Density matrix, change of basis, I don't understand the basics

[Matterwave]

Yeah I must find the rotation matrix and apply it to diagonalize rho.

I've just found the matrix $\hat n \cdot \vec S$. It's worth $\frac{\hbar}{2} \begin{bmatrix} \cos \theta && \sin (\theta) e^{- i \phi} \\ \sin (\theta )e^{i \phi} && - \cos \theta \end{bmatrix}$ where I've used the spherical coordinates, theta is the zenith angle and phi is the azimuth one.

I just looked at this matrix again, and I'm sorry but you made a really confusing use of the angle θ here. The θ here is not the same as the θ you had in your previous equation for the rotation matrix $U$. The θ in $U=\cos(\theta/2)\hat{I}+\sin(\theta/2)\hat{n}\cdot \vec{S}$ is the rotation around the unit normal vector. The θ you have in this matrix is the polar angle of the unit normal itself.

A rotation in 3-space requires 3 angles, not 2! I got confused on this point for a minute.

So once you can relate $\alpha,\beta$ which has 3 independent components, since they are 2 complex numbers with 1 constraint, with $\theta,\phi,\eta$, where $\eta$ will replace one of the $\theta$'s that you used, you will have the final answer.

Whew! That was confusing!

 

[fluidistic]

I just looked at this matrix again, and I'm sorry but you made a really confusing use of the angle θ here. The θ here is not the same as the θ you had in your previous equation for the rotation matrix $U$. The θ in $U=\cos(\theta/2)\hat{I}+\sin(\theta/2)\hat{n}\cdot \vec{S}$ is the rotation around the unit normal vector. The θ you have in this matrix is the polar angle of the unit normal itself.

A rotation in 3-space requires 3 angles, not 2! I got confused on this point for a minute.

So once you can relate $\alpha,\beta$ which has 3 independent components, since they are 2 complex numbers with 1 constraint, with $\theta,\phi,\eta$, where $\eta$ will replace one of the $\theta$'s that you used, you will have the final answer.

Whew! That was confusing!

I am sorry to have caused confusion MatterWave. I am aware that the theta's are different. I never used the formula for U so far, I only wrote it down in the first post as an idea that I could use later. But so far I don't see how it could help me.
Also I did not wrote it down the way you did because it seems you took $\hat u$ as $\hat n$ which is supposed to be the axis of rotation and I would not have thought about rotating around chi.
I am stuck on post 28. 2 equations with 4 unknown. I guess the idea is to express both alpha and beta in terms of both theta and phi. It doesn't look that easy at first glance but I'll try.Edit: I can't seem to express neither alpha nor beta in terms of only theta and phi.

 

[Matterwave]

Maybe this information can help a little. Any arbitrary unitary matrix (in suggestive notation!) can be expressed as:

$$U=\begin{bmatrix} \alpha && -\bar{\beta} \\ \beta && \bar{\alpha} \end{bmatrix}$$

With the special unitary matrices satisfying the additional condition that the determinant is 1, or $|\alpha|^2+|\beta|^2=1$.

Notice that this matrix rotates $\left|z_+\right>$ into $\left|\chi\right>$ already... and the left column is orthogonal to the right column...

 

[fluidistic]

Thanks MatterWave, I'll think about it. Meanwhile I think I found a trick to deal with my 2 equations. I can multiply the first one by the complex conjugate of alpha for example. I'm busy right now but I'll get back to it as soon as I can.Edit:If I didn't do any algebra mistake, I reached that $|\alpha|^2=\frac{\cos (\theta ) +1}{2}$ which would mean that $|\beta |^2=\frac{1-\cos \theta }{2}$ and for some values of theta this is negative...
Also I see no dependence on phi whatsoever.

 

[Matterwave]

I'm actually not sure myself how finding the matrix $\hat{n}\cdot\vec{S}$ for which $\left|\chi\right>$ is an eigen-vector is going to get you to your goal, since I'm not sure that it's guaranteed that a rotation $U$ with that matrix is going to get you the right answer. I am especially confused because I don't see how you can get the angle $\eta$ in $U=\cos(\eta/2)\hat{I}+\sin(\eta/2)\hat{n}\cdot\vec{S}$ from this method. Perhaps Frederick can expand on what he meant when he mentioned that.

