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

[fluidistic]

Homework Statement

Hello people,
I am trying to understand a problem statement as well as the density operator, but I still don't get it, desperation is making me posting here.
The problem comes as

We would like to describe N non interacting particles of spin one half.
Calculate the density matrix in a basis of eigenfunctions common to ##S_z## and ##S^2## for one spin 1/2 in a state ##|\chi>##, and use the rotation operator to transform that matrix into the basis in which it is diagonal.

The problem then continues with other questions but I'm having troubles with the very first one, for now.

Homework Equations

Diagonalization of the density matrix: ##\hat \rho _{\text{diagonal}} =U^* \rho U## where U is the rotation matrix, it's a unitary operator.

The Attempt at a Solution

First, I do not really know how to form the density matrix. The definition I have is "##\hat \rho = \sum _i p_i \left | \phi _i \right > \left < \phi _i \right |##" where ##p_i## is the probability to find the system in the state ##|\phi _i>##.
I see this as a sum of many operators whose matricial form I am not sure about. So I don't know how to get the matrix representation of the density operator from there.

Another approach is, ##\hat \rho = \begin{bmatrix} \left < \psi _1 \right | \hat \rho \left | \psi _1 \right > && \left < \psi _1 \right | \hat \rho \left | \psi _2 \right > \\ \left < \psi _2 \right | \hat \rho \left | \psi _1 \right > && \left < \psi _2 \right | \hat \rho \left | \psi _2 \right > \end{bmatrix}## but my problem is that I don't know how the density operator acts on psi_1 and psi_2, the eigenfunctions of ##S_z##.
I know that written in the basis that diagonalizes ##S_z## and ##S^2##, ##\psi_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix}## and ##\psi_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix}##. But still, this doesn't get me any further.

Another thoughts: since ## \left | \chi \right >## is arbitrary I can write it as a linear combinations of psi_1 and psi_2 such as ## \left | \chi \right > =\alpha \left | \psi _1 \right > + \beta \left | \psi _2 \right >## where alpha and beta are in principle complex valued.
Maybe I can also think of ## \left | \chi \right >## as a rotation of ## \left | \psi _1 \right >## by arbitrary angles theta and phi if I work in spherical coordinates.
The matrix that rotates from an angle phi and theta can be written under the form ##U=\cos \left ( \frac{\theta}{2} \right ) \hat 1 -i \sin \left ( \frac{\theta}{2} \right ) \hat u \cdot \vec \sigma## where ##\hat u## is the unit vector of the axis of rotation and the sigma is the vector of the Pauli matrices.

But overall I'm still stuck at even going into the right direction to solve the problem.

The answer (for the diagonal matrix) is, if I'm not mistaken, ##\hat \rho _{\text{diagonalized}}=\begin{bmatrix} 1 &&0 \\ 0&&0 \end{bmatrix}##, that is, if I remember well the solution I've seen.

Any tip, help and explanation is appreciated! Thanking you.

 

[Orodruin]

Do you have any additional information about the state ##|\chi\rangle##? Are you to assume that it is a general state? What is the density matrix if the particles are all in this state?

 

[fluidistic]

Do you have any additional information about the state ##|\chi\rangle##? Are you to assume that it is a general state? What is the density matrix if the particles are all in this state?

No, I do not have any more information about that state. As I understand it, it's a general state, a linear combination of ##|\uparrow >## and ##|\downarrow >##.
I don't know the answer to your question about the density matrix. I think the answer will depend upon the chosen basis.

 

[Matterwave]

Alright, I'm not sure where this problem is really going, but let's write ##\left|\chi\right>=\alpha\left|z_+\right>+\beta\left|z_- \right>## for complex ##\alpha,\beta## such that ##|\alpha|^2+|\beta|^2=1##, which is the most general pure state one can write down in a 2 state system.

How would I write down the density matrix in the ##\left|z_+ \right>,\left|z_- \right>## basis?

 

[Orodruin]

Start from the definition of the density matrix and the assumption that the particle is in the state ##|\chi\rangle## with probability 1. What is the density matrix then?

Since you need to express the state in some way, start by assuming that it is a general linear combination of up and down eigenstates.

 

[Fredrik]

The general rule to get the matrix of components of an operator T, with respect to a particular ordered basis ##(e_i)_{i=1}^n## is ##T_{ij}=(Te_j)_i##. The right-hand side is the ith component of the vector ##Te_j##. This is equal to ##\langle e_i,Te_j\rangle##, so in bra-ket notation, we have ##T_{ij}=\langle i|T|j\rangle##, where ##|j\rangle## is the ket notation for the basis vector ##e_j##. See the https://www.physicsforums.com/showthread.php?t=694922 if you need more information about this.

