Why two s=1/2 states can be linked by a rotation, but not two s=1 states?

[wdlang]

given two arbitrary states of a spin of 1/2

we can always find a rotation to link these two states

however, given two arbitrary states of a spin of 1

this is not so

for example, (0,1,0) and (1,0,0) can not be linked by a rotation

the former has vanishing expectations of sx, sy, sz

so in rotation, the expectations values are always vanishing

<sx>^2+<sy>^2+<sz>^2=0

however, the latter has <sx>=<sy>=0, <sz>=1;

in rotation, <sx>^2+<sy>^2+<sz>^2=1

this indicates the two can not be linked by a rotation

 

[kof9595995]

Of course these two can be related by a rotation operator. Your reasoning is fallacious,<sx>^2+<sy>^2+<sz>^2 has no meaning, or at least it's not a norm of the vector space. However <sx^2>+<sy^2>+<sz^2>=1 always, and it represents <S^2>.

 

[wdlang]

Of course these two can be related by a rotation operator. Your reasoning is fallacious,<sx>^2+<sy>^2+<sz>^2 has no meaning, or at least it's not a norm of the vector space. However <sx^2>+<sy^2>+<sz^2>=1 always, and it represents <S^2>.

no

<sx>^2+<sy>^2+<sz>^2 is a conserved quantity

(<sx>,<sy>,<sz>) transforms in the same way as (sx,sy,sz), and actually as the coordinates of a vector in 3d.

 

[tom.stoer]

We start with the two vectors $$\vec{e}_x$$ and $$\vec{e}_y$$. They can be transformed into each other via

$$\vec{e}_y = D_{x \to y} \; \vec{e}_x$$

with

$$D_{x \to y} = \left(\begin{array}{ccc}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{array}\right)$$

What's wrong with this transformation applied to spin-1 states?

 

[jostpuur]

We start with the two vectors $$\vec{e}_x$$ and $$\vec{e}_y$$. They can be transformed into each other via

$$\vec{e}_y = D_{x \to y} \; \vec{e}_x$$

with

$$D_{x \to y} = \left(\begin{array}{ccc}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{array}\right)$$

What's wrong with this transformation applied to spin-1 states?

What's wrong is that it's wrong. You cannot transform spinors with ordinary SO(3)-matrices, not even in the spin-1 case where the spinor coincidentally has three components.

 

[tom.stoer]

For spin 1 the "spinor" becomes a vector and the spin 1 representation of SU(2) is identical with the SO(3) vector representation.

 

[Dickfore]

The $\mathbf{J}^{2}$ operator commutes with all the components of $J_{i}, \ (i = x, y, z)$, which, in turn, means that it commutes with any operator $D(R)$ corresponding to an arbitrary rotation $R$. That means that:

$$\mathbf{J}^{2} (D(R) \, |j, m>) = D(R) \, (\mathbf{J}^{2} |j, m>) = j(j + 1) \, (D(R) \, |j, m>)$$

i.e. $D(R) \, |j, m>$ is a state corresponding to the same eigenvalue $\mathbf{J}^{2} = j(j + 1)$. However, $J_{z}$ does not necessarily have a definite eigenvalue in the state $D(R) \, |j, m>$.

 

[tom.stoer]

... $J_{z}$ does not necessarily have a definite eigenvalue in the state $D(R) \, |j, m>$.

Your formulas are completely right, but that's not what was asked for. Of course if you rotate a state then your new state is no longer an eigenstate for $J_{z}$ but for some rotated $J_{z^\prime}$.

The original question was whether two arbitrary states of spin 1 can be linked by a rotation. First of all the D matrices (Wigner functions) for j=1 are a faithful representation of SO(3) which is nothing else but the rotation group. So given an arbitrary rotation R specified by an axis and an angle one can always construct a qm operator D[R] as follows

$$R[\vec{n}, \phi] \to D[R] = e^{-i\vec{J}\vec{n}\phi}$$

The matrix elements

$$D^{(j)}_{m^\prime m}[R] = <j,m^\prime | D[R] | j, m>$$

obey the SO(3) group multiplication

$$\sum_{m^\prime} D^{(j)}_{m^{\prime\prime} m^\prime}[R_1]\,D^{(j)}_{m^\prime m}[R_2] = D^{(j)}_{m^{\prime\prime} m}[R_1 R_2]$$

The rotation of an arbitrary state vector is given by

$$D[R] | j, m> = \sum_{m^\prime} |j,m^\prime> D^{(j)}_{m^\prime m}[R]$$

 

[Dickfore]

Your formulas are completely right, but that's not what was asked for. Of course if you rotate a state then your new state is no longer an eigenstate for $J_{z}$ but for some rotated $J_{z^\prime}$.

