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Do Lx and Ly have a same eigenstate?

  • Thread starter Haorong Wu
  • Start date
103
13

Homework Statement:

##\left [ L_x , L_y \right ] \neq 0##, then do they have a same eigenstate?

Homework 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.
 

Answers and Replies

hilbert2
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The only common eigenstate is the one where ##L^2## and all of ##L_x ,L_y ,L_z## have eigenvalue zero.
 
PeroK
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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.
 
103
13
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!
 
103
13
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
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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.
 

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