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

1. Jul 4, 2010

### 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

Last edited: Jul 5, 2010
2. Jul 4, 2010

### 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>.

3. Jul 4, 2010

### wdlang

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.

4. Jul 4, 2010

### 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?

5. Jul 4, 2010

### jostpuur

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.

6. Jul 4, 2010

### 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.

7. Jul 4, 2010

### 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 $D(R) \, |j, m>$.

8. Jul 4, 2010

### tom.stoer

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

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

Last edited: Jul 4, 2010
9. Jul 4, 2010

### Dickfore

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.

10. Jul 4, 2010

### 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).

11. Jul 4, 2010

### 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:
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.

12. Jul 4, 2010

### tom.stoer

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

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

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 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.

Last edited: Jul 4, 2010
13. Jul 4, 2010

### jostpuur

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

14. Jul 4, 2010

### 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 $j = 1/2$.

15. Jul 4, 2010

### tom.stoer

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 wich 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.

Last edited: Jul 5, 2010
16. Jul 5, 2010

### Dickfore

Nice post.

17. Jul 5, 2010

### tom.stoer

corrected :-)

18. Jul 5, 2010

### 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.

19. Jul 5, 2010

### 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?

20. Jul 5, 2010

### 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.

21. Jul 5, 2010

### jostpuur

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:

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 $\textrm{Im}(\phi)$.

22. Jul 5, 2010

### 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:

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

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?

Last edited: Jul 5, 2010
23. Jul 5, 2010

### jostpuur

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.

24. Jul 5, 2010

### 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.

25. Jul 5, 2010

### 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?