5
$\begingroup$

The electron spin can be described using two orthonormal state vectors in two dimensional Hilbert space. If we assume the two orthonormal vectors as $|+\rangle$ and $|-\rangle$, we can represent the general spin state as $$|S\rangle = c_1 |+\rangle + c_2 |-\rangle$$ with condition that $|c_1|^2 + |c_2|^2 = 1$.

We now determine components $S_x$, $S_y$ and $S_z$ of spin state $|S\rangle$ along the three axes of the Cartesian coordinate system. This is done by using Pauli matrices which serve as the operators for observables $S_x$, $S_y$ and $S_z$.

Now my question is this: While studying Pauli matrices, it appears to me that state vector $|+\rangle$ lies along $+z$ axis while $|-\rangle$ lies along $-z$ axis. Is this by assumption or is there any mathematical reason why the orthogonal vectors $|+\rangle$ and $|-\rangle$ model directions that are at 180 degree with each other in three dimensional space?

In other words, although $|+\rangle$ and $|-\rangle$ are at 90 degree (orthogonal) in Hilbert space, they model $+z$ and $-z$ directions that are at 180 degree in three dimensional space. Is this by assumption or is there any mathematical reason for it?

If the answer requires mathematics, please keep it as simple as possible.

ABOUT THE EDIT

Edited the question a little bit to reflect that I am not asking why orthogonal spin states 'point towards' directions that are at 180 degree but, instead, why orthogonal spin states are chosen to 'model' directions that are at 180 degree. Is it by assumption or is there any deeper meaning?

$\endgroup$
3
  • 3
    $\begingroup$ Very similar question: Up and down spin are orthogonal, not antiparallel $\endgroup$ Commented Oct 30, 2022 at 14:57
  • $\begingroup$ @ThomasFritsch Thanks for the link. I did not see that question before. But, even though your answer there is great, I do not get answer to my question. Can you please provide your answer here focused on my question? $\endgroup$
    – t2m
    Commented Oct 30, 2022 at 15:43
  • 4
    $\begingroup$ Bloch sphere half angle. $\endgroup$ Commented Oct 30, 2022 at 16:11

2 Answers 2

11
$\begingroup$

Because quantum mechanical states do not live in ordinary space, the notion of orthogonality of quantum mechanical states cannot be understood as orthogonality in 3d space.

In quantum mechanics, states are orthogonal when they correspond to distinct outcomes of an observable. This orthogonality is a mathematical property of eigenstates of Hermitian operators (which represent observables). In particular, the dimension of the space is the number of possible distinct outcomes of the observable.

There are two possible outcomes of measuring spin along any direction so the space will be 2-dimensional for any direction of spin, and the states corresponding to spin-up and spin-down will be orthogonal in this 2-dimensional space.

In the same way, there are three angular momentum states with $L=1$, and they have projections $m=+1,0,-1$. The states with outcomes $m=+1$ and $m=-1$ are orthogonal in the space of states even if they can be at $180^\circ$ in 3d space. There are 5 orthogonal angular momentum states with $L=2$, even if these "orbitals" live in 3d space. The solutions to the particle in the box problems are orthogonal in the (infinite dimensional) space of states even if all the particle "lives" in 1d. Ditto for the harmonic oscillator.

