Eigenstates and Angular Momentum

[IHateMayonnaise]

[SOLVED] Eigenstates and Angular Momentum

Homework Statement

At a given instant, a rigid rotor is in the state:

$$\Psi(\theta,\phi)=\sqrt{\frac{3}{4\pi}}Sin(\phi) Sin(\theta)$$

If the z component of the orbital angular momentum is measured, what are the possible values of $$<\hat{L_{z}}>$$, and with what probability will they occur?

Homework Equations

These are the equations that I think are relevant:

$$\hat{L_z}=\frac{\hbar}{i}\frac{\partial}{\partial \theta}$$

$$L_z = m_l \hbar$$

$$|m_l| \leq l$$

$$<\hat{L_z}>=\int_0^{2\pi} \Psi^*(\theta, \phi)\left(\frac{\hbar}{i}\frac{\partial}{\partial \theta}\right)\Psi(\theta, \phi) d\phi$$

$$Y_1^1=-\sqrt{\frac{3}{8\pi}}Sin(\theta)e^{i\theta}$$

$$Y_1^{-1}=\sqrt{\frac{3}{8\pi}}Sin(\theta)e^{-i\theta}$$

...and also the exponential identities for Sine and Cosine.

The Attempt at a Solution

I suppose my question really has to do with the nature of the wavefunctions defined by Spherical Harmonics. Since the solution to The Schrodinger Equation in spherical coordinates has solutions that correspond to Spherical Harmonics (i.e., $$Y_l^{m_l}$$ corresponds to $$\Psi(\theta,\phi)$$ above), it seems like we should only be able to get solutions (ie wavestates) that are in this form! And while $$\Psi(\theta,\phi)$$ is close to both $$Y_1^1$$ and $$Y_1^{-1}$$, it isn't the same. Furthermore, it looks like it could be a superposition of the two, but it's not!

So, really, I need to find $$l$$ and $$m_l$$. Since our $$\phi$$ dependence in our wavefunction has a coefficient of 1 (i.e. $$Sin(\phi)$$ corresponds to $$e^{im_l\phi}$$ where $$m_l=1$$). So, $$l$$ can have the possible values of $$+1$$ or $$-1$$. But, like I stated in the above paragraph, neither $$Y_1^1$$ or $$Y_1^{-1}$$ correspond to our wavefunction! So what the hell is $$m_l$$ and $$l$$?

 

[nrqed]

Homework Statement

At a given instant, a rigid rotor is in the state:

$$\Psi(\theta,\phi)=\sqrt{\frac{3}{4\pi}}Sin(\phi) Sin(\theta)$$

If the z component of the orbital angular momentum is measured, what are the possible values of $$<\hat{L_{z}}>$$, and with what probability will they occur?

Homework Equations

These are the equations that I think are relevant:

$$\hat{L_z}=\frac{\hbar}{i}\frac{\partial}{\partial \theta}$$

$$L_z = m_l \hbar$$

$$|m_l| \leq l$$

$$<\hat{L_z}>=\int_0^{2\pi} \Psi^*(\theta, \phi)\left(\frac{\hbar}{i}\frac{\partial}{\partial \theta}\right)\Psi(\theta, \phi) d\phi$$

$$Y_1^1=-\sqrt{\frac{3}{8\pi}}Sin(\theta)e^{i\theta}$$

$$Y_1^{-1}=\sqrt{\frac{3}{8\pi}}Sin(\theta)e^{-i\theta}$$

...and also the exponential identities for Sine and Cosine.

The Attempt at a Solution

I suppose my question really has to do with the nature of the wavefunctions defined by Spherical Harmonics. Since the solution to The Schrodinger Equation in spherical coordinates has solutions that correspond to Spherical Harmonics (i.e., $$Y_l^{m_l}$$ corresponds to $$\Psi(\theta,\phi)$$ above), it seems like we should only be able to get solutions (ie wavestates) that are in this form!

why? A general solution is a linear combination of the eigenstates. Nothing says that a physical system must absolutely be an eigenstate of an operator (unless this operator was measured and we consider the state just after the measurement)

And while $$\Psi(\theta,\phi)$$ is close to both $$Y_1^1$$ and $$Y_1^{-1}$$, it isn't the same. Furthermore, it looks like it could be a superposition of the two, but it's not!

Of course it's a superposition. Look at it carefully and you can write it as a linear combination.

 

[IHateMayonnaise]

why? A general solution is a linear combination of the eigenstates. Nothing says that a physical system must absolutely be an eigenstate of an operator (unless this operator was measured and we consider the state just after the measurement)

Of course it's a superposition. Look at it carefully and you can write it as a linear combination.

Okay, so it is a superposition of $$l=1$$ and $$l=-1$$, so those are the two possible values of the orbital angular momentum, right?

If I wanted to find the probability of each wouldn't I just compute:

$$|a_1|^2=<\Psi(\phi),e^{i\phi}>^2$$?

(here $$\Psi(\phi)$$ is equal to $$\Psi(\theta, \phi)$$ except without the theta dependence?)

 

[nrqed]

Okay, so it is a superposition of $$l=1$$ and $$l=-1$$, so those are the two possible values of the orbital angular momentum, right?

If I wanted to find the probability of each wouldn't I just compute:

$$|a_1|^2=<\Psi(\phi),e^{i\phi}>^2$$?

No you would compute

$$|<\Psi(\theta,\phi),Y_1^1>|^2$$ and so on. (Of course if you write the explicit expansion in term of the Y_m^l, you can also read of directly the coefficients.)

 

[IHateMayonnaise]

Ok so I actually don't see how it's a linear combination of Y_1^1 and Y_1^-1...because every time I do the computation I get:

$$-i\sqrt{\frac{3}{4\pi}}Sin(\theta)Sin(\phi)$$

I can't get rid of that -i! Any thoughts?

 

[nrqed]

Ok so I actually don't see how it's a linear combination of Y_1^1 and Y_1^-1...because every time I do the computation I get:

$$-i\sqrt{\frac{3}{4\pi}}Sin(\theta)Sin(\phi)$$

I can't get rid of that -i! Any thoughts?

just multiply your coefficients by +i !
Nothing says that the coefficients have to be real.

 

[IHateMayonnaise]

just multiply your coefficients by +i !
Nothing says that the coefficients have to be real.

I'm sorry but I don't understand! Why can you just "multiply by i" ?

 

[nrqed]

I'm sorry but I don't understand! Why can you just "multiply by i" ?

What I mean is that if you have

$$c_1 Y_1^1 + c_2 y_1^{-1} = -i\sqrt{\frac{3}{4\pi}}Sin(\theta)Sin(\phi)$$


for some c_1 and c_2, then obviously



$$i c_1 Y_1^1 + i c_2 y_1^{-1} = \sqrt{\frac{3}{4\pi}}Sin(\theta)Sin(\phi)$$

So if you use i c_1 and i c_2 as your coefficients, you get the correct answer!

 

[IHateMayonnaise]

thanks :)