Do Lx and Ly have a same eigenstate?

[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.

 

[hilbert2]

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

 

[PeroK]

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:

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.

 

[Haorong Wu]

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!

 

[Haorong Wu]

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]

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.