$\endgroup$
8
  • $\begingroup$ You seem to be explaining the premises around my question but not answering it. I understand that orthogonality in Hilbert space does not imply orthogonality in 3d space. But when we are using orthogonal vectors from Hilbert space to find spin components towards the axes of real 3d space, which axes can be modeled by them? Do we simply 'assume' that they model +z and -z axes or is there any deeper reason? $\endgroup$
    – t2m
    Commented Oct 30, 2022 at 15:38
  • 1
    $\begingroup$ The measurement for $L=2$ can also be oriented in any direction. In this case $m=2$ and $m=-1$ states are still orthogonal even if they are not at $90^\circ$ or $180^\circ$ in 3d space. You're trying to tie the notion of angles in 3d space with orthogonality in state space and there is no such connection. Neither the relative angle nor the "orientation" matters. $\endgroup$ Commented Oct 30, 2022 at 15:49
  • 1
    $\begingroup$ no you need pick one observable, find its eigenvectors and these become $\vert +\rangle$ and $\vert -\rangle$. The observable can be $\sigma_x$ or $\sigma_z$, or for that matter a combination of them, but the outcomes of this one observable will be $\pm\hbar /2$ and the states will be $\vert \pm\rangle$. $\endgroup$ Commented Oct 30, 2022 at 15:52
  • 2
    $\begingroup$ The historical choice of $\hat z$ is because $z=r\cos\theta$ has a simpler expression in spherical than $x$ or $y$. $\endgroup$ Commented Oct 30, 2022 at 15:57
  • 3
    $\begingroup$ @Aarone The state space is abstract, it doesn't by itself encode the directions, or even what the thing it is representing actually is. That information is extrinsic, it comes from the person doing the modelling. You're not observing the abstract objects $|+\rangle$ or $|-\rangle$ (the states in Hilbert space), you're measuring a different quantity (spin) which can either take the value +ℏ/2 (corresponding to $|+\rangle$) or -ℏ/2 (corresponding to $|-\rangle$). $|+\rangle$ is not the spin itself, the $+$ is just a label; $|+\rangle$ is really $|\text{"state corresponding to spin up"}\rangle$ $\endgroup$ Commented Oct 31, 2022 at 0:28
9
$\begingroup$

The state vectors ($|+\rangle$, $|-\rangle$, and their superpositions $|c_1 |+\rangle + c_2 |-\rangle$) are members of the Hilbert space $\mathbb{C}^2$.

The two states $|+\rangle$ and $|-\rangle$ are said to be orthogonal to each other just because their scalar product is zero, $\langle +|-\rangle=0$. Only in this sense the "angle" between them is $90°$. This may seem counter-intuitive. But this is only because human intuition isn't strong in the $\mathbb{C}^2$ space. Our intuition is much stronger in our every-day space $\mathbb{R}^3$ (containing vectors with 3 real components in $x$, $y$ and $z$ direction).

There is a mathematical procedure using the Pauli matrices to get a vector $\vec{n}$ in $\mathbb{R}^3$ from a state vector $|\chi\rangle$ in $\mathbb{C}^2$. The physical meaning of this is, that for a state $|\chi\rangle$ the vector $\vec{n}$ points along the spinning axis of this state. $$\vec{n} = \langle\chi|\vec{\sigma}|\chi\rangle$$ or written in components $$\begin{pmatrix}n_x\\n_y\\n_z\end{pmatrix}= \begin{pmatrix} \langle\chi|\sigma_x|\chi\rangle \\ \langle\chi|\sigma_y|\chi\rangle \\ \langle\chi|\sigma_z|\chi\rangle \end{pmatrix}$$

Note, there is a correspondence between $|\chi\rangle$ and $\vec{n}$, but they are not the same thing.

For your two special states you get $$|+\rangle=\begin{pmatrix}1\\0\end{pmatrix} \Rightarrow \vec{n}_+ =\langle +|\vec{\sigma}|+\rangle =\begin{pmatrix}0\\0\\1\end{pmatrix}$$ $$|-\rangle=\begin{pmatrix}0\\1\end{pmatrix} \Rightarrow \vec{n}_- =\langle -|\vec{\sigma}|-\rangle =\begin{pmatrix}0\\0\\-1\end{pmatrix}$$

Here you see: The angle between $\vec{n}_+$ and $\vec{n}_-$ (two vectors in $\mathbb{R}^3$) is $180°$.

Do not confuse the $\mathbb{C}^2$-vectors $|+\rangle$ and $|-\rangle$ with the $\mathbb{R}^3$-vectors $\vec{n}_+$ and $\vec{n}_-$. They live in two very different vector spaces. And therefore it should not come as a surprise that the angles between these two pairs of vectors are different.

$\endgroup$
0

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.