Unclear passage, autoket and eigenvalues

Question

I am not understanding a passage that our professor wrote: those are the lines.

$e_0 \cdot \hat\sigma$ is an operator, whose eigenvalues are $\pm 1$. He applied this to a ket $|e_0, \pm 1 \rangle$:

$$e_0 \cdot \hat\sigma |e_0, \pm 1 \rangle = |e_0, \pm 1 \rangle (\pm 1)$$ multiply by $-1$

$$-e_0 \cdot \hat\sigma |e_0, \pm 1 \rangle = |e_0, \pm 1 \rangle (\mp 1)$$

From here he just wrote

$$|e_0, \pm 1 \rangle = |-e_0, \mp 1 \rangle$$

I have not clear why the operator $e_0 \cdot \hat\sigma $ vanished.

He wanted to show that indeed $|e_0, \pm 1 \rangle = |-e_0, \mp 1 \rangle$, where $e_0$ is a unit vector and $\pm 1$ in the ket are quantum numbers.

So where did the operator end up?

Answer 1

The cleaner notation would be that a ket $|A,\lambda^{(A)}_i\rangle$ is an eigenstate of operator $A$ with eigenvalue $\lambda^{(A)}_i$. This general notation would tell us that an operator $A$ might have a whole bunch of different eigenstates with different eigenvalues $\lambda^{(A)}_i$, where $i$ indexes the different eigenstates; we also see that the eigenvalues $\lambda^{(A)}_i$ ``belong to'' the operator $A$, because we have no idea a priori what happens to these states if you operator on them by some other operator $B$.

The calculation shown by the professor tells us that the eigenstates of two different operators $A=\mathbf{e}\cdot\boldsymbol{\sigma}$ and $B=-\mathbf{e}\cdot\boldsymbol{\sigma}$ are the same, except the corresponding eigenvalues are different. This means that the professor found a relationship between $|A,\lambda^{(A)}_i\rangle$ and $|B,\lambda^{(B)}_j\rangle$. Some of these eigenstates are exactly equal to each other, without having to operator on them. For example, the professor showed that $A|A,+1\rangle=|A,+1\rangle$ and $B|A,+1\rangle=-|A,+1\rangle$, meaning that $|A,+1\rangle$ is also an eigenstate of $B$ with eigenvalue $-1$, and our notation says we'd write such an eigenstate of $B$ as $|B,-1\rangle$, so we have identified that these two states are the same: $|A,+1\rangle=|B,-1\rangle$. A similar calculation tells us that the remaining eigenstates of each operator are also the same as each other $|A,-1\rangle=|B,+1\rangle$.

The next step is to write the operators $A$ and $B$ directly into the expressions: $$|\mathbf{e}\cdot\boldsymbol{\sigma},+1\rangle=|-\mathbf{e}\cdot\boldsymbol{\sigma},-1\rangle\quad\mathrm{and}\quad |\mathbf{e}\cdot\boldsymbol{\sigma},-1\rangle=|-\mathbf{e}\cdot\boldsymbol{\sigma},+1\rangle;$$ collectively, we can write this as $$|\mathbf{e}\cdot\boldsymbol{\sigma},\pm1\rangle=|-\mathbf{e}\cdot\boldsymbol{\sigma},\mp 1\rangle.$$

Finally, your professor decided it was too much hassle to carry around the operator $\boldsymbol{\sigma}$, because it will always be there for all of the types of states you are considering (not all of the states possible in the world, but all two-level quantum systems), so instead of writing kets with the full operators $A$ and $B$ we just write what unit vector is in the dot product with $\boldsymbol{\sigma}$: for $A$ it is $\mathbf{e}$ and for $B$ it is the unit vector pointing in the opposite direction, $-\mathbf{e}$.

Overall, this is all cleared up if you can keep track of which states are eigenstates of which operators with which eigenvalues. The professor showed a general trend that a whole bunch of different eigenstates of different operators are somehow related to each other, and because this is so general a trend they were able to write it without specifying the direction of $\mathbf{e}$.

Answer 2

Start with $$ -e_0\cdot \vec\sigma\vert e_0,\pm 1\rangle = \mp \vert e_0,\pm 1\rangle $$ and make the substitution $e_0\to -e_0$ on the LHS: $$ e_0\cdot\vec \sigma\vert -e_0,\pm 1\rangle = \mp \vert e_0,\pm 1\rangle\, . $$ It follows that $\vert -e_0,\pm 1\rangle$ are eigenstates of your operator $e_0\cdot\vec\sigma$.

Thus, the state $\vert -e_0,+1\rangle$ must be proportional to $\vert e_0,-1\rangle$ since there is only one eigenstate of $e_0\cdot\vec \sigma$ with eigenvalue $-1$. The same argument holds for $\vert -e_0,-1\rangle$.

The physics here is that $e_0$ is a unit vector in a particular direction - say $+z$ to connect with what you already know. “Spin up” is now in the direction of the North Pole. Then $e_0\cdot \vec\sigma=\sigma_z$, and $\vert e_0,+ 1\rangle$ correspond to the “North Pole” state.

Now flip $e_0\to -e_0$. “Spin up” is now in the direction of the South Pole, and clearly now $\vert -e_0,+1\rangle$ will point to the South Pole.

This is the same story but repeated in a general direction $\hat e_0$ rather than $\hat z$.