The original question was whether two arbitrary states of spin 1 can be linked by a rotation. First of all the D matrices (Wigner functions) for j=1 are a faithful representation of SO(3) which is nothing else but the rotation group. So given an arbitrary rotation R specified by an axis and an angle one can always construct a qm operator D[R] as follows

$$R[\vec{n}, \phi] \to D[R] = e^{-i\vec{J}\vec{n}\phi}$$

The matrix elements

$$D^{(j)}_{m^\prime m}[R] = <j,m^\prime | D[R] | j, m>$$

obey the SO(3) group multiplication

$$\sum_{m^\prime} D^{(j)}_{m^{\prime\prime} m^\prime}[R_1]\,D^{(j)}_{m^\prime m}[R_2] = D^{(j)}_{m^{\prime\prime} m}[R_1 R_2]$$

The rotation of an arbitrary state vector is given by

$$D^{(j)}_{m^\prime m}[R] | j, m> = \sum_{m^\prime} |j,m^\prime> D^{(j)}_{m^\prime m}[R]$$

The way I understood the question was why a j = 1/2 state can always be connected to any other j = 1/2 state by a rotation, but not to a j = 1 state.

 

[tom.stoer]

I think this was not the question.

But the answer is simple: a rotation does not change the value of j (= a rotation always stays within a given j-representation; it does not connect two different j-representations).

 

[jostpuur]

It's long time since I thought about spinor stuff last time, so I thought I would first check some equations from books before engaging more seriously. Perhaps I'll return after few days again.

But even before recalling any details from books, I might quickly point out two things.

1: The matrix in tom.stoer's wasn't in SO(3), but only in O(3). Is this relevant for some argument? I'm not sure. Anyway, noticing the sign of the determinant may help avoiding some confusion.

2:

The original question was whether two arbitrary states of spin 1 can be linked by a rotation. First of all the D matrices (Wigner functions) for j=1 are a faithful representation of SO(3) which is nothing else but the rotation group.

So we are dealing with some mapping $\textrm{SO}(3)\to \textrm{U}(3)$. The relevant question is that what's the image of this mapping? You pointed out that it's injective (faithful), but it probably is not surjective.

 

[tom.stoer]

1: The matrix in tom.stoer's wasn't in SO(3), but only in O(3). Is this relevant for some argument? I'm not sure. Anyway, noticing the sign of the determinant may help avoiding some confusion.

:-) The sign is simply missing! It should read

$$D_{x \to y} = \left(\begin{array}{ccc}0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{array}\right)$$

2: So we are dealing with some mapping ...

The mapping should read $\textrm{SO}(3)\to \textrm{SU}(2)$. This is one-to-two, the SU(2) is the double cover of SO(3). That's related to the fact that SU(2) has half-integer representations, whereas SO(3) hasn't.

You pointed out that it's injective (faithful), but it probably is not surjective.

You are talking about the mapping $\textrm{SO}(3)\to \textrm{SU}(2)$ whereas I am talking about $R \to D[R]$. I am constructing a qm operator D for each SO(3) rotation matrix R; this mapping is one-to-one in the j=1 representation.

Again this is not relevant as there are no two states which cannot be rotated into each other.

 

[jostpuur]

The mapping should read $\textrm{SO}(3)\to \textrm{SU}(2)$. This is one-to-two, the SU(2) is the double cover of SO(3).

You are confusing spin-1/2 and spin-1 representations. We need 3x3-matrices to transform spin-1 states.

 

[Dickfore]

The reason behind it is because a that a general $N \times N$ special unitary matrix is determined by $N^{2} - 1$ parameters. When this number gets bigger than 3 (the number of Euler angles uniquely determining a rotation in 3D), i.e., when:

$$N^{2} - 1 > 3 \Leftrightarrow N > 2$$

we cannot find a mapping from the set of rotation matrices to the set of $SU(N)$ matrices that has an image for each $SU(N)$ matrix. In the j-representation, the matrices have dimensionality $N = 2 j + 1$. This means that whenever:

$$2j + 1 > 2 \Rightarrow j > 1/2$$

we cannot find this kind of mapping. We conclude that the mapping is only possible in the case $j = 1/2$.

 

[tom.stoer]

:-)
You are talking about the mapping $\textrm{SO}(3)\to \textrm{SU}(2)$ whereas I am talking about $R \to D[R]$. I am constructing a qm operator D for each SO(3) rotation matrix R; this mapping is one-to-one in the j=1 representation.

