I Angular Momentum Vector and Its Magnitude

[hokhani]

TL;DR

The relation between the quantized values of xyz components of angular momentum and its magnitude

For the quantum state $|l,m\rangle= |2,0\rangle$ the z-component of angular momentum is zero and $|L^2|=6 \hbar^2$. According to uncertainty it is impossible to determine the values of $L_x, L_y, L_z$ simultaneously. However, we know that $L_x$ and $L_y$, like $L_z$, get the values $(-2,-1,0,1,2) \hbar$. In other words, for the state $|2,0\rangle$ we have $\vec{L}=(L_x, L_y,0)$ with $L_x$ and $L_y$ one of the values $(-2,-1,0,1,2) \hbar$. But none of these configurations gives $|L^2|=6\hbar^2$!? I would appreciate if help me with this problem.

 

[PeroK]

Well, exactly! There is no realistic configuration that satisfies the requirements.

I'm suprised this example isn't shown more often. It's a good example of why QM isn't just classical mechanics with a bit of uncertainty thrown in. Spin values really, truly do not have well-defined values in a non-eigenstate.

 

[PeterDonis]

none of these configurations gives $|L^2|=6\hbar^2$!?

Right, which means, to amplify on @PeroK's answer a bit, that the system cannot be in any of these configurations. It must be in a quantum superposition of them. There is no such thing in classical physics, which is one of the main reasons QM is so counterintuitive.

 

[hokhani]

It must be in a quantum superposition of them.

Even if the system be in an equal superposition of all of these 25 configurations, we have $\langle L^2 \rangle =4 \hbar^2$ which again doesn't result in the expected value.

 

[PeroK]

Even if the system be in an equal superposition of all of these 25 configurations, we have $\langle L^2 \rangle =4 \hbar^2$ which again doesn't result in the expected value.

In general an expected value does not itself need to be one of the possible values. For example, the expected value of a roll of a standard 6-sided die is 3.5. This comes from a uniform distribution of the numbers 1-6.

In this case, you could achieve the expected value of $\langle L^2_x \rangle = 3\hbar^2$ and $\langle L^2_y \rangle = 3\hbar^2$from a weighted superposition of the relevant eigenstates.

You could look up online (or calculate for yourself) the decomposition of the $L_z = 0$ eigenstate into the $L_x$ eigenstates. Then calculate the expected value of $L_x^2$ and check that it comes to $3\hbar^2$.

For example, you could achieve this with the state $\frac 1 {\sqrt 3} |L_x = 1\rangle + \sqrt {\frac 2 3}|L_x =2\rangle$.

This is the same idea as you have for a spin 1/2 particle, except that is it clear that when $L_z = 0$, then $L_x$ and $L_y$ cannot be definitely one eigenstate or the other. It must be a superposition.

This again emphasises that the superposition is the state of the system. It's not just a distribution based on ignorance of what the underlying state really is.

 

[JimWhoKnew]

Even if the system be in an equal superposition of all of these 25 configurations, we have $\langle L^2 \rangle =4 \hbar^2$ which again doesn't result in the expected value.

What are these 25 configurations?

From the eigenvalues $~(-2,-1,0,1,2)\hbar~$ we can infer that the result of a measurement of $~L_x^2~$ (or $~L_y^2~$) alone can't exceed $~4\hbar^2~$. In the state $~|2,0\rangle~$, the sum $~L_x^2+L_y^2~$ is equal to $~\mathbf{L}^2~$, so the latter must be bound by $~8\hbar^2~$ (compatible with the true value $~6\hbar^2~$).

Note that in the state $~|2,0\rangle~$ we have $~\langle \mathbf{L} \rangle=0~$.

At least in principle, we should be able to measure $~\mathbf{L}^2~$ without perturbing the state $~|2,0\rangle~$. If we attempt to measure it "by parts", we may encounter the following problem: a measurement of $~L_z^2~$ will yield 0, so we are good for now. Suppose we now measure $~L_x^2~$ and get $~1\hbar^2~$. Naively, we should now expect a further measurement of $~L_y^2~$ to yield the impossible result $~5\hbar^2~$ (because $~\mathbf{L}^2 |2,0\rangle=(L_x^2+L_y^2+L_z^2)|2,0\rangle=6\hbar^2|2,0\rangle~$). Can you spot what's wrong with this argument?

 

[PeterDonis]

an equal superposition

Who said the superposition had to be equal?

 

[hokhani]

What are these 25 configurations?

All the configurations of $(L_x,L_y,0)$ are as follows;
4 configurations with $(L_x,L_y)=(\pm2, \pm2)$ and $L^2=8 \hbar^2$
4 configurations with $(\pm1, \pm1)$ and $L^2=2 \hbar^2$
4 configurations with $(\pm1, \pm2)$ and $L^2=5 \hbar^2$
4 configurations with $(\pm2, \pm1)$ and $L^2=5 \hbar^2$
2 configurations with $(0, \pm2)$ and $L^2=4 \hbar^2$
2 configurations with $(\pm2,0)$ and $L^2=4 \hbar^2$
2 configurations with $(\pm1,0)$ and $L^2= \hbar^2$
2 configurations with $(0,\pm1)$ and $L^2= \hbar^2$
1 configurations with $(0,0)$ and $L^2= 0$
So, $\langle L^2 \rangle =\hbar^2 \frac{100}{25}$.