 

[Fredrik]

I haven't actually solved this problem. All I said was this:

I haven't worked out the solution to this problem, so I don't know what exactly will be useful. But the first thing that comes to mind is that every vector in this space is an eigenstate of an operator of the form $\vec n\cdot\vec S$, where $\vec n$ is some unit vector in $\mathbb R^3$. This seems relevant, since you're asked to use a rotation operator.

I'll give it a shot and see if I can think of something.

 

[fluidistic]

I've edited my last post, I actually found the modulus squared of alpha in terms of theta, but I am not sure I didn't make any algebra mistake.

 

[Matterwave]

Working with the modulus squared will lose you any phase information. Really the point I was trying to make earlier about the equality $|\alpha|^2+|\beta|^2=1$ was I was hoping you would realize that it looks awfully similar to the trigonometric identity $\sin^2(\theta)+\cos^2(\theta)=1$.

 

[fluidistic]

Working with the modulus squared will lose you any phase information. Really the point I was trying to make earlier about the equality $|\alpha|^2+|\beta|^2=1$ was I was hoping you would realize that it looks awfully similar to the trigonometric identity $\sin^2(\theta)+\cos^2(\theta)=1$.

I see but that's basically what I reached then.
Because $|\alpha | ^2=\frac{1+ \cos \theta }{2}=\cos ^2 \left ( \frac{\theta}{2} \right )$ and $|\beta |^2 =\frac{1-\cos \theta}{2} = \sin ^2 \left ( \frac{\theta}{2} \right )$.

 

[Matterwave]

Yes, but that form is not conducive to writing out $\alpha,\beta$.

 

[fluidistic]

Yes, but that form is not conducive to writing out $\alpha,\beta$.

hmm right I lose the phase information as you said.
Edit: I'll try to see if I can manage to get alpha and beta in terms of the angles, knowing their modulus squared.

 

[Matterwave]

Make sure you keep track of the different angles that appear in this calculation. For one, the $\theta$ I defined in order to parametrize $\alpha,\beta$ will not necessarily have anything to do with the (2 other ) $\theta$'s previously defined.

The point I was trying to get at is since you noticed that $\sin^2(\theta)+\cos^2(\theta)=1$ shouldn't I be able to write my $\alpha,\beta$ as $\alpha=\sin(\theta')e^{i\phi'}$ and $\beta=\cos(\theta')e^{i\eta'}$? If I put this in my standard special unitary matrix form, how is this special unitary matrix decomposed into $U=\cos(\theta/2)\hat{I}+\sin(\theta/2)\hat{n}\cdot\vec{S}$?

 

[Fredrik]

I found a solution, sort of. But maybe you guys are already past this stage. I haven't been following the thread closely.

Let a,b be the unique complex numbers such that $|\chi\rangle=a|+\rangle+b|-\rangle$. We have $\rho=|\chi\rangle\langle\chi|$. Let $|\phi\rangle$ be any unit vector that's orthogonal to $|\chi\rangle$. We'll be working with the two ordered bases $B=\big(|+\rangle,|-\rangle\big)$ and $B'=\big(|\chi\rangle,|\phi\rangle\big)$.

It's straightforward to calculate $[\rho]_B$. That's one of the things the problem asks us to do. I see that you've already done that, so I'll write down the result
$$[\rho]_B=\begin{pmatrix}|a|^2 & ab^*\\ a^*b & |b|^2\end{pmatrix}.$$

Regarding the rest of the problem: We can immediately tell that
$$[\rho]_{B'}=\begin{pmatrix}1 & 0\\ 0 & 0\end{pmatrix}.$$ So why don't we just define the rotation operator by $U|+\rangle=|\chi\rangle$ and $U|-\rangle=|\phi\rangle$, and then use this to determine ##_B## in terms of a and b?

This worked out nicely for me. The only thing that bugs me about this is that the problem statement seems to require you to use the rotation operator to diagonalize $\rho$. In my approach, we use the assumptions that $\rho$ is diagonal in the B' basis and that the B' basis is obtained by applying U to the elements of B, to find U. So U is the last matrix we find of all the matrices we're interested in. When we find it, there's nothing left to do.

 

[Matterwave]

I found a solution, sort of. But maybe you guys are already past this stage. I haven't been following the thread closely.

