Angular momentum raising/lowering operators

[dyn]

Hi. I have come across the following statement - the eigenvalue equation for J₊ is given by

J₊ | j m > = ħ √{(j+1)-m(m+1)} | j , m+1>

My question is this - how can this be an eigenvalue equaton as the ket | j, m> has changed to | j , m+1> ? Surely the raising/lowering operators don't have eigenkets as they always change the eigenket ?

 

[blue_leaf77]

Yeah it's indeed not an eigenvalue equation.

 

[PeterDonis]

Surely the raising/lowering operators don't have eigenkets as they always change the eigenket ?

This is not correct. Raising/lowering operators will always change kets that are eigenstates of the number operator. But they will not necessarily always change other kets.

In fact, it is straightforward to construct a state that is an eigenstate of the lowering operator. These states are called "coherent states". See here:

https://en.wikipedia.org/wiki/Coherent_states#Quantum_mechanical_definition

There is no similar construction for the raising operator, however. One way of seeing why is to look at how a coherent state is expressed when you use the eigenstates of the number operator as the basis. In this basis, a coherent state has just the right coefficients for each number eigenstate so that, when the lowering operator is applied, each number eigenstate is "lowered" into the one below it and has just the right coefficient for that new number eigenstate. The "lowest" number eigenstate in the chain, the state with zero particles, gets annihilated, so it just "moves out of the way", so to speak. But this trick won't work for the raising operator, because there is no state that you can apply the raising operator to to get the number eigenstate with zero particles.

 

[Misha]

In fact, it is straightforward to construct a state that is an eigenstate of the lowering operator. These states are called "coherent states". See here:

OP is talking about lowering operators for angular momentum. In the Schwinger representation, they are

$J_- = a_1^\dagger a_2$

and they don't have eigenstates.

Coherent states that appear in this context are rather spin coherent states, which were introduced by Radcliffe and were put into a proper context by Perelomov. Generally, such coherent states are not eigenstates of lowering operators.

 

[stevendaryl]

Hi. I have come across the following statement - the eigenvalue equation for J₊ is given by

J₊ | j m > = ħ √{(j+1)-m(m+1)} | j , m+1>

That should actually be
$J_+ |j m \rangle = \hbar \sqrt{j(j+1) - m (m+1)}$

My question is this - how can this be an eigenvalue equaton as the ket | j, m> has changed to | j , m+1> ? Surely the raising/lowering operators don't have eigenkets as they always change the eigenket ?

The point of that equation is that it is a method for solving a different set of equations:

$J^2 |j, m \rangle = j (j+1) |j, m \rangle$ [edited]
$J_z |j, m\rangle = m |j, m \rangle$

Usually, the way it goes is that the extreme case where $m = -j$ is easy to solve. Then you can use the equation

$J_+ |j m \rangle = \hbar \sqrt{j(j+1) - m (m+1)}$

repeatedly to find the other cases. For example, for pure orbital angular momentum of one particle (no spin), the state $|j -j \rangle$ is the wave function $\psi(\theta, \phi) = e^{-i j \phi}$. Then you apply the operator $J_+$ to find the solution to the equations

$J^2 |j -(j-1)\rangle = j(j+1) |j -(j-1)\rangle$
$J_z |j -(j-1)\rangle = -(j-1) |j -(j-1)\rangle$

 

[vanhees71]

The first eigenvalue equation should of course be
$$J^2 |j, m \rangle = j (j+1) |j, m \rangle.$$

 

[dyn]

Thanks for all replies