# Homework Help: Proving an Eigenfunction of Momentum Operators

1. Feb 20, 2010

### Hart

1. The problem statement, all variables and given/known data

2. Relevant equations

Stated in the question.

3. The attempt at a solution

It is a eigenfunction of L_z as it has no dependance on Z? Not sure if I can just state this, I do need to actually prove it but I can't get the calculations to work.

I managed a similar problem earlier but I just cant understand how to do it with these operators.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Feb 20, 2010

### Staff: Mentor

Isn't an eigenfunction of a transformation a function that satisfies
$$L_Z\psi = \lambda \psi$$ for some constant lambda?

and similarly for $$L_x^2 + L_y^2 + L_z^2$$?

3. Feb 20, 2010

### Hart

Indeed it is.. I just forgot to include that in my post.

So using that, I have rearranged to get:

$$\lambda = ( L_Z . \psi) / \psi$$

And then if $$\lambda$$ has no dependance on Z, then the eigenfunction will be proved. And similarly for the other variations, $$L_y$$ and $$L_z$$

Problem is I don't know how to actually to that calculation.

4. Feb 20, 2010

Well, what is $$\hat{L}_z[/itex] in Cartesian coordinates? 5. Feb 20, 2010 ### Matterwave All you have to do is to act on the function with the operator, and if it returns a function which is a constant times the original function, it should be an eigenfunction. The calculations should not be difficult. As for the second question, If that function is an eigenfunction of Lz, then it shouldn't be an eigenfunction of Lx or Ly since Lz and Lx and Ly don't commute, and therefore do not share eigenfunctions. 6. Feb 20, 2010 ### Hart [tex]L_{z} = -ih(x\frac{\partial}{\partial y}-y\frac{\partial}{\partial z})$$

This is $$L_z$$ in cartesian coordinates? So I use this?

Last edited: Feb 20, 2010
7. Feb 20, 2010

### diazona

You've got a little typo there near the end, but once you fix that: yes, you just apply that operator to the wavefunction.

8. Feb 20, 2010

### gabbagabbahey

Close, its actually

$$\hat{L}_{z} = -ih(x\frac{\partial}{\partial y}-y\frac{\partial}{\partial x})$$

And yes, apply this operator to the wavefunction you are given and see what you get

9. Feb 20, 2010

### Hart

Here's my calculations:

$$\frac{\partial((x+iy)^{2})}{\partial y} = 2ix-2y$$

$$\frac{\partial((x+iy)^{2})}{\partial x} = 2x+2iy$$

Therefore:

$$x\frac{\partial((x+iy)^{2})}{\partial y} = 2ix^{2}-2xy$$

$$y\frac{\partial((x+iy)^{2})}{\partial x} = 2xy+2iy^{2}$$

Therefore:

$$x\frac{\partial((x+iy)^{2})}{\partial y} - y\frac{\partial((x+iy)^{2})}{\partial x} = 2ix^{2}-2xy - 2xy+2iy^{2} = 2(ix^{2}-2xy-iy^{2})$$

Hence:

$$L_{z}\psi = -2i\hbar(ix^{2}-2xy-iy^{2})$$

Then I need to divide this by $$\psi = (x+iy)^{2}$$ but that looks messy at the moment..

.. I'm guessing I must have done something wrong for this not to work out!

10. Feb 20, 2010

### gabbagabbahey

$$ix^2-2xy-iy^2=i(x^2+2ixy-y^2)$$...what do you get when you factor the quantity in brackets?

11. Feb 20, 2010

### Hart

.. oh yeah how did I not spot that!

Right, so then:

$$L_{z}\psi = 2\hbar\psi$$

Hence the multiplicative value $$\lambda = 2\hbar$$ , which has no dependance on Z, so therefore have proven that $$\psi = (x+iy)^{2}$$ is an eigenfunction.

Am I right to say then that it is also a eigenfunction of $$L = L_{x} + L_{y} + L_{z}$$, but not of $$L_{y}$$ or $$L_{x}$$ individually?

12. Feb 20, 2010

### gabbagabbahey

First,

$$\hat{\mathbf{L}}=\hat{L}_x\mathbf{i}+\hat{L}_y\mathbf{j}+\hat{L}_z\mathbf{k}$$

is a vector operator.

Second, you are asked to show that the given wavefunction is an eigenfunction of $\hat{L}^2$, not $\hat{\mathbf{L}}$.

