}) & \cos(\frac{\phi}{2}) \end{bmatrix} \ \begin{bmatrix} 1 \\ 0 \end{bmatrix}$$

But if we consider a quantum system for say $s = 1$ which is intially in state $\ \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$ how would we then write the matrix which corresponds to the rotation $\text{exp}(-i\frac{S_y \phi}{\hbar})$ so that we could evaluate the rotation $\text{exp}(-\frac{-iS_y \phi}{2})\ \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$?

Thanks for any assistance.

• The answer is a straightforward application of the Rodrigues' rotation formula for vectors once you understand the language. In that language, the vector aligned with the z axis (so left invariant by a z-rotation ---check it!) is (0,0,1). – Cosmas Zachos Jun 21 '17 at 22:14
• @CosmasZachos Thanks for your response. You mean the vector aligned with the z axis is $(0,0,1)$ as opposed to $(1,0,0)$ as I have it? – user110903 Jun 21 '17 at 22:24
• Yes, in that basis. It is the only one invariant under k aligned along that axis, no? – Cosmas Zachos Jun 21 '17 at 22:59
• @CosmasZachos Thanks. Could I just confirm that for $s =1$ you also get $$R_{z}(\phi) =\begin{bmatrix} -i\sin \phi + cos \phi & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & i\sin \phi + cos \phi \end{bmatrix}= \text{exp}(-iJ_z \phi)$$? – user110903 Jun 24 '17 at 18:07
• Absolutely! Right you are. For arbitrary s you simply plug in here. – Cosmas Zachos Jun 24 '17 at 18:21

You are going to need unitary matrices, i.e. matrices $R$ such that

$$R^+R=I\\ \det R = 1.$$

Note that these matrices can and often do contain complex entries.

For two-dimensional space, you can get such matrices by exponentiating Pauli matrices. This means that you simply write down the Taylor series of the exponential function, taking the matrix you wish to exponentiate as an argument. See more: https://en.wikipedia.org/wiki/Matrix_exponential.

For three-dimensional linear space, there is another set of matrices called Gell-Mann matrices: https://en.wikipedia.org/wiki/Gell-Mann_matrices. The rotation matrices are formed by the exact same procedure: writing matrix exponentials of those matrices.

P.S. The solution suggested in the comments is not sufficient, as Rodrigues' rotation formula only creates real-valued matrices.

EDIT okay so I was apparenty wrong about Rodrigues' formula, and the correct application for quantum mechanics can be found in Pedro's answer to this question: What is the spin rotation operator for spin > 1/2?

• @АлексейУваровA Thanks for the answer. Could you please give an example of say how we could go about computing a rotation of a spin $1$ system starting in the state $|s = 1, m = 1 \rangle$ which we want to rotate around the $x$-axis? For a $s= \frac{1}{2}$ system, it has a clear matrix representation after using the exponential matrix as I showed above. In the spin $s = 1/2$ case, the exponential representation reduces to a 2x2 matrix again, is this not the case for higher spin? – user110903 Jun 22 '17 at 20:35
• @JohnJack On exponential representation: correct, for any nxn matrix, the exponent is again an nxn matrix. As for rotation of a spin 1 system, my answer may be somewhat misleading, I'll try to elaborate in a moment – Алексей Уваров Jun 22 '17 at 20:42
• Okay thanks that would be great. One thing, to do the rotation that I mentioned, for $|s=1, m=1 \rangle$, could I apply the exponential operator in Pedro's answer of this post to $\begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}$ where I consider $\vec{J} \cdot \hat{n} = \hat{J_{x}}$? – user110903 Jun 22 '17 at 20:45
• @JohnJack Pedro is right, I corrected my answer – Алексей Уваров Jun 22 '17 at 21:00
• Thanks, so would I be right in stating that in his answer the $1$ is the identity operator? And that if I wanted the rotation of $|s = 1, m = 1 \rangle$ about the $x$ axis I would simply apply his formula to $\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$ where I take $\hat{n} \cdot \vec{J}= \hat{J_x}$? – user110903 Jun 22 '17 at 21:22