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

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  • #51
Fredrik
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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'.
 
  • #52
Matterwave
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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?
 
  • #53
Orodruin
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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.
 
  • #54
Matterwave
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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.
 
  • #55
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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.
 
  • #56
Fredrik
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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.
 
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  • #57
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Perhaps it's worth pointing out that that a general density operator for this situation takes the form
[tex]\rho = \frac{1}{2}({\bf 1} + {\bf a} \cdot {\bf \sigma})[/tex]
where [itex]{\bf a}[/itex] 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 [itex]{\bf a}[/itex] into the z-axis.
 
  • #58
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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/
 
  • #59
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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)##.
 
  • #60
Fredrik
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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.
 
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  • #61
Matterwave
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##\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.

I see what you mean. So to clarify for my own sake, the matrix I wrote down can ONLY be used to specify a SU(2) matrix, and not a U(2) one because it is either SU(2) (given ##|\alpha|^2+|\beta|^2=1##) or not unitary at all (if the previous condition is violated), am I right? A general U(2) matrix has the form of the matrix I wrote down, along with the requirement ##|\alpha|^2+|\beta|^2=1##, AND multiplied by an overall phase ##e^{i\varphi}## correct?
 
  • #62
Matterwave
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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.

I was not really arguing "against" your approach, so to speak. In fact, your approach was the way I was trying to lead fluid in the first place (and also why I wrote in my post #33 such a matrix in the first place, and remarked that the first column was basically ##\left|\chi\right>## and that it was orthogonal to the second column). I was hoping that fluid would realize that such a matrix solves his problem (I could not figure out a way to guide him to that matrix...).

But what I did mean was that such an approach might not be enough for what the question was asking for, since it asked for a "rotation" (otherwise, I would just be giving fluid the answer to his homework problem).

Of course, a "rotation" on 2-component spinor space IS just some matrix in SU(2). But, I thought the question also wanted you to correspond such a rotation with a rotation in real space. I.e. I assumed the question also wanted you to make a correspondence with a matrix (find a matrix) in SO(3) as well.

So I would do everything with your approach, and then construct also the SU(2) matrix ##U=\cos(\eta/2)\hat{I}-i\sin(\eta/2)\hat{u}\cdot\vec{\sigma}## and make a correspondence between the two matrices.
 
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  • #63
Fredrik
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Of course, a "rotation" on 2-component spinor space IS just some matrix in SU(2).
Right, so maybe the word "rotation" in the problem statement is just telling us to choose that ##r## in post #56 so that the matrix gets the determinant 1.

But, I thought the question also wanted you to correspond such a rotation with a rotation in real space. I.e. I assumed the question also wanted you to make a correspondence with a matrix (find a matrix) in SO(3) as well.

So I would do everything with your approach, and then construct also the SU(2) matrix ##U=\cos(\eta/2)\hat{I}-i\sin(\eta/2)\hat{u}\cdot\vec{\sigma}## and make a correspondence between the two matrices.
Have you found a way to do that (other than the complicated method that I described but didn't carry out)? I haven't, but I haven't really tried yet.
 
  • #64
Matterwave
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Right, so maybe the word "rotation" in the problem statement is just telling us to choose that ##r## in post #56 so that the matrix gets the determinant 1.


Have you found a way to do that (other than the complicated method that I described but didn't carry out)? I haven't, but I haven't really tried yet.

I mean at the end of that calculation you have a matrix exactly of the form in post #33, so you can make an easy identification with ##\alpha,\beta##. Frist we want to construct ##U=\cos(\eta/2)\hat{I}-i\sin(\eta/2)\hat{u}\cdot\vec{\sigma}##. We can use spherical coordinates for ##\hat{u}## so that it has components ##\hat{u}=(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta)## where ##\theta## is the azimuthal angle, and ##\phi## is the polar angle.

Thus, assuming I did all the calculations right, the final matrix you get at the end is:

$$U=\begin{pmatrix} \cos(\eta/2)-i\cos(\theta)sin(\eta/2) & -i\sin(\eta/2)\sin(\theta)e^{-i\phi} \\ -i\sin(\eta/2)\sin(\theta)e^{i\phi} & \cos(\eta/2)+i\cos(\theta)\sin(\eta/2)\end{pmatrix}$$

Therefore, we simply make the identifications:

$$\alpha=\cos(\eta/2)-i\cos(\theta)sin(\eta/2),~~\beta= -i\sin(\eta/2)\sin(\theta)e^{-i\phi}$$

So that now we know how a real space rotation given by a rotation of angle ##\eta## around an axis of rotation given by ##\theta,\phi## corresponds to a rotation in spin space from an eigenstate ##\left|z_+\right>## to ##\left|\chi\right>##, some arbitrary spin state.
 

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