Let a,b be the unique complex numbers such that $|\chi\rangle=a|+\rangle+b|-\rangle$. We have $\rho=|\chi\rangle\langle\chi|$. Let $|\phi\rangle$ be any unit vector that's orthogonal to $|\chi\rangle$. We'll be working with the two ordered bases $B=\big(|+\rangle,|-\rangle\big)$ and $B'=\big(|\chi\rangle,|\phi\rangle\big)$.

It's straightforward to calculate $[\rho]_B$. That's one of the things the problem asks us to do. I see that you've already done that, so I'll write down the result
$$[\rho]_B=\begin{pmatrix}|a|^2 & ab^*\\ a^*b & |b|^2\end{pmatrix}.$$

Regarding the rest of the problem: We can immediately tell that
$$[\rho]_{B'}=\begin{pmatrix}1 & 0\\ 0 & 0\end{pmatrix}.$$ So why don't we just define the rotation operator by $U|+\rangle=|\chi\rangle$ and $U|-\rangle=|\phi\rangle$, and then use this to determine ##_B## in terms of a and b?

This worked out nicely for me. The only thing that bugs me about this is that the problem statement seems to require you to use the rotation operator to diagonalize $\rho$. In my approach, we use the assumptions that $\rho$ is diagonal in the B' basis and that the B' basis is obtained by applying U to the elements of B, to find U. So U is the last matrix we find of all the matrices we're interested in. When we find it, there's nothing left to do.

Yes...I was trying to guide fluid to this conclusion... I also assumed that what the problem meant by asking him to find the rotation matrix, is that he had to find the correspondence between the $U$ and a rotation in real space given by $U=\cos(\theta/2)\hat{I}+\sin(\theta/2)\hat{n}\cdot\vec{S}$. So that's just one more step where after you find $U$ you have to decompose it in this way.

 

[Fredrik]

I said earlier that every vector in this space is an eigenvector of $\vec n\cdot\vec S$, where $\vec n$ is some unit vector in $\mathbb R^3$. I just took a peek at some notes I made at least 5 years ago. I will give you some of the highlights, but no guarantee that the result is correct.

If we assume that $|\chi\rangle =a|+\rangle+b|-\rangle$ is a normalized eigenstate of $\vec n\cdot\vec S$, with eigenvalue $\lambda$, and that $\vec n$ is a unit vector, we can prove the following results:
\begin{align}
\lambda a&=\frac 1 2 (n^*b+n_3a)\\
\lambda b&=\frac 1 2 (na-n_3b)\\
\lambda &=\pm\frac 1 2.
\end{align} The $n$ in the first two equalities is defined by $n=n_1+in_2$. My starting point was just to write down the eigenvalue equation with $S_1$ and $S_2$ expressed in terms of the raising and lowering operators $S_\pm=S_1\pm iS_2$.

After deriving the results above, I verified that when those conditions are satisfied, $|\chi\rangle$ is an eigenstate with eigenvalue $\frac 1 2$ or $-\frac 1 2$. So there are two eigenstates. I denoted them by $|\vec n+\rangle$ and $|\vec n-\rangle$. The last thing I did was to find that
$$|\vec n\pm\rangle =N\left(|+\rangle+\frac{n_1+in_2}{n_3\pm 1}|-\rangle\right),$$ where N is a normalization constant that I didn't bother to determine.

 

[Matterwave]

I said earlier that every vector in this space is an eigenvector of $\vec n\cdot\vec S$, where $\vec n$ is some unit vector in $\mathbb R^3$. I just took a peek at some notes I made at least 5 years ago. I will give you some of the highlights, but no guarantee that the result is correct.

If we assume that $|\chi\rangle =a|+\rangle+b|-\rangle$ is a normalized eigenstate of $\vec n\cdot\vec S$, with eigenvalue $\lambda$, and that $\vec n$ is a unit vector, we can prove the following results:
\begin{align}
\lambda a&=\frac 1 2 (n^*b+n_3a)\\
\lambda b&=\frac 1 2 (na-n_3b)\\
\lambda &=\pm\frac 1 2.
\end{align} The $n$ in the first two equalities is defined by $n=n_1+in_2$. My starting point was just to write down the eigenvalue equation with $S_1$ and $S_2$ expressed in terms of the raising and lowering operators $S_\pm=S_1\pm iS_2$.