You are confusing spin-1/2 and spin-1 representations. We need 3x3-matrices to transform spin-1 states.

Hey jostpuur,

you should read more carefully what I am writing.

You write something regarding a mapping between SO(3) and U(3). I corrected that because U(3) has nothing to do with spins. I said it must read SU(2).

Then I said that I construct a qm operator. R is clearly always a 3*3 rotation matrix, but D[R] is a qm operator valid for all j (!), both integer and half integer. It's matrix elements (the D-matrices) are again 3*3 matrices for j=1. For higher j you get larger matrices. So I know perfectly well that we need 3*3 matrices for j=1; and I gave one specific example a couple of posts before.

What Dickfore is writing is correct, but again it's not relevant for our discussion here, because we are concentrating on j=1 for which we can restrict to SO(3) - even if SU(2) is still valid.

I can only repeat what I wrote before: you can rotate between arbitrary states |j,m> with fixed j within one j representation using the the D[R] operator.

 

[Dickfore]

aaaaaaaaa

Nice post.

 

[tom.stoer]

corrected :-)

 

[Dickfore]

A general rotation matrix in j = 1 representation is given by:

$$D^{(1)}_{m m'}(\alpha, \beta, \gamma) = \left[\begin{array}{ccc} \left(\frac{1}{2}\right) \, (1 + \cos \beta) \, e^{-i(\alpha + \gamma)} & -\left( \frac{1}{\sqrt{2}}\right) \, \sin \beta \, e^{-i \alpha} & \left(\frac{1}{2}\right) \, (1 - \cos \beta) \, e^{-i(\alpha - \gamma)} \\ \left( \frac{1}{\sqrt{2}}\right) \, \sin \beta \, e^{-i \gamma} & \cos \beta & -\left( \frac{1}{\sqrt{2}}\right) \, \sin \beta \, e^{i \gamma} \\ \left(\frac{1}{2}\right) \, (1 - \cos \beta) \, e^{i(\alpha - \gamma)} & -\left( \frac{1}{\sqrt{2}}\right) \, \sin \beta \, e^{i \alpha} & \left(\frac{1}{2}\right) \, (1 + \cos \beta) \, e^{i(\alpha + \gamma)}\right] \end{array}\right]$$

The OP ask for such a matrix that will bring (0, 1, 0) to (1, 0, 0). But, acting with $D^{(1)}$ from above on (0. 1, 0) gives the second column:

$$\left[\begin{array}{c} \left( \frac{1}{\sqrt{2}}\right) \, \sin \beta \, e^{-i \alpha} \\ \cos \beta \\ -\left( \frac{1}{\sqrt{2}}\right) \, \sin \beta \, e^{i \alpha} \end{array}\right]$$

If we want this to be equal to (1, 0, 0), the second and third term have to be equal to 0 simultaneously, i.e. we have to have:

$$\sin \beta = \cos \beta = 0$$

This is impossible, so we see that (0, 1, 0) cannot be transformed to (1, 0, 0) with any rotation.

 

[tom.stoer]

What do you mean by (1,0,0)? shall this be a spin-1 state? or an ordinary vector in R³?

In spin 1 space you are acting on spin states |j=1,m> using D[R]. So to which |j,m> does your (1,0,0) correspond?

 

[Dickfore]

$$\begin{array}{rcl} (1, 0, 0)^{\top} & \equiv & |j = 1, m = 1 \rangle \\ (0, 1, 0)^{\top} & \equiv & |j = 1, m = 0 \rangle \\ (0, 0, 1)^{\top} & \equiv & |j = 1, m = -1 \rangle \end{array}$$

are the standard notations for the basis column vectors in the $|j, m \rangle$ representation.

 

[jostpuur]

So we are dealing with some mapping $\textrm{SO}(3)\to \textrm{U}(3)$. The relevant question is that what's the image of this mapping?

The mapping I'm talking about exists. I'll denote it with $\phi:\textrm{SO}(3)\to\textrm{U}(3)$ now. It's a kind of mapping that when you're given a rotation $A\in\textrm{SO}(3)$, then $\phi(A)\in\textrm{U}(3)$ will tell how to act on the spin-1 state. I don't have a formula for this mapping now, but if Dickfore got his details correct, then the image $\textrm{Im}(\phi)\subsetneq\textrm{U}(3)$ consists of these matrices:

