Why Are There Additional Eigenvalues and Eigenvectors of J3?

[StenEdeback]

Homework Statement

Eigenvalues and eigenvectors of J3

Relevant Equations

Eigenvectors

The J3 matrix of two dimensional SU2 consists of two row vectors (1 0) and (0 -1). When I calculate the eigenvalues of an eigenvector v the usual way with J3v=kv then I find eigenvalues +-1 and eigenvectors (1 0) and (0 1). But how is it possible to say that there are other eigenvectors and eigenvalues of J3, like (j(j+1) m+-1) ?

 

[PeroK]

Homework Statement:: Eigenvalues and eigenvectors of J3
Relevant Equations:: Eigenvectors

The J3 matrix of two dimensional SU2 consists of two row vectors (1 0) and (0 -1). When I calculate the eigenvalues of an eigenvector v the usual way with J3v=kv then I find eigenvalues +-1 and eigenvectors (1 0) and (0 1). But how is it possible to say that there are other eigenvectors and eigenvalues of J3, like (j(j+1) m+-1) ?

If we take $j = \frac 1 2$, then we have:
$$J_3 = \frac{\hbar}{2}
\begin{bmatrix}
1& 0\\
0&-1
\end{bmatrix}
$$
And:
$$J^2= \frac{3\hbar^2}{4}
\begin{bmatrix}
1& 0\\
0&1
\end{bmatrix}
$$
The shared eigenvectors of these two matrices are $(1, 0)$ and $(0,1)$ with eigenvalues $\pm \frac \hbar 2$ (which correspond to $m\hbar$, where $m = \pm j = \pm \frac 1 2$) and $\frac{3\hbar^2}{4}$ (which corresponds to $j(j+1)\hbar^2$, where $j = \frac 1 2$).

If we have $j > \frac 1 2$, then the rotation operators would be represented by higher-dimensional matrices with the appropriate eigenvalues. E.g. for $j = 1$ we need $3 \times 3$ matrices to represent $J_3$ and $J^2$.

 

[Gaussian97]

What you have computed are the eigenvalues and eigenvectors of the 2-dimensional representation of SU(2), which happens to be the fundamental one. If you compute the eigenvalues and eigenvectors of the matrices of another representation you will get other values. For example, the 3-dimensional representation is given by the matrices
$$J_1=\frac{1}{\sqrt{2}}\begin{pmatrix}0&1&0\\1&0&1\\0&1&0\end{pmatrix}$$
$$J_2=\frac{1}{\sqrt{2}}\begin{pmatrix}0&-i&0\\i&0&-i\\0&i&0\end{pmatrix}$$
$$J_3=\begin{pmatrix}1&0&0\\0&0&0\\0&0&-1\end{pmatrix}$$
which have different eigen values and eigenvectors.
SU(2) has exactly one (irreducible) representation for each dimension so you can find the 4x4 representation and so on...
The representation of matrices $N\times N$ describes a system of spin $j=\frac{N-1}{2}$.

 

[StenEdeback]

If we take $j = \frac 1 2$, then we have:
$$J_3 = \frac{\hbar}{2}
\begin{bmatrix}
1& 0\\
0&-1
\end{bmatrix}
$$
And:
$$J^2= \frac{3\hbar^2}{4}
\begin{bmatrix}
1& 0\\
0&1
\end{bmatrix}
$$
The shared eigenvectors of these two matrices are $(1, 0)$ and $(0,1)$ with eigenvalues $\pm \frac \hbar 2$ (which correspond to $m\hbar$, where $m = \pm j = \pm \frac 1 2$) and $\frac{3\hbar^2}{4}$ (which corresponds to $j(j+1)\hbar^2$, where $j = \frac 1 2$).

If we have $j > \frac 1 2$, then the rotation operators would be represented by higher-dimensional matrices with the appropriate eigenvalues. E.g. for $j = 1$ we need $3 \times 3$ matrices to represent $J_3$ and $J^2$.

Thank you!
I understand this, but my question is why it is often written in textbooks for instance that J3/j(j+1),m)=m/j(j+1),m) ? I do not understand how this can be correct. The vector acted upon, (j(j+1),m) is not as far as I can see an eigenvector of J3. Left and right sides of the equation yield results with opposite signs for m?

 

[Gaussian97]

Thank you!
I understand this, but my question is why it is often written in textbooks for instance that J3/j(j+1),m)=m/j(j+1),m) ? I do not understand how this can be correct. The vector acted upon, (j(j+1),m) is not as far as I can see an eigenvector of J3. Left and right sides of the equation yield results with opposite signs for m?