The problem asks for the density operator of a single particle in the pure state ##|\chi\rangle##. I don't know how to answer that in the form of a hint, so I'll just say it: ##\rho=|\chi\rangle\langle\chi|##.

That matrix near the end of your post #1 is the matrix of components of ##|\uparrow\rangle\langle\uparrow|## with respect to the ordered basis ##(|\uparrow\rangle,|\downarrow\rangle)##. (You should use the formula in the first paragraph of this post to verify this). So it's also the matrix of components of ##|\chi\rangle\langle\chi|## with respect to the ordered basis ##(|\chi\rangle,|\psi\rangle)##, where ##|\psi\rangle## is some unit vector in the orthogonal complement of ##\{|\chi\rangle\}##.

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.

A rotation around the z axis is probably not going to be sufficient.

 

[fluidistic]

Thank you very much guys so far.
Using Fredrik's notation, I do not know how to find the density matrix because it seems to be necessary to know how T acts on ##\left | j \right >##.
Or using the notation in my first post, how ##\hat \rho## acts on ##\left | \psi _i \right >##.

 

[Matterwave]

Thank you very much guys so far.
Using Fredrik's notation, I do not know how to find the density matrix because it seems to be necessary to know how T acts on ##\left | j \right >##.
Or using the notation in my first post, how ##\hat \rho## acts on ##\left | \psi _i \right >##.

Frederik gave you the answer in his post basically, but I can expand on that (in my notation since this is a 2 state problem, I think the ijk indices are overboard here). We have ##\rho=\left|\chi\right>\left<\chi\right|##. Given that ##\left|\chi\right>=\alpha\left|z_+\right>+\beta\left|z_-\right>## what do you think ##\rho## equals in terms of ##\left|z_+\right>,\left|z_-\right>##?

 

[Fredrik]

Thank you very much guys so far.
Using Fredrik's notation, I do not know how to find the density matrix because it seems to be necessary to know how T acts on ##\left | j \right >##.
Or using the notation in my first post, how ##\hat \rho## acts on ##\left | \psi _i \right >##.

The ##|\psi_i\rangle## are just ##|\uparrow\rangle## and ##|\downarrow\rangle##, and ##\rho## is just ##|\chi\rangle\langle\chi|## where ##|\chi\rangle## is a linear combination of those two vectors.

 

[fluidistic]

I'm very slow and sloppy with Dirac's notation (our professor never taught it to us).
Here is an attempt of calculating the first entry of the matrix: ##\left < \uparrow \right | \rho \left | \uparrow \right >##.
I do it by parts: ##\rho \left | \uparrow \right > = \left | \chi \right > \left < \chi | \uparrow \right > =\left | \chi \right > (\overline \alpha \left < \uparrow \right | +\overline \beta \left < \downarrow \right | ) \left | \uparrow \right >=\left | \chi \right > \overline \alpha \left < \uparrow | \uparrow \right > = \left | \chi \right > \overline \alpha##.
So that ##\left < \uparrow \right | \rho \left | \uparrow \right > =\left < \uparrow | \chi \right > \overline \alpha =\left < \uparrow \right | (\alpha \left | \uparrow \right > + \beta \left | \downarrow \right > ) \overline \alpha =\overline \alpha ^2 ##.
Is this correct so far?
Edit: Can't be correct I guess... the entry has to be real...

 

[Matterwave]

You can do it that way, but it seems quite convoluted. Your calculation is correct except for the fact that ##\alpha\bar{\alpha}=|\alpha|^2## so you did get a real number at the end. But not all entries in the density matrix have to be real anyways, it is a Hermitean matrix, so it is in fact a complex matrix. An easier way would be to perhaps just write ##\left|\chi\right>\left<\chi\right|=(\alpha\left|z_+\right>+\beta\left|z_-\right>)(\bar{\alpha}\left<z_+\right|+\bar{\beta}\left<z_-\right|)##.

 

[fluidistic]

You can do it that way, but it seems quite convoluted. Your calculation is correct except for the fact that ##\alpha\bar{\alpha}=|\alpha|^2## so you did get a real number at the end. But not all entries in the density matrix have to be real anyways, it is a Hermitean matrix, so it is in fact a complex matrix. An easier way would be to perhaps just write ##\left|\chi\right>\left<\chi\right|=(\alpha\left|z_+\right>+\beta\left|z_-\right>)(\bar{\alpha}\left<z_+\right|+\bar{\beta}\left<z_-\right|)##.

