# Determining Eigenfunction of Operator

1. Feb 20, 2010

### Hart

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

Determing the constant c such that $$\psi_{c}(x,y,z) = x^{2}+cy^{2}$$ is an eigenfunction of $$\hat{L_{z}}$$

2. Relevant equations

$$\hat{L_{z}} = -i \hbar (x\frac{\partial \psi}{\partial y} - y\frac{\partial \psi}{\partial x}$$

3. The attempt at a solution

$$x\frac{\partial \psi}{\partial y} - y\frac{\partial \psi}{\partial x}) = 2x^{2}-2y^{2}c$$

Therefore:

$$\hat{L_{z}} \psi = -i \hbar (2x^{2}-2y^{2}c) = -2i \hbar (x^{2}-y^{2}c)$$

.. and now I'm stuck. :|

2. Feb 21, 2010

### gabbagabbahey

You need to pay more attention to your notation. You can either say,

$$\hat{L_{z}}\psi(x,y,z)= -i \hbar \left(x\frac{\partial \psi}{\partial y} - y\frac{\partial \psi}{\partial x}\right)$$

or

$$\hat{L_{z}} \longrightarrow -i \hbar \left(x\frac{\partial}{\partial y} - y\frac{\partial}{\partial x}\right)$$

But you can't equate an abstract differential operator to a scalar function like you did above

You need to double check this.

3. Feb 21, 2010

### Hart

I was meant to state this:

$$\hat{L_{z}}\psi(x,y,z)= -i \hbar \left(x\frac{\partial \psi}{\partial y} - y\frac{\partial \psi}{\partial x}\right)= -i \hbar (2x^{2}-2y^{2}c) = -2i \hbar (x^{2}-y^{2}c)$$

[/tex]

I don't know how to rearrange that the find the value of C.

4. Feb 21, 2010

### gabbagabbahey

$$\left(x\frac{\partial \psi}{\partial y} - y\frac{\partial \psi}{\partial x}\right)\neq (2x^{2}-2y^{2}c)$$

5. Feb 21, 2010

### Hart

I have:

$$(x\frac{\partial \psi}{\partial y}) = x(2cy) = 2cxy$$

and:

$$(y\frac{\partial \psi}{\partial x}) = y(2x + c) = 2cxy$$

so:

$$\hat{L_{z}}\psi(x,y,z) = -i \hbar \left(x\frac{\partial \psi}{\partial y} - y\frac{\partial \psi}{\partial x}\right) = -i \hbar (2cxy - 2cxy) = 0$$

?

6. Feb 21, 2010

### gabbagabbahey

Where's the $c$ in this part coming from?

$$\frac{\partial\psi}{\partial x}=\frac{\partial}{\partial x}\left(x^2+cy^2\right)=2x[/itex] 7. Feb 21, 2010 ### Hart It should be: [tex](y\frac{\partial \psi}{\partial x}) = y(2x + 0) = 2xy$$

then.

Ok, so:

$$\hat{L_{z}}\psi(x,y,z)= -i \hbar \left(x\frac{\partial \psi}{\partial y} - y\frac{\partial \psi}{\partial x}\right)= -i \hbar (2cxy - 2xy)= -2i \hbar xy(c-1)$$ ??

8. Feb 21, 2010

Yup, so if $\psi$ is an eigenfunction of $$\hat{L}_z[/itex], what equation must be true? 9. Feb 21, 2010 ### Hart [tex]\lambda = \frac{-2i \hbar xy(c-1)}{x^{2}+cy^{2}}?$$

10. Feb 21, 2010

### gabbagabbahey

Sure, but I'd write this as

$$\lambda\left(x^2+cy^2\right)=-2i \hbar xy(c-1)[/itex] At first, it may not look like there are any values of $c$ that will make this true for all $x$ and $y$, but what if $\lambda=0$? 11. Feb 21, 2010 ### Hart [tex]\lambda\left(\psi_{c}\right)=-2i \hbar xy(c-1)$$

It would be true if $$\lambda = 0$$?

12. Feb 21, 2010

### gabbagabbahey

You tell me....if $\lambda=0$ what does that equation become? Are there any values of $c$ that make that equation true?

13. Feb 21, 2010

### Hart

$$\lambda\left(\psi_{c}\right)=-2i \hbar x y (c-1) = -2i \hbar x y c + 2i \hbar x y$$

$$0 = -2i \hbar x y c + 2i \hbar x y$$

$$2i \hbar x y c = 2i \hbar x y$$

$$c = 1$$

14. Feb 21, 2010

### gabbagabbahey

Right, so for c=1, $\psi$ is an eigenfunction of $\hat{L}_z$ with corresponding eigenvalue $\lambda=0$.

If $\lambda\neq 0$ are there any values of $c$ which satisfy the eigenvalue equation for all $x$ and $y$?

15. Feb 21, 2010

### Hart

Um.. c can be any integer value? it's just a scaling value.

16. Feb 26, 2010

### gabbagabbahey

So