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?

  • 3
    $\begingroup$ .Why don't you ask your professor what the notation means?! $\endgroup$ Commented Mar 2, 2023 at 13:16
  • $\begingroup$ @TobiasFünke Trust me, you wouldn't want him as a professor. His replies are of the type: "it's obvious", or "there is nothing to understand here", or "get a book". $\endgroup$
    – Heidegger
    Commented Mar 2, 2023 at 13:24
  • 2
    $\begingroup$ Well, for me the notation is not clear at all (but perhaps its only me). So the most obvious person to ask about clarification is your professor. Anyway, you should give some context: Where do these calculations appear and what do the symbols mean; e.g. a spin 1/2 particle with Pauli matrices?! $\endgroup$ Commented Mar 2, 2023 at 13:25
  • $\begingroup$ @TobiasFünke It's about Berry phrase, and yes $\sigma$ is a Pauli matrix! $\endgroup$
    – Heidegger
    Commented Mar 2, 2023 at 13:50
  • $\begingroup$ @Numb3rs although the tone of the reply would matter quite a bit, the answer itself is what it should be. I probably would not answer directly this question either unless the student showed me their work; this being said, my initial reply would be more constructive than dismissive. $\endgroup$ Commented Mar 2, 2023 at 15:32

2 Answers 2


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 $\mathbf{e}$.


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 $\hat z$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.