The problem is that I do not get any alpha. I only get the complex conjugate of it (twice). The diagonal entries must be real I think. The off-diagonal need not be real as you said.
And about the suggestion that's basically what I've used. But by parts.Nevermind, I do get the modulus of alpha squared. I got confused about the convention of the inner product. Thanks I will proceed with the other entries.

 

[Matterwave]

Why do you have a conjugate in the second part of your calculation? I don't see a conjugate there.

Also, my suggestion was to simply foil it out and then you can basically read off the entries of the matrix, without having to go through these calculations.

 

[Orodruin]

I suggest that you do it the way Matterwave proposed even if you want to get it in matrix representation rather than in braket notation. Multiplying the parentheses will allow you to simply read off the matrix elements. The next step is to relate the absolute values of alpha and beta upon which you will get a relation they must fulfill. I am sure you know how to parametrize this relation ...

 

[fluidistic]

I just calculated the matrix rho the hard way: ##[\hat \rho ] =\begin{bmatrix} |\alpha | ^2 && \alpha \overline \beta \\ \overline \alpha \beta && |b| ^2 \end{bmatrix}##.
Orodruin, are you suggesting me to replace ##|\beta |^2## by ##1- |\alpha | ^2##?

Edit: I've just tried your way MatterWave. Indeed much better! I get rho as a sum of operators and I can see that the coefficients are the entries of the matrix.

 

[Matterwave]

You could make that substitution if you wanted, but I don't see how it will really help you. The next part of the problem is to rotate that matrix so it becomes diagonal. To do that though, one should use a little ingenuity instead of brute force. In fact it is much easier to figure out the result of this rotation than the rotation matrix itself.

Frederick already gave you the answer in his post earlier to this question, but, in which basis will the density matrix be diagonal? And what will the density matrix be in such a basis? (Hint: a pure state has what kind of a diagonal density matrix?)

 

[fluidistic]

You could make that substitution if you wanted, but I don't see how it will really help you. The next part of the problem is to rotate that matrix so it becomes diagonal. To do that though, one should use a little ingenuity instead of brute force. In fact it is much easier to figure out the result of this rotation than the rotation matrix itself.

Frederick already gave you the answer in his post earlier to this question, but, in which basis will the density matrix be diagonal? And what will the density matrix be in such a basis? (Hint: a pure state has what kind of a diagonal density matrix?)

A pure state has a 1 somewhere in the diagonal, all the other entries are 0.

 

[Matterwave]

Yeah, so you already know the answer. Do you still need to find the rotation matrix? The problem did not specify what exactly you were looking for.

 

[fluidistic]

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.

 

[Orodruin]

I just calculated the matrix rho the hard way: ##[\hat \rho ] =\begin{bmatrix} |\alpha | ^2 && \alpha \overline \beta \\ \overline \alpha \beta && |b| ^2 \end{bmatrix}##.
Orodruin, are you suggesting me to replace ##|\beta |^2## by ##1- |\alpha | ^2##?

Exactly, do you know of two functions of one single parameter that fulfill this relation? To completely parametrize the density matrix you will need this single parameter and the phase of ##\alpha\bar\beta##.

After that, think about how you can remove the phase by multiplying by a unitary matrix on each side. Once that is done you are almost done.

 

[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 ##\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.

How are your ##\theta,\phi## related to ##\alpha,\beta##? And how did you come up with this matrix?

 

[fluidistic]

Exactly, do you know of two functions of one single parameter that fulfill this relation? To completely parametrize the density matrix you will need this single parameter and the phase of ##\alpha\bar\beta##.

After that, think about how you can remove the phase by multiplying by a unitary matrix on each side. Once that is done you are almost done.

Hmm I'll have to think about it, I'm not sure as of now.

How are your ##\theta,\phi## related to ##\alpha,\beta##? And how did you come up with this matrix?

I considered ##\hat n## as an arbitrary unit vector in spherical coordinates. Namely ##(\cos \phi \sin \theta, \sin \theta \sin \phi , \cos \theta )##. I considered ##\vec S=\frac{\hbar}{2} \vec \sigma## where ##\vec \sigma =(\sigma _x, \sigma _y , \sigma _z)##. Then did the dot product and replaced the sigma's by the Pauli matrices.
P.S.:I don't know the relation between theta and phi with alpha and beta.

 

[Orodruin]

Hmm I'll have to think about it, I'm not sure as of now.

Hint:
Think about a unit circle.