After deriving the results above, I verified that when those conditions are satisfied, $|\chi\rangle$ is an eigenstate with eigenvalue $\frac 1 2$ or $-\frac 1 2$. So there are two eigenstates. I denoted them by $|\vec n+\rangle$ and $|\vec n-\rangle$. The last thing I did was to find that
$$|\vec n\pm\rangle =N\left(|+\rangle+\frac{n_1+in_2}{n_3\pm 1}|-\rangle\right),$$ where N is a normalization constant that I didn't bother to determine.

But I'm not sure how you go from this solution to finding a unitary rotation matrix $U$ since $\vec{n}\cdot\vec{S}$ is a Hermitian matrix, not a unitary one.

EDIT: After thinking for a minute, I see where you're going with this. You want to at the end rotate the $z$-axis into $\vec{n}$ with some $SO(3)$ matrix to find "the rotation matrix"?

 

[Fredrik]

I haven't actually worked that part of the problem, and I don't have a perfect understanding of rotation operators, but I'll offer some of my thoughts: If $e^{i\theta S_z}$ is a rotation by $\theta$ around the z axis (am I off by a factor of 2 or something?), then $e^{i\vec\theta\cdot\vec S}$ should be a rotation by $|\vec\theta|$ around the $\vec\theta$ axis. We can write $\theta=|\vec\theta|$, $\hat\theta=\vec\theta/\theta$ and $e^{i\vec\theta\cdot\vec S} =e^{i\theta(\hat\theta\cdot\vec S)}$.

So now it's a matter of determining $\hat\theta$ and then $\theta$. We should be able to express $\vec n$ in terms of a,b. Then we can determine the plane that contains $\vec 0$, $\vec n$ and $\vec e_3$. The unit vector $\hat\theta$ should be orthogonal to that plane. $\theta$ is the angle between $\vec n$ and $\vec e_3$.

This seems like a lot of work, and I doubt that the person who came up with the problem intended us to do something like this.

 

[Matterwave]

I haven't actually worked that part of the problem, and I don't have a perfect understanding of rotation operators, but I'll offer some of my thoughts: If $e^{i\theta S_z}$ is a rotation by $\theta$ around the z axis (am I off by a factor of 2 or something?), then $e^{i\vec\theta\cdot\vec S}$ should be a rotation by $|\vec\theta|$ around the $\vec\theta$ axis. We can write $\theta=|\vec\theta|$, $\hat\theta=\vec\theta/\theta$ and $e^{i\vec\theta\cdot\vec S} =e^{i\theta(\hat\theta\cdot\vec S)}$.

Actually, you're not missing a factor of 2. The factor of 2 will come when you take $\vec{S}=\frac{\hbar}{2}\vec{\sigma}$. The method in your post seems too complicated though.

I also noticed that I have been writing the wrong $U$ this whole time...the formula for $U$ in fluid's original post is what I meant...and nobody corrected me even!

...In fact I must apologize, since I have been guiding fluid into a longer way to solve this problem. I never bothered to solve this problem out myself, and now that I have, I found a much simpler way...

Fluid, start with the $\hat{u}\cdot\vec{\sigma}$ part of the equation, which you basically already got in post #19 (forget the factor of $\hbar/2$). Now find $U$ (by that I mean construct $U=\cos(\eta/2)\hat{I}-i\sin(\eta/2)\hat{u}\cdot\vec{\sigma}$). Now identify the left column of that final matrix with the column vector for $\left|\chi\right>$ and you're done. Because that left column will take $\left|z_+\right>$, which in that basis is of course the column vector $(1,0)$, to the most general spin state constructable, which for our problem is $\left|\chi\right>$. I verified this since it is equation 14-23 of Baym's lectures on quantum mechanics.

We've been working way too hard to try to find the answer.

 

[Fredrik]

Let a,b be the unique complex numbers such that $|\chi\rangle=a|+\rangle+b|-\rangle$. We have $\rho=|\chi\rangle\langle\chi|$. Let $|\phi\rangle$ be any unit vector that's orthogonal to $|\chi\rangle$. We'll be working with the two ordered bases $B=\big(|+\rangle,|-\rangle\big)$ and $B'=\big(|\chi\rangle,|\phi\rangle\big)$.