Can you spot what's wrong with this argument?

In this approach you neglect the uncertainty.

 

[hokhani]

Right,

Who said the superposition had to be equal?

Right, but how to realize which configuration is not preferred?

 

[JimWhoKnew]

All the configurations of $(L_x,L_y,0)$ are as follows;
4 configurations with $(L_x,L_y)=(\pm2, \pm2)$ and $L^2=8 \hbar^2$
4 configurations with $(\pm1, \pm1)$ and $L^2=2 \hbar^2$
4 configurations with $(\pm1, \pm2)$ and $L^2=5 \hbar^2$
4 configurations with $(\pm2, \pm1)$ and $L^2=5 \hbar^2$
2 configurations with $(0, \pm2)$ and $L^2=4 \hbar^2$
2 configurations with $(\pm2,0)$ and $L^2=4 \hbar^2$
2 configurations with $(\pm1,0)$ and $L^2= \hbar^2$
2 configurations with $(0,\pm1)$ and $L^2= \hbar^2$
1 configurations with $(0,0)$ and $L^2= 0$
So, $\langle L^2 \rangle =\hbar^2 \frac{100}{25}$.

You can take a superposition of states, but what is the meaning of a superposition of "configurations"?
If you are approaching it as a statistical mechanics problem, note that you have only 5 independent ("orthogonal") configurations.

I'm not sure I understand your question, so please point out what I'm missing: by $~(L_x,L_y,0)~$ you seem to mean that the system is in the state $~|2,L_z=0\rangle~$. If so, let's look at any of the "configurations" listed in #8. Suppose we start with (2,2,0) for example. We can calculate the probabilities $~|\langle 2,L_x=2|2,L_z=0\rangle|^2~$ and $~|\langle 2,L_y=2|2,L_z=0\rangle|^2~$. Because of the uncertainty, I don't know how to calculate the ("mutual") probability for the "configuration" (2,2,0), and I don't know how to relate it to the constraints in the previous sentence, but are we free to arbitrarily assign 1/25 to it, as if it is unconstrained (by the given state) at all? (I don't know whether that's what @PeterDonis meant in #7, but the questions seem similar at the bottom line)

In this approach you neglect the uncertainty.

Correct.

 

[PeroK]

All the configurations of $(L_x,L_y,0)$ are as follows;
4 configurations with $(L_x,L_y)=(\pm2, \pm2)$ and $L^2=8 \hbar^2$
4 configurations with $(\pm1, \pm1)$ and $L^2=2 \hbar^2$
4 configurations with $(\pm1, \pm2)$ and $L^2=5 \hbar^2$
4 configurations with $(\pm2, \pm1)$ and $L^2=5 \hbar^2$
2 configurations with $(0, \pm2)$ and $L^2=4 \hbar^2$
2 configurations with $(\pm2,0)$ and $L^2=4 \hbar^2$
2 configurations with $(\pm1,0)$ and $L^2= \hbar^2$
2 configurations with $(0,\pm1)$ and $L^2= \hbar^2$
1 configurations with $(0,0)$ and $L^2= 0$
So, $\langle L^2 \rangle =\hbar^2 \frac{100}{25}$.

In this approach you neglect the uncertainty.

I'm not convinced you've understood superposition and the linear algebra that describes them. $L_z = 0$ is a complete description of a state. In this state, there are definite no values for $L_x$ and $L_y$. In this sense, your "configurations" are not quantum mechanical descriptions.

Instead, the state $L_z = 0$ can also be written in terms of eigenstates of $L_x$. My guess is that it would be:
$$|L_z = 0\rangle = \sqrt {\frac 2 6}|L_x = -2\rangle + \sqrt {\frac 1 6}|L_x = -1\rangle + \sqrt {\frac 1 6}|L_x = 1\rangle + \sqrt {\frac 2 6}|L_x = 2\rangle$$(You can do some research and see whether my guess is correct or not. But, the precise expansion is not crucial to the argument.)

And this state would also have a similar expansion in terms of $L_y$ eigenstates. The modulus of the coeffcients would be the same but the coefficients themselves would be complex.

The "configuration" $L_x = 1, L_y = 2, L_z = 0$ has no meaning in QM, as these three observables are mutually incompatible, and only at most one can have a definite value in any given state. That "configuration", therefore, does not describe any possible state.

 

[PeterDonis]

Right, but how to realize which configuration is not preferred?

Who cares? The system isn't in a "configuration", it's in a quantum state. Write down a quantum state that satisfies your constraint and see what its expectation value is for the observable you're interested in. So far you haven't done that, which means everything you've posted is irrelevant to answering the question you want to answer.