 

[fluidistic]

I've reread Fredrik's post. Apparently ##\left | \chi \right >## is an eigenstate of ##\hat n \cdot \vec S##. Which means that ##\hat n \cdot \vec S \left | \chi \right > = \pm \frac{\hbar}{2}\left | \chi \right > ##. By writting chi has a linear combination of "up" and "down" states I think I could get the relation between alpha, beta and theta, phi.

 

[Matterwave]

Well, once you relate ##\alpha,\beta## to ##\theta,\phi## then you're done (there will be 1 phase that is left arbitrary however, as an overall phase that could be multiplied to the density matrix).

Alternatively, you can try to think about this physically, and Oro's suggestion will make the calculations a lot easier.

Here's a hint on how to implement Oro's suggestion. Every complex number can be described as a (real) magnitude and a phase: ##\alpha = |\alpha|e^{i\phi},~\beta=|\beta|e^{i\delta}##. Which trig identity does the restriction ##|\alpha|^2+|\beta|^2=1## look like?

 

[fluidistic]

Well, once you relate ##\alpha,\beta## to ##\theta,\phi## then you're done (there will be 1 phase that is left arbitrary however, as an overall phase that could be multiplied to the density matrix).

Alternatively, you can try to think about this physically, and Oro's suggestion will make the calculations a lot easier.

Here's a hint on how to implement Oro's suggestion. Every complex number can be described as a (real) magnitude and a phase: ##\alpha = |\alpha|e^{i\phi},~\beta=|\beta|e^{i\delta}##. Which trig identity does the restriction ##|\alpha|^2+|\beta|^2=1## look like?

I'll do both ways. The last thing of your post looks like a unit circle.

I reach some non sense via the road ##\hat n \cdot \vec S \left | \chi \right > = \pm \frac{\hbar}{2}\left | \chi \right >##. A relation impossible to satisfy.

Edit: I know why, the eingenvalues of S_z are not ##\pm \hbar /2##. The 1/2 shouldn't be there.

 

[Orodruin]

I'll do both ways. The last thing of your post looks like a unit circle.

And how do we normally parametrize the unit circle?

The final U will be arbitrary up to two phases. If you have found one U you can always multiply it by a diagonal matrix containing phases only (which means two phases since it is a 2x2 matrix) from the right and the relation will still be fulfilled.

 

[fluidistic]

And how do we normally parametrize the unit circle?

The final U will be arbitrary up to two phases. If you have found one U you can always multiply it by a diagonal matrix containing phases only (which means two phases since it is a 2x2 matrix) from the right and the relation will still be fulfilled.

I'd say ##(\cos t , \sin t)##.

I got 2 equations using ##\hat n \cdot \vec S \left | \chi \right > = \hbar \left | \chi \right >##. I reached that ##\alpha (\cos (\theta ) -1 ) + \beta \sin (\theta ) e^{-i\phi } =0## and ##\alpha \sin ( \theta ) e^{i \phi} - \beta (1+ \cos \theta ) =0##.

Edit: I can replace 1- cos theta and 1+ cos theta by sine squared and cosine squared of theta over two. But I'm getting lost.

 

[Matterwave]

Why is the relation impossible to satisfy? ##\hat{n}\cdot\vec{S}## is a trace-less hermitian matrix, and definitely has eigenvalues and eigenvectors. But it seems to me that this is the harder method, since after getting ##\hat{n}\cdot\vec{S}## you still have to get the rotation matrix ##U##. I'm not sure I can at the moment figure out how to do it this way. I would definitely think of it in a more physical way.

What's the equation for the unit circle? How do you parametrize the unit circle?

Frederick told you that the basis in which ##\rho## is diagonal is the basis ##(\chi,\psi)## where we must have ##\left<\psi|\chi\right>=0## (the orthogonal compliment). So what we need is a rotation that rotates ##\left|z_+\right>## to ##\left|\chi\right>## and ##\left|z_-\right>## to ##\left|\psi\right>##. What does ##\left|\psi\right>## look like in the ##\left|z_+\right>,\left|z_-\right>## basis? When you have the parameterization down, ##\left|\psi\right>## will look a lot better.

Remember though that ##\left|\psi\right>## will have an arbitrary phase in it that is not important.

EDIT: I myself am getting a little confused on all the phases floating around lol. I think Oro is right in his count of 2 arbitrary phases total left in the rotation matrix.

 

[Orodruin]

EDIT: I myself am getting a little confused on all the phases floating around lol. I think Oro is right in his count of 2 arbitrary phases total left in the rotation matrix.

I am a neutrino phenomenologist, counting physical phases of unitary matrices is part of my job ...

Since the density matrix contains only one phase, it would be strange if you needed more than that to diagonalize it.

 

[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.