Find the spinor-state for a given expectation value

[Markus Kahn]

Homework Statement

Let $\vec{e}\in\mathbb{R}^3$ be any unit vector. A spin $1/2$ particle is in state $|\chi \rangle$ for which
$$\langle\vec{\sigma}\rangle =\vec{e},$$
where $\vec{\sigma}$ are the Pauli-Matrices. Find the state $|\chi\rangle$

Homework Equations

:[/B] are all given above.

The Attempt at a Solution

Well, I honestly have trouble even understanding what exactly the exercise is about. First of all, I'm really confused about the fact that the expectation value of an operator is supposed to be a vector. The only explanation I have for this kind of notation is
$$\langle\vec{\sigma}\rangle =\vec{e} \Leftrightarrow \sigma_i|\chi\rangle = e_i|\chi\rangle \hspace{0.5cm} i=1,2,3$$
Is this the right interpretation of the notation?
If this was true my approach was as follows: I assumed a basis of the corresponding Hilber space to be $\mathcal{H}=\operatorname{span}\left(|0\rangle ,|1\rangle \right)$ and therefore $a,b\in\mathbb{C}^2$ exist such that $|\chi\rangle = a|0\rangle + b |1\rangle$. Now applying $\sigma_1$ to $|\chi\rangle$ results in the equation $e_i \cdot (a|0\rangle + b|1\rangle )= b|0\rangle + a|1\rangle \Leftrightarrow a=e_i b$ and $b = e_i a$, which doesn't make any sense at all considering that I want to find $a,b$ as functions of $e_1,e_2,e_3$.

I really don't know how to approach this problem and would be thankful for any suggestions.

 

[PeroK]

Homework Statement

Let $\vec{e}\in\mathbb{R}^3$ be any unit vector. A spin $1/2$ particle is in state $|\chi \rangle$ for which
$$\langle\vec{\sigma}\rangle =\vec{e},$$
where $\vec{\sigma}$ are the Pauli-Matrices. Find the state $|\chi\rangle$

Homework Equations

:[/B] are all given above.

The Attempt at a Solution

Well, I honestly have trouble even understanding what exactly the exercise is about. First of all, I'm really confused about the fact that the expectation value of an operator is supposed to be a vector. The only explanation I have for this kind of notation is
$$\langle\vec{\sigma}\rangle =\vec{e} \Leftrightarrow \sigma_i|\chi\rangle = e_i|\chi\rangle \hspace{0.5cm} i=1,2,3$$
Is this the right interpretation of the notation?
If this was true my approach was as follows: I assumed a basis of the corresponding Hilber space to be $\mathcal{H}=\operatorname{span}\left(|0\rangle ,|1\rangle \right)$ and therefore $a,b\in\mathbb{C}^2$ exist such that $|\chi\rangle = a|0\rangle + b |1\rangle$. Now applying $\sigma_1$ to $|\chi\rangle$ results in the equation $e_i \cdot (a|0\rangle + b|1\rangle )= b|0\rangle + a|1\rangle \Leftrightarrow a=e_i b$ and $b = e_i a$, which doesn't make any sense at all considering that I want to find $a,b$ as functions of $e_1,e_2,e_3$.

I really don't know how to approach this problem and would be thankful for any suggestions.

First, if you measure an observable that is a vector, then the expectation (average) of your measurement must be a vector. Expecation value of position or momentum in 3D, for example.

That said, that's not quite what's asked here. In general, if you have three operators (normally one will be associated with $x$, one with $y$ and one with $z$), then you can define the vector operator:

$\vec{Q} = (Q_x, Q_y, Q_z)$

That's what you have in this case with:

$\vec{\sigma} = (\sigma_x, \sigma_y, \sigma_z)$

Then, the expected value of this is defined to be:

$\langle \vec{\sigma} \rangle = (\langle \sigma_x \rangle,\langle \sigma_y \rangle, \langle \sigma_z \rangle)$

In terms of solving this problem, you might like to consider the expected value of $\vec{\sigma}$ on eigenstates of spin in the x, y and z directions.

 

[PeroK]

The only explanation I have for this kind of notation is
$$\langle\vec{\sigma}\rangle =\vec{e} \Leftrightarrow \sigma_i|\chi\rangle = e_i|\chi\rangle \hspace{0.5cm} i=1,2,3$$
Is this the right interpretation of the notation?

This won't work because here you have assumed that $\chi$ is an eigenstate of $\sigma_i$ with eigenvalue $e_i$. But, that's not right at all.

 

[Markus Kahn]

Thank you very much for your response!

As you mentioned in your first answer I tried to calculate the expectation values in the basis of the corresponding Puali matrix. Let us again assume that $|\chi\rangle = a|0\rangle + b |1\rangle$ with $a,b\in\mathbb{C}$. The results of this are

• z-direction: This was the easiest since the eigenbasis of $\sigma_z$ is already the basis in which we write $|\chi\rangle$. We therefore find $\langle\sigma_z\rangle = |a|^2-|b|^2\overset{!}=e_3$.
• x-direction: we define $|0_x\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$ and $|1_x\rangle = \frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)$ and note that we can write $|\chi\rangle =\frac{a}{\sqrt{2}}(|0_x\rangle+|1_x\rangle)+\frac{b}{\sqrt{2}}(|0_x\rangle-|1_x\rangle)$. We therefore than find $\langle\sigma_x\rangle = a^*b+b^*a\overset{!}{=}e_1$.
• y-direction: In a similar fashion to the x coordinate we find here $\langle\sigma_2\rangle=i (ab^*-a^*b)\overset{!}{=}e_2$.