I'm not sure if I understand what you write, do you mean something like
$$J_3 \left|j(j+1), m\right>=m\left|j(j+1),m\right>$$
? If that's what you mean, how do you define the states $\left|j(j+1),m\right>$ to say that they are not eigenvectors of $J_3$? Because usually these states are defined precisely as the eigenvectors of $J_3$...

 

[PeroK]

Thank you!
I understand this, but my question is why it is often written in textbooks for instance that J3/j(j+1),m)=m/j(j+1),m) ? I do not understand how this can be correct. The vector acted upon, (j(j+1),m) is not as far as I can see an eigenvector of J3. Left and right sides of the equation yield results with opposite signs for m?

In general, we have a state vector $|j, m \rangle$, which is a shared eigenstate/eigenvector of $J_3$ and $J^2$ satisfying: $$J_3 |j, m \rangle = m|j, m \rangle, \ \text{and} \ J^2 |j, m \rangle = j(j+1)|j, m \rangle$$ Assuming units where $\hbar = 1$.

For $j =\frac 1 2$, we have $m = \pm 1$, hence two state vectors satisfying:
$$J_3 |\frac 1 2 , \pm \frac 1 2 \rangle = \pm \frac 1 2 |\frac 1 2, \pm \frac 1 2 \rangle, \ \text{and} \ J^2 |\frac 1 2, \pm \frac 1 2 \rangle = \frac 3 4 |\frac 1 2, \pm \frac 1 2 \rangle$$
In the appropriate basis, these operators and eigenvectors are represented by the matrices given above. It should be clear that the vector $(1,0)$ is en eigenvector of
$$J_3 = \frac{1}{2}
\begin{bmatrix}
1& 0\\
0&-1
\end{bmatrix}
$$
with eigenvalue $\frac 1 2$. And the vector $(0, 1)$ is an eigenvector with eigenvalue $-\frac 1 2$.

If you don't get that, then you must be doing matrix multiplication incorrectly.

 

[PeroK]

I'm not sure if I understand what you write, do you mean something like
$$J_3 \left|j(j+1), m\right>=m\left|j(j+1),m\right>$$

I must admit, I've never seen this notation. It's always been $\left|j, m\right>$.

 

[Gaussian97]

I must admit, I've never seen this notation. It's always been $\left|j, m\right>$.

Yeah, it's really not standard, but since the OP used this notation I decided to follow it...
I think I've seen this notation somewhere else, but I have no idea where.

 

[StenEdeback]

I'm not sure if I understand what you write, do you mean something like
$$J_3 \left|j(j+1), m\right>=m\left|j(j+1),m\right>$$
? If that's what you mean, how do you define the states $\left|j(j+1),m\right>$ to say that they are not eigenvectors of $J_3$? Because usually these states are defined precisely as the eigenvectors of $J_3$...

Yes, I meant what you wrote but did not have the text facilities to write it as clearly as you did. But I think that an eigenvector v is defined as satisfying the equation J3v = mv , that is when operated on by J3 equalling multiplied by the constant m. And I cannot see that (j(j+1),m) is such a vector.

 

[PeroK]

Yes, I meant what you wrote but did not have the text facilities to write it as clearly as you did. But I think that an eigenvector v is defined as satisfying the equation J3v = mv , that is when operated on by J3 equalling multiplied by the constant m. And I cannot see that (j(j+1),m) is such a vector.

You mean that, in general, you don't believe eigenvectors can be found?

 

[Gaussian97]

Yes, I meant what you wrote but did not have the text facilities to write it as clearly as you did. But I think that an eigenvector v is defined as satisfying the equation J3v = mv , that is when operated on by J3 equalling multiplied by the constant m. And I cannot see that (j(j+1),m) is such a vector.

Then I ask you the same question, how do you define the vector $\left|j(j+1),m\right>$? Because for me, and for most of the textbooks I've looked, the definition of such a vector is one that fulfills the two properties:
$$J_3\left|j(j+1),m\right>=m\left|j(j+1),m\right>, \qquad J^2\left|j(j+1),m\right> = j(j+1)\left|j(j+1),m\right>$$
(which can be proved to exist).
So yes, of course you can see that with this definition $\left|j(j+1),m\right>$ is an eigenvector of $J_3$ by definition... See that both conditions are eigenvector equations.

So please, tell us how do you define such a vector $\left|j(j+1),m\right>$ and then we may be able to help you further.

 

[StenEdeback]

Then I ask you the same question, how do you define the vector $\left|j(j+1),m\right>$? Because for me, and for most of the textbooks I've looked, the definition of such a vector is one that fulfills the two properties:
$$J_3\left|j(j+1),m\right>=m\left|j(j+1),m\right>, \qquad J^2\left|j(j+1),m\right> = j(j+1)\left|j(j+1),m\right>$$
(which can be proved to exist).
So yes, of course you can see that with this definition $\left|j(j+1),m\right>$ is an eigenvector of $J_3$ by definition... See that both conditions are eigenvector equations.

