# Do Lx and Ly have a same eigenstate?

• Haorong Wu
In summary: If ##[A, B]## is an operator and ##[A, B] = \hat 0 ##, then ##[A, B]## will always equate to the identity matrix.
Haorong Wu
Homework Statement
##\left [ L_x , L_y \right ] \neq 0##, then do they have a same eigenstate?
Relevant Equations
None
Certainly, ##\left [ A ,B \right ] \neq 0## does not mean that they do not have a same eigenstate.

But how to construct a same eigenstate for ##L_x## and ##L_y## if it exists?

Since ##L_x Y_l^m = \frac \hbar 2 \left ( \sqrt { l \left ( l+1 \right ) -m \left ( m+1 \right )} Y_l^{m+1} + \sqrt {l \left ( l+1 \right ) -m \left ( m-1 \right ) } Y_l^{m-1} \right )##,
and ##L_y Y_l^m = \frac \hbar {2i} \left ( \sqrt { l \left ( l+1 \right ) -m \left ( m+1 \right )} Y_l^{m+1} - \sqrt {l \left ( l+1 \right ) -m \left ( m-1 \right ) } Y_l^{m-1} \right )##, I have troubles when constructing the eigenstate with ##Y_l^m##.

At the same time, I am wondering if the following statements are true.

if ##\left [ A ,B \right ] \neq 0## but ## \left < \left [ A ,B \right ] \right > =0##, then they have same eigenstates;
if ##\left [ A ,B \right ] \neq 0## and ## \left < \left [ A ,B \right ] \right > \neq 0##, then they do not have same eigenstates.

If these statements are true, then ##L_x## and ##L_y## do not have same eigenstates since ## \left < \left [ L_x ,L_y \right ] \right > = i {\hbar}^2 m \neq 0##.

Thank you for your time.

The only common eigenstate is the one where ##L^2## and all of ##L_x ,L_y ,L_z## have eigenvalue zero.

In general, you have three cases:

1) ##A, B## commute, hence you can find a basis of common eigenstates.

2) ##A, B## do not commute and have no common eigenstates.

3) ##A, B## do not commute but have at least one common eigenstate.

Note that when you ask:

Haorong Wu said:
At the same time, I am wondering if the following statements are true.

if ##\left [ A ,B \right ] \neq 0## but ## \left < \left [ A ,B \right ] \right > =0##, then they have same eigenstates;
if ##\left [ A ,B \right ] \neq 0## and ## \left < \left [ A ,B \right ] \right > \neq 0##, then they do not have same eigenstates.

You need to be careful. ##[A, B]## is an operator and ##[A, B] = \hat 0 ## is an operator equation - which I've emphasised by the hat on the zero.

This equation is equivalent to saying that the operator ##[A, B]## maps all states/vectors to zero.

But, ## \langle [ A ,B] \rangle =0 ## is a numerical equation, for an implied state which is not specified here. This would be true for a common eigentstate, but it might be true otherwise. The sufficient condition is that:

##[A, B] \psi## is orthogonal to ##\psi## for some state ##\psi##.

If the equation holds for all states ##\psi##, then you should be able to show that ##[A, B] = \hat 0##.

I see that @hilbert2 has answered the specific question in this case.

hilbert2 said:
The only common eigenstate is the one where ##L^2## and all of ##L_x ,L_y ,L_z## have eigenvalue zero.

Ah. Thank you. I forget ##L_x Y_0^0=0##. Thanks!

PeroK said:
In general, you have three cases:

1) ##A, B## commute, hence you can find a basis of common eigenstates.

2) ##A, B## do not commute and have no common eigenstates.

3) ##A, B## do not commute but have at least one common eigenstate.

Note that when you ask:
You need to be careful. ##[A, B]## is an operator and ##[A, B] = \hat 0 ## is an operator equation - which I've emphasised by the hat on the zero.

This equation is equivalent to saying that the operator ##[A, B]## maps all states/vectors to zero.

But, ## \langle [ A ,B] \rangle =0 ## is a numerical equation, for an implied state which is not specified here. This would be true for a common eigentstate, but it might be true otherwise. The sufficient condition is that:

##[A, B] \psi## is orthogonal to ##\psi## for some state ##\psi##.

If the equation holds for all states ##\psi##, then you should be able to show that ##[A, B] = \hat 0##.

I see that @hilbert2 has answered the specific question in this case.

Thanks, PeroK. I did misunderstand the meaning of ##[A, B] = \hat 0 ##. Thanks for pointing it out.

PeroK said:
If the equation holds for all states ##\psi##, then you should be able to show that ##[A, B] = \hat 0##.

In fact, even this isn't true.

## 1. What is an eigenstate?

An eigenstate is a state in which a physical system can exist with a definite value for a particular observable quantity, such as energy or momentum. In other words, it is a state in which the system is in a specific, well-defined state.

## 2. What are Lx and Ly?

Lx and Ly are operators that represent different physical observables, specifically the x and y components of angular momentum. These operators are part of the larger set of angular momentum operators known as the Pauli matrices.

## 3. Why is it important to determine if Lx and Ly have the same eigenstate?

Determining if Lx and Ly have the same eigenstate is important because it allows us to understand the relationship between the x and y components of angular momentum. It also helps us to better understand the behavior of a physical system and make predictions about its behavior.

## 4. How do you determine if Lx and Ly have the same eigenstate?

To determine if Lx and Ly have the same eigenstate, we can use the commutator [Lx,Ly] = iħLz, where Lz is the z component of angular momentum. If the commutator is equal to zero, then Lx and Ly have the same eigenstate. If the commutator is not equal to zero, then Lx and Ly do not have the same eigenstate.

## 5. What are the implications if Lx and Ly have the same eigenstate?

If Lx and Ly have the same eigenstate, it means that the x and y components of angular momentum are not independent of each other. This can have implications for the behavior of a physical system and may lead to new insights and discoveries in the field of quantum mechanics.

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