So we have now found the system of equations:
$$\begin{align}|a|^2-|b|^2&=e_3\\a^*b+b^*a&=e_1\\ i (ab^*-a^*b)&=e_2\\ |\vec{e}|=1.\end{align}$$
Does this work up until this point? If it does, I'm currently struggling to solve the system of equations above... Any helpful suggestions how to approach that?

 

[PeroK]

Looks good so far. It might help to note that $(a^*b + b^*a) = 2Re(a^*b)$ etc.

Remember that you also have $|a|^2 + |b|^2 = 1$.

 

[Markus Kahn]

Thanks for the suggestion but I just couldn't figure anything out in that form so I defined $a:=(w+i x)$ and $b:=(y+i z)$ which leads me to the following system of equations:
$$\begin{align*}w^2+x^2-y^2-z^2&=e_3\\
2wy + 2xz &= e_1\\
2wz - 2xy &= e_2\\
|\vec{e}|&=1\\
w^2+x^2+y^2+z^2 &= 1\end{align*}$$
I also tried adding the first and last equation to get something a bit simpler, but I'm not really sure if this is leading anywhere...

 

[PeroK]

There's a general trick. If $a, b$ is a solution, then so is $\alpha a, \alpha b$, where $\alpha$ is any complex number of modulus 1. So, you can take $a$ to be real, without loss of generality.

 

[PeroK]

Here's another approach, which has two nice ideas.

First, we can express $e_1, e_2, e_3$ in spherical coordinates: $\sin \theta \cos \phi, \sin \theta \sin \phi, \cos \theta$

Also, whenever you have $|a|^2 + |b|^2 = 1$, you can find $\alpha, \beta, \gamma$ such that $a = (\cos \alpha) e^{i\beta}, \ b = (\sin \alpha) e^{i\gamma}$. It's easy enough to prove.

In this case, taking $a$ to be real, we have $a = \cos \alpha, \ b = (\sin \alpha) e^{i\gamma}$.

Then, you can match up the angles of $a, b$ with the spherical representation of $e_1, e_2, e_3$.

PS This approach has a lot to recommend it. The answer reveals an extraordinary aspect of rotation in QM.

 

[Markus Kahn]

Thank you very much for the last suggestion. The transformation did the trick for me. Assuming that $\vec{e}$ corresponds to the spherical coordinates $(r=1,\theta,\phi)$ I find
$$|\chi\rangle = \cos\frac{\theta}{2}|0\rangle + \sin\frac{\theta}{2}e^{i\phi}|1\rangle.$$
I'm am not very good when it comes to visualizing things, so I'm not really sure what this results really implies (your PS seems to indicate that here something special should happen)...

Even though I was able to solve it by switching to spherical coordinates I am still interested in how one would solve this in cartesian coordinates. Could you maybe help me here a bit more... Sorry for the bother..

 

[Markus Kahn]

There's a general trick. If $a, b$ is a solution, then so is $\alpha a, \alpha b$, where $\alpha$ is any complex number of modulus 1. So, you can take $a$ to be real, without loss of generality.

If possible could you point me to a prove of this statement, or maybe explain it a bit more... I'm not really sure what the explanation for this is

 

[PeroK]

If possible could you point me to a prove of this statement, or maybe explain it a bit more... I'm not really sure what the explanation for this is

First, what we had was:

$|a|^2 - |b|^2 = e_3$, which gives $2|a|^2 = e_3 + 1$

$2Re(a^*b) = e_1$

$2Im(a^*b) = e_2$

If we replace $a, b$ with $\alpha a, \alpha b$, then none of the quantities on the left-hand side changes. $\alpha$ is called a "phase factor" and, in general in QM solutions are up to an arbitrary phase factor - in terms of having the same expectation values. The same applies to a position-space wave-function, for example.

Taking $a$ to be real and $a \ne 0$ gives:

$b = \frac{1}{2a}(e_1 + ie_2)$

And, you can solve for $a$ using the first equation. Note the special case when $e_3 = -1$ and $a = 0$.

The moral is that identifying the Real and Imaginary parts of a complex number is a good idea. Always prefer that to equations in lots of real variables like you had!

On the second point, you have a state that yields a physical expectation value of $\vec{e}$, which has an azimuthal angle of $\theta$.

But, the quantum state $|\chi \rangle$ is associated with an angle of $\frac{\theta}{2}$. That's interesting! Does that mean that if you rotate your system by 360°, then you only rotate the quantum state by 180°; and, to get back to your original state, you must rotate a physical system by 720°? Do you think there might be experimental evidence of that?

 

[Markus Kahn]

Do you think there might be experimental evidence of that?

I'm honestly still puzzled by the idea that rotating a state by $2\pi$ doesn't result in the same state, so I really can't think of a way to experimentally show something like that... After a quick search on Google this actually seems to be true for Fermions but not for Bosons.. Could one do the same procedure as described here and show that this doesn't happen for particles with integer spin?

PS. Thank you for the explanation regarding the Cartesian coordinates.

 

[PeroK]

I'm honestly still puzzled by the idea that rotating a state by $2\pi$ doesn't result in the same state, so I really can't think of a way to experimentally show something like that... After a quick search on Google this actually seems to be true for Fermions but not for Bosons.. Could one do the same procedure as described here and show that this doesn't happen for particles with integer spin?

PS. Thank you for the explanation regarding the Cartesian coordinates.

Rotating the state by $2\pi$ results in the same state. But, rotating the physical system by $2\pi$ rotaes the state by only $\pi$.