A general rotation matrix in j = 1 representation is given by:

$$D^{(1)}_{m m'}(\alpha, \beta, \gamma) = \left[\begin{array}{ccc} \left(\frac{1}{2}\right) \, (1 + \cos \beta) \, e^{-i(\alpha + \gamma)} & -\left( \frac{1}{\sqrt{2}}\right) \, \sin \beta \, e^{-i \alpha} & \left(\frac{1}{2}\right) \, (1 - \cos \beta) \, e^{-i(\alpha - \gamma)} \\ \left( \frac{1}{\sqrt{2}}\right) \, \sin \beta \, e^{-i \gamma} & \cos \beta & -\left( \frac{1}{\sqrt{2}}\right) \, \sin \beta \, e^{i \gamma} \\ \left(\frac{1}{2}\right) \, (1 - \cos \beta) \, e^{i(\alpha - \gamma)} & -\left( \frac{1}{\sqrt{2}}\right) \, \sin \beta \, e^{i \alpha} & \left(\frac{1}{2}\right) \, (1 + \cos \beta) \, e^{i(\alpha + \gamma)}\right] \end{array}\right]$$

The original problem is related to the fact, mentioned by me, that $\phi$ is not surjective. The image is so small, that (1,0,0) and (0,1,0) cannot be connected with a transform in $\textrm{Im}(\phi)$.

 

[tom.stoer]

I don't doubt that the mapping exists, but I think U(3) is not relevant here.

Anyway - in the meantime I am no longer sure that my statement that there are no two states which cannot be rotated into each other is true!

Perhaps I overlook something but I tried a construction which after some thinking results in using Dickfore's representation of the D-matrix in post #18 https://www.physicsforums.com/showpost.php?p=2787710&postcount=18

What I did was the following:

I start with an arbitrary normalized state

$$|j,\xi> = \sum_{m^{\prime\prime}}\xi_{m^{\prime\prime}} |j,{m^{\prime\prime}}>$$

with

$$\sum_{m^{\prime\prime}}|\xi_{m^{\prime\prime}}|^2 = 1$$

My thesis is that for arbitrary m and $$\xi_{m^{\prime\prime}}$$ one can always find a rotation $$R(\alpha, \beta, \gamma)$$ such that the corresponding operator $$D[R]$$ fulfills

$$|j,\xi> = D[R]|j,m>$$

Projecting on $$<j,m^\prime|$$ one finds

$$\xi_{m^{\prime}} = D_{m^\prime m}^{(j=1)}(\alpha, \beta, \gamma)$$

In order to show that this is not always possible I use the state $$|j,m> = |1,0>$$. That means I have to construct the matrix element $$D_{m^\prime 0}^{(1)}(\alpha, \beta, \gamma)$$ for some angles $$\alpha, \beta, \gamma$$.

Looking at $$m^{\prime} = 0$$ and using a representation (e.g. from Wikipedia http://en.wikipedia.org/wiki/Wigner_D-matrix) of $$D_{m^\prime 0}^{(1)}$$ one can easily construct a contradiction. For $$D_{00}^{(1)}$$ one gets

$$D_{00}^{(1)}(\alpha, \beta, \gamma) = d_{00}(\beta) = \cos\beta$$

That means that if the imaginary part $$Im\,\xi_0 \neq 0$$ the construction fails!

This comes really as a surprise because it means that all vectors in R³ can be rotated into each other using rotation matrices with appropriate Euler angles, but that this does not carry over to spin in general. Is there some deep reason behind this consclusion? Or are there any flaws in my argument?

 

[jostpuur]

I don't doubt that the mapping exists, but I think U(3) is not relevant here.

I was considering about talking about a mapping $\textrm{SO}(3)\to\mathbb{C}^{3\times 3}$ too, but then I thought that my claim "the mapping is not surjective" would have been too trivial, so I shrank the codomain a little bit to emphasise the smallness of the image.

 

[DrDu]

Tom, there exists a theorem due to Wigner which states that by selecting functions of the appropriate phase one can allways achieve that the representations of a group are either unitary or anti-unitary. That means that the multiplication of a wavefunction with some phase factor is not included in the rotation group representation as it would make the representation non-unitary.

 

[tom.stoer]

OK, let me repeat in own words to see if I understand: Using an arbitrary state means that I include an overall phase factor which prevents me from constructing an appropriate D matrix. Is this correct?

 

[tom.stoer]

OK, let me repeat in own words to see if I understand: Using an arbitrary state means that I include an overall phase factor which prevents me from constructing an appropriate D-matrix. Is this correct?

Of course I can eliminate an U(1) phase factor; this is rather trivial, nevertheless I messed it up :-)

In order to fix this it is sufficient to use a slightly more general mapping

$$\xi_{m^{\prime}} = e^{i\phi_\xi} D_{m^\prime m}^{(j=1)}(\alpha, \beta, \gamma)$$

I will continue to investigate this further.