It's straightforward to calculate $[\rho]_B$. That's one of the things the problem asks us to do. I see that you've already done that, so I'll write down the result
$$[\rho]_B=\begin{pmatrix}|a|^2 & ab^*\\ a^*b & |b|^2\end{pmatrix}.$$

Regarding the rest of the problem: We can immediately tell that
$$[\rho]_{B'}=\begin{pmatrix}1 & 0\\ 0 & 0\end{pmatrix}.$$ So why don't we just define the rotation operator by $U|+\rangle=|\chi\rangle$ and $U|-\rangle=|\phi\rangle$, and then use this to determine ##_B## in terms of a and b?

A few more words about this approach, which I believe is the simplest one. The matrix equalities corresponding to $U|+\rangle=|\chi\rangle$ and $U|-\rangle=|\phi\rangle$ in the B basis are
\begin{align}&_B\begin{pmatrix}1\\ 0\end{pmatrix} =\begin{pmatrix}a \\ b\end{pmatrix}\\
&_B\begin{pmatrix}0\\ 1\end{pmatrix} =\text{ something that's orthogonal to }\begin{pmatrix}a \\ b\end{pmatrix}.
\end{align} This allows us to almost immediately write down ##_B$in terms of a,b. There's no need to figure out what the rotation axis is. We can use the formula$\big([\rho]_B\big)_{ij}=\langle e_i,\rho e_j\rangle$and the corresponding one for B' to prove that$[\rho]_{B'}=[U^*]_B[\rho]_B _B##. We already know that all of these matrices look like, so we won't learn anything new if we perform the matrix multiplication on the right, unless we've made mistakes along the way.

 

[Matterwave]

A few more words about this approach, which I believe is the simplest one. The matrix equalities corresponding to $U|+\rangle=|\chi\rangle$ and $U|-\rangle=|\phi\rangle$ in the B basis are
\begin{align}&_B\begin{pmatrix}1\\ 0\end{pmatrix} =\begin{pmatrix}a \\ b\end{pmatrix}\\
&_B\begin{pmatrix}0\\ 1\end{pmatrix} =\text{ something that's orthogonal to }\begin{pmatrix}a \\ b\end{pmatrix}.
\end{align} This allows us to almost immediately write down ##_B$in terms of a,b. There's no need to figure out what the rotation axis is. We can use the formula$\big([\rho]_B\big)_{ij}=\langle e_i,\rho e_j\rangle$and the corresponding one for B' to prove that$[\rho]_{B'}=[U^*]_B[\rho]_B _B##. We already know that all of these matrices look like, so we won't learn anything new if we perform the matrix multiplication on the right, unless we've made mistakes along the way.

But solving it this way just gives you an arbitrary unitary matrix, you get the same exact matrix as how any arbitrary special unitary matrix can be written...which I posted in post #33. I don't see how this actually gives anything new to the problem, so I assumed that the problem was asking how does this unitary matrix correspond to a rotation in 3-space. o.o

Indeed this seems like a weird problem to me since $\left|\chi\right>$ is arbitrary.

 

[Fredrik]

It's arbitrary only in the sense that $|\chi\rangle$ is arbitrary. The components are completely determined by $|\chi\rangle$. If we're asked to find the rotation operator that diagonalizes $\rho$, then this is the answer.

I think this is a plausible interpretation of the problem: Find a unitary U such that $U|+\rangle=|\chi\rangle$. Then find ##_B$and verify that$[U^*]_B[\rho]_B _B$is diagonal. That second part can be done by explicitly doing the matrix multiplication, or by proving the formula$[\rho]_{B'}=[U^*]_B[\rho]_B _B## for arbitrary bases B and B'.

 

[Matterwave]

It's arbitrary only in the sense that $|\chi\rangle$ is arbitrary. The components are completely determined by $|\chi\rangle$. If we're asked to find the rotation operator that diagonalizes $\rho$, then this is the answer.

I think this is a plausible interpretation of the problem: Find a unitary U such that $U|+\rangle=|\chi\rangle$. Then find ##_B$and verify that$[U^*]_B[\rho]_B _B$is diagonal. That second part can be done by explicitly doing the matrix multiplication, or by proving the formula$[\rho]_{B'}=[U^*]_B[\rho]_B _B## for arbitrary bases B and B'.