13. Feb 20, 2010

### Hart

I've just calculated the result:

$$\hat{L} = 4\hbarz(2x+iy)+2\hbar\psi$$

Basically by calculating $$\hat{L_{x}},\hat{L_{y}},$$ and $$\hat{L_{z}}$$ seperately using their cartesian forms;

$$\hat{L_{x}} = 2ixz-2yz$$

$$\hat{L_{y}} = 2xz+2iyz$$

$$\hat{L_{z}} = 2ix^{2}-4xy-2iy^{2}$$

.. and then adding the three results together. Not forgetting the $$-i\hbar$$ factor of course.

Don't really know where to go with this now, it doesn't seem to simplify nicely like for $$\hat{L}_{z}$$.

As need to find $$\hat{L^{2}}$$ so I cud now square the result, see if that makes a difference, but it looks like it won't be any nicer... nope :|

* * Edit Post * *

I think if I work them out one at a time fullly first that might be better..

From calculations:

$$\hat{L_{x}^{2}} = \hat{L_{y}^{2}} = \hat{L_{z}^{2}} = -4 \hbar^{2} z^{2} \psi$$

Hence:

$$\hat{L^{2}} = -12.\hbar^{2}.z^{2}.\psi$$

Therefore:

$$\lambda = -12 \hbar^{2} z^{2}$$

Last edited: Feb 20, 2010
14. Feb 27, 2010

### gabbagabbahey

Again, $$\mathbf{\hat{L}}=\hat{L}_x\mathbf{i}+\hat{L}_y\mathbf{j}+\hat{L}_z\mathbf{k}[/itex] is a vector operator. There is no such operator as $\hat{L}=\hat{L}_x+\hat{L}_y+\hat{L}_z$. You are being incredibly sloppy with your notation and your calculations here. [tex]\hat{L_{x}}=i\hbar\left(z\frac{\partial}{\partial y}-y\frac{\partial}{\partial z}\right)\neq 2ixz-2yz$$

It is only when you operate $\hat{L}_x$ on the wavefunction you are given that you get something like $2ixz-2yz$. What you should say is that $\hat{L}_x\psi=-2\hbar z(x+iy)$

I'll say this again just to drive home the point; you can't just add these together, they are components of a vector.

I agree, however...

No, you've clearly made some mistakes in your calculations (and you are again failing to distinguish $L^2$ from $L^2\psi$....one is an abstract operator, the other is what you get after you act on the wavefunction $\psi(x,y,z)$ with that abstract operator....which of these are you really calculating here?).

For starters,

$$\hat{L}_{z}^2\psi=\hat{L}_z(\hat{L}_z\psi)=\hat{L}_z(2\hbar\psi)=2\hbar\hat{L}_z\psi=4\hbar^2\psi$$

Second, $\hat{L}_x^2\psi\neq\hat{L}_y^2\psi$ and neither are equal to $-4\hbar^2z^2\psi$

Third;

How is it that you have gotten an eigenvalue which depends on $z$, and hence is not a constant (as an eigenvalue should be); and have not realized that something must be wrong with your calculations?

15. Feb 27, 2010

### Hart

I've think realised the mistakes (hopefully all of them!) , let me try again..

.. after some revised calculations..

$$\hat{L_{z}} = 2\hbar$$

$$\hat{L_{x}} = -2\hbarz(x+iy)$$

$$\hat{L_{y}} = -2\hbarz(x+iy)$$

hence:

$$\hat{L_{z^{2}}} = 4\hbar^{2}$$

$$\hat{L_{x^{2}}} = 4\hbar^{2}z^{2}(x+iy)^{2} = 4\hbar^{2}z^{2}\psi$$

$$\hat{L_{y^{2}}} = 4\hbar^{2}z^{2}(x+iy)^{2} = 4\hbar^{2}z^{2}\psi$$

so:

$$\hat{L^{2}} = 4\hbar^{2} + 2\left(4\hbar^{2}z^{2}\right)$$..

16. Feb 27, 2010

### gabbagabbahey

Again, what you are essentially claiming with this equation is that the z-component of the angular momentum operator always has the value $2\hbar$ times the identity matrix no matter which state it operates on. But that simply isn't true. $$\hat{L_z}\psi=2\hbar\psi$$ does not mean that $$\hat{L_z}=2\hbar$$.

Surely this fixed quote is what you really mean, right?

Again, you can say $$\hat{L}_z^2\psi=4\hbar^2\psi$$, but you cannot claim $$\hat{L}_z^2=4\hbar^2$$. An operator is not equal to one of its eigenvalues.

Even ignoring the missing $\psi$'s on the LHS of these equations, you have some calclulation errors. Keep in mind that both [tex]\hat{L}_x\psi[/itex] and [tex]\hat{L}_y\psi[/itex] have a factor of $z$ in them and so $\frac{\partial}{\partial z}$ of them is not zero....You will need to use the product rule and your results will be messier than that of [tex]L_{z}^2\psi[/itex].