What do you think? Does the extra phase factor allow for a construction of the D-matrix for arbitrary state defined by $$\xi_{m^{\prime}}$$ for any $$|j,m>$$?

 

[kof9595995]

A general rotation matrix in j = 1 representation is given by:

$$D^{(1)}_{m m'}(\alpha, \beta, \gamma) = \left[\begin{array}{ccc} \left(\frac{1}{2}\right) \, (1 + \cos \beta) \, e^{-i(\alpha + \gamma)} & -\left( \frac{1}{\sqrt{2}}\right) \, \sin \beta \, e^{-i \alpha} & \left(\frac{1}{2}\right) \, (1 - \cos \beta) \, e^{-i(\alpha - \gamma)} \\ \left( \frac{1}{\sqrt{2}}\right) \, \sin \beta \, e^{-i \gamma} & \cos \beta & -\left( \frac{1}{\sqrt{2}}\right) \, \sin \beta \, e^{i \gamma} \\ \left(\frac{1}{2}\right) \, (1 - \cos \beta) \, e^{i(\alpha - \gamma)} & -\left( \frac{1}{\sqrt{2}}\right) \, \sin \beta \, e^{i \alpha} & \left(\frac{1}{2}\right) \, (1 + \cos \beta) \, e^{i(\alpha + \gamma)}\right] \end{array}\right]$$

The OP ask for such a matrix that will bring (0, 1, 0) to (1, 0, 0). But, acting with $D^{(1)}$ from above on (0. 1, 0) gives the second column:

$$\left[\begin{array}{c} \left( \frac{1}{\sqrt{2}}\right) \, \sin \beta \, e^{-i \alpha} \\ \cos \beta \\ -\left( \frac{1}{\sqrt{2}}\right) \, \sin \beta \, e^{i \alpha} \end{array}\right]$$

If we want this to be equal to (1, 0, 0), the second and third term have to be equal to 0 simultaneously, i.e. we have to have:

$$\sin \beta = \cos \beta = 0$$

This is impossible, so we see that (0, 1, 0) cannot be transformed to (1, 0, 0) with any rotation.

I got the same result by the explicit calculation, sorry about my previous unthoughtful post, although still not convinced by OP's reasoning :)

 

[tom.stoer]

@Dickfore & kof9595995:

I still do not see what your (1,0,0) vector is.

But what I see is that your negative result aplies directly to my construction. If I chose my vector to be

$$(\xi_{-1}, \xi_0, \xi_{+1}) = (\xi_{-1}, \xi_0, 0)$$ this cannot be solved. By setting $$\xi_{+1} = 0$$ it follows that $$\sin\beta = 0$$ which immediately results in $$\xi_{-1} = 0$$, too.

Therefore even with my relaxed construction introducting a free phase the mapping is not possible. That means that all vectors in R³ can be rotated into each other using rotation matrices with appropriate Euler angles, but that this does not carry over to spin in general.

That comes as a surprise as it means that there exists properties of vectors in R³ which do not carry over to the j=1 vector representation of SO(3). How can it be that this representation is faithful on the level of R => D[R] but not on the level of the vector space the group acts on?

 

[kof9595995]

$$\[\left[ \begin{array}{l} 1 \\ 0 \\ 0 \\ \end{array} \right]\]$$ is just the m=1 state, and $$\[\left[ \begin{array}{l} 0 \\ 1 \\ 0 \\ \end{array} \right]\]$$is the m=0 state. And I'm also quite surprised after the calculation :)

 

[kof9595995]

I find there's an intuition here, if you draw the vetors S and Sz by "line and arrow" on the paper, for |j,m>=|1,1> you get a cone, while for |j,m>=|1,0> you get a flat disc, and we know a rotation can never change the shape of a figure, so |1,1> can not be rotated to |1,0>. If this intuition holds for j=2, I suppose |2,2> cannot be rotated to |2,1> or |2,0> either.

 

[tom.stoer]

I agree ...

.. but my intuition told me that "faithful" means "faithful including aspects regarding the underlying vector spaces". And this intuition is not bad either ...

 

[DrDu]

I also don't see how to rotate z into x+iy.

 

[DrDu]

The mathematical term for the behaviour you we are discussing is probably that the rotation group does not act transitively on the complex vector space spanned by the |j,m> with fixed j.

 

[tom.stoer]

So that means that there are "equivalence classes" in the 2j+1 dim. Hilbert space which are defined by equivalence regarding D-rotation.

The next questions are: how many equivalence classes are there? And do they have a mathematical or physical meaning?

 

[DrDu]

These equivalence classes are called orbits, I think.