Well if that's the case, then isn't the answer just the matrix in post 33?

 

[Orodruin]

Maybe this information can help a little. Any arbitrary unitary matrix (in suggestive notation!) can be expressed as:

$$U=\begin{bmatrix} \alpha && -\bar{\beta} \\ \beta && \bar{\alpha} \end{bmatrix}$$

With the special unitary matrices satisfying the additional condition that the determinant is 1, or $|\alpha|^2+|\beta|^2=1$.

Notice that this matrix rotates $\left|z_+\right>$ into $\left|\chi\right>$ already... and the left column is orthogonal to the right column...

Speaking of which, yes this essentially the solution to the problem. Note that what you are giving is already a special unitary matrix since its determinant is real and positive. The condition $|\alpha|^2+|\beta|^2=1$ is the condition that it is unitary at all. For an arbitrary unitary matrix, you need to multiply by an overall phase. In addition, one other phase may also be removed.

 

[Matterwave]

Speaking of which, yes this essentially the solution to the problem. Note that what you are giving is already a special unitary matrix since its determinant is real and positive. The condition $|\alpha|^2+|\beta|^2=1$ is the condition that it is unitary at all. For an arbitrary unitary matrix, you need to multiply by an overall phase. In addition, one other phase may also be removed.

But if $\alpha,\beta$ are just 2 arbitrary complex numbers, it seems to me then that matrix should have 4 degrees of freedom, but SU(2) is of dimension 3. Where was 1 degree of freedom removed? I do see your point, but I'm not seeing where one degree of freedom was removed in writing the matrix in that form.

 

[Orodruin]

But if $\alpha,\beta$ are just 2 arbitrary complex numbers, it seems to me then that matrix should have 4 degrees of freedom, but SU(2) is of dimension 3. Where was 1 degree of freedom removed? I do see your point, but I'm not seeing where one degree of freedom was removed in writing the matrix in that form.

$\alpha$ and $\beta$ are not completely arbitrary if the matrix is to be unitary. For the matrix to be unitary at all, you have the condition on their square modulus, which reduces the degrees of freedom to 3. The point being that the 4-degrees-of-freedom matrix is not unitary without this condition. However, if the matrix is unitary, then it is also special unitary and you need to add the overall phase to get a general unitary matrix.

 

[Fredrik]

Well if that's the case, then isn't the answer just the matrix in post 33?

Yes, but you also need to mention that the $\alpha$ and $\beta$ in that matrix are the components of $|\chi\rangle$ in the B basis. $|\chi\rangle=\alpha|+\rangle+\beta|-\rangle$.

I used the notation $|\chi\rangle=a|+\rangle+b|-\rangle$, so my a,b are your $\alpha,\beta$. I didn't just write down the most general form of a unitary matrix. I used the following argument:

The matrix equalities corresponding to $U|+\rangle=|\chi\rangle$ and $U|-\rangle=|\phi\rangle$ in the B basis are
\begin{align}&_B\begin{pmatrix}1\\ 0\end{pmatrix} =\begin{pmatrix}a \\ b\end{pmatrix}\\
&_B\begin{pmatrix}0\\ 1\end{pmatrix} =\text{ something that's orthogonal to }\begin{pmatrix}a \\ b\end{pmatrix}.
\end{align}

This gave me the result
$$_B=\begin{pmatrix}a & b^*\\ b & -a^*\end{pmatrix}.$$ You put the minus sign on the upper right instead of the lower right. That works just as well. The quoted argument says only that we need to choose the second column of ##_B$so that it's orthogonal to the first. Since (b*,-a*) and (-b*,a*) are both orthogonal to (a,b), we can take either one as the second column of$_B$. I should also have mentioned that the second column must be normalized (because unitary operators take elements of an orthonormal set (like {(1,0),(0,1)}) to elements of an orthonormal set). We can use any$e^{ir}(b^*,-a^*)$with$r\in\mathbb R$as the second column. My choice corresponds to r=1, yours to r=$\pi##.