So please, tell us how do you define such a vector $\left|j(j+1),m\right>$ and then we may be able to help you further.

As far as I can see, the only eigenvectors v for J3 that satisfy the equation J3v = Cv where C is a constant, the eigenvalue of J3, according to the definition of an eigenvector, are A(1,0) and B(0,1) where A and B are arbitrary constants. So (j,m) cannot be an eigenvector of J3, but (j,0) and (0,m) could? J2 (the Casimir operator) has eigenvectors (j(j+1),m), I agree with that.

 

[PeroK]

As far as I can see, the only eigenvectors v for J3 that satisfy the equation J3v = Cv where C is a constant, the eigenvalue of J3, according to the definition of an eigenvector, are A(1,0) and B(0,1) where A and B are arbitrary constants. So (j,m) cannot be an eigenvector of J3, but (j,0) and (0,m) could?

I suspect you have completely misunderstood the notation. The notation $|j, m\rangle$ denotes the eigenvector (independent of any basis). It is the normalised vector (that we know exsts) that is a simultaneous eigenvector of $J^2$ with eigenvalue $j(j+1)$ and $J_3$ with eigenvalue $m$.

If $j = \frac 1 2$, then we have two orthogonal eigenvectors $|\frac 1 2, \frac 1 2 \rangle$ and $|\frac 1 2, -\frac 1 2 \rangle$.

We then choose these to be our basis vectors, so that $|\frac 1 2, \frac 1 2 \rangle$ has components $(1,0)$ and $|\frac 1 2, -\frac 1 2 \rangle$ has components $(0, 1)$.

 

[StenEdeback]

I am trying to understand, but if you multiply J3(1/2,1/2) then you get 1/2(1/2,-1/2), so how can (1/2,1/2) be an eigenvector of J3? And J3(1/2,-1/2) = 1/2(1/2,1/2), so it seems to me that (1/2,-1/2) is not either an eigenvector of J3? (The first 1/2 in the results of the multiplications coming from J3). I may misunderstand the difference between eigenvectors and basis vectors?

 

[PeroK]

I am trying to understand, but if you multiply J3(1/2,1/2) then you get 1/2(1/2,-1/2), so how can (1/2,1/2) be an eigenvector of J3? And J3(1/2,-1/2) = 1/2(1/2,1/2), so it seems to me that (1/2,-1/2) is not either an eigenvector of J3? (The first 1/2 in the results of the multiplications coming from J3). I may misunderstand the difference between eigenvectors and basis vectors?

Look at post #13. When you see $|j, m \rangle$, $j$ and $m$ are NOT the components of a vector. The whole expression is an abstract description of the vector, in terms of its eigenvalues.

You cannot treat $|\frac 1 2, \frac 1 2 \rangle$ and $(\frac 1 2, \frac 1 2)$ as the same thing! That's completely wrong!

 

[PeroK]

PS Another example of this idea is the spherical harmonics. These are the eigenfunctions of the orbital angular momentum operators $L^2$ and $L_3$ and are written $$Y^{m}_{l}$$ Which denotes the simultaneous eigenfunction of $L^2$ and $L_3$ with eigenvalues $l(l+1)\hbar^2$ and $m\hbar$ respectively.

 

[StenEdeback]

Look at post #13. When you see $|j, m \rangle$, $j$ and $m$ are NOT the components of a vector. The whole expression is an abstract description of the vector, in terms of its eigenvalues.

You cannot treat $|\frac 1 2, \frac 1 2 \rangle$ and $(\frac 1 2, \frac 1 2)$ as the same thing! That's completely wrong!

Thank you! I will contemplate this and try to understand it. I am an old man at 77, reviving the interest in theoretical physics that I had in my youth, now privately studying mathematical group theory, quantum physics, an general theory of relativity. Mostly when I have questions I can find the answer by googling, but sometimes I get stuck and then it is very valuable to have PhysicsForums with its many helpful and knowledgeable members as a last resort. I am very grateful for that, and for your patience.

 

[StenEdeback]

PS Another example of this idea is the spherical harmonics. These are the eigenfunctions of the orbital angular momentum operators $L^2$ and $L_3$ and are written $$Y^{m}_{l}$$ Which denotes the simultaneous eigenfunction of $L^2$ and $L_3$ with eigenvalues $l(l+1)\hbar^2$ and $m\hbar$ respectively.

I have studied those things in David J Griffiths' excellent book "Introduction to Quantum Mechanics" and I had no problem then. Perhaps I got confused when I now read Jakob Schwichtenberg's book "Physics from Symmetry" which treats this from a group theoretical point of view. The thing that tends to confuse me most is Dirac notation, which I never seem to get on good terms with.