# Homework Help: Physical interpretation of the rotation of a ket

1. Jun 29, 2011

### bjnartowt

physical interpretation of the "rotation" of a ket

1. The problem statement, all variables and given/known data

No problem statement. I'm just having trouble imagining what it means to "rotate" a quantum -ket, especially since not all -kets are eigenstates of position. I know what the rotation operator is. I also can picture rotating a coordinate system in my head--that's easy of course. But what does it mean when a -ket is rotated?

2. Relevant equations

see attached .pdf. they are notes i've taken from shankar p. 306-306.

3. The attempt at a solution

see attached .pdf. they are notes i've taken from shankar p. 306-306.

#### Attached Files:

• ###### 306 - rotations in two dimensions.pdf
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2. Jun 30, 2011

### vela

Staff Emeritus
Re: physical interpretation of the "rotation" of a ket

Take the spin of an electron, for example. Suppose it's in the state
$$\vert \alpha \rangle = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 1 \end{pmatrix}$$
which is the +-eigenstate of Sx. If you rotate it by 90 degrees about the z-axis, you'd expect it to be in the +-eigenstate of Sy. The rotation operator about the z-axis for spin-1/2 is
$$R(\phi)=\exp\left(-\frac{i S_z \phi}{\hbar}\right) = \begin{pmatrix}e^{-i\phi/2} & 0 \\ 0 & e^{i\phi/2}\end{pmatrix}$$
For $\phi=\pi/2$, you get
$$\vert \alpha_R \rangle = R(\pi/2)\vert \alpha \rangle = \frac{1}{\sqrt{2}} \begin{pmatrix}e^{-i\pi/4} \\ e^{i\pi/4}\end{pmatrix} = \frac{e^{-i\pi/4}}{\sqrt{2}} \begin{pmatrix} 1 \\ i \end{pmatrix}$$
which is indeed the +-eigenstate of Sy.