Edit: As I was writing my next post below, I realized that the determinant of ##_B$is determined by the "arbitrary" parameter r. We have$\det _B=e^{-ir}(-|a|^2-|b|^2)$. My choice of r makes this equal to -1. Your choice makes it equal to 1. So if we just want to diagonalize$\rho##, either choice is fine, but if we want to do it with a rotation operator (as the problem says), we have to go with your choice. (Because rotation operators are by definition elements of SU(2)). So I chose the wrong sign for the second column of my ##_B$. This is the correct$_B##:
$$_B=\begin{pmatrix}a & -b^*\\ b & a^*\end{pmatrix}.$$

But if $\alpha,\beta$ are just 2 arbitrary complex numbers, it seems to me then that matrix should have 4 degrees of freedom, but SU(2) is of dimension 3. Where was 1 degree of freedom removed? I do see your point, but I'm not seeing where one degree of freedom was removed in writing the matrix in that form.

They're not arbitrary. They're defined by $|\chi\rangle=\alpha|+\rangle+\beta|-\rangle$. (OK, so they're arbitrary in the sense that $|\chi\rangle$ is arbitrary, but they're completely determined by $|\chi\rangle$). 1 degree of freedom is removed by the requirement that $|\chi\rangle$ is normalized.

 

[Oxvillian]

Perhaps it's worth pointing out that that a general density operator for this situation takes the form
\rho = \frac{1}{2}({\bf 1} + {\bf a} \cdot {\bf \sigma})
where {\bf a} is a unit vector on the Bloch sphere.

In this language, the problem reduces to simply finding the SO(3) rotation matrix that takes the 3-vector {\bf a} into the z-axis.

 

[Oxvillian]

btw fluidistic - this online lecture should tell you everything you need to know about density operators, traces, Bloch sphere etc etc...

http://pirsa.org/10090019/

 

[Orodruin]

They're not arbitrary. They're defined by $|\chi\rangle=\alpha|+\rangle+\beta|-\rangle$. (OK, so they're arbitrary in the sense that $|\chi\rangle$ is arbitrary, but they're completely determined by $|\chi\rangle$). 1 degree of freedom is removed by the requirement that $|\chi\rangle$ is normalized.

In that particular part of this discussion, $\alpha$ and $\beta$ were not related to any state but used to parametrize a general special unitary matrix so the argument should not be based on the state $|\chi\rangle$. The question was why the matrix
$$
\begin{pmatrix} \alpha & - \bar \beta \\ \beta & \bar\alpha\end{pmatrix}
$$
with 4 degrees of freedom could only be a parametrization of the special unitary group (3 dof) and not a general unitary matrix (4 dof). The answer to this was given in my last post: Being unitary at all requires that the modulus of the determinant is one:
$$
\left| |\alpha|^2 + |\beta|^2\right| = |\alpha|^2 + |\beta|^2 = 1,
$$
which happens to be the same as the requirement of the special unitary group, i.e., if we call the set of matrices of the above form $A$, then $A \cap U(2) = SU(2)$.

 

[Fredrik]

In that particular part of this discussion, $\alpha$ and $\beta$ were not related to any state but used to parametrize a general special unitary matrix so the argument should not be based on the state $|\chi\rangle$.

He asked if what I had said meant that simply writing down the general form of a unitary matrix solves the problem. I'm assuming that this was meant to be an argument against my approach, since it clearly doesn't solve the problem. So I explained how and why my approach solves the problem.

The question was why the matrix
$$
\begin{pmatrix} \alpha & - \bar \beta \\ \beta & \bar\alpha\end{pmatrix}
$$
with 4 degrees of freedom could only be a parametrization of the special unitary group (3 dof) and not a general unitary matrix (4 dof). The answer to this was given in my last post: Being unitary at all requires that the modulus of the determinant is one:
$$
\left| |\alpha|^2 + |\beta|^2\right| = |\alpha|^2 + |\beta|^2 = 1,
$$
which happens to be the same as the requirement of the special unitary group, i.e., if we call the set of matrices of the above form $A$, then $A \cap U(2) = SU(2)$.

OK, I see that that part of my reply to Matterwave looks weird, since I quoted the same question and gave him a different answer. Your answer is appropriate in a discussion about Matterwave's approach. I gave him the answer that's appropriate in a discussion about my approach.

Thinking about what you said also made me see that there's a detail in my previous post that I need to elaborate on. I have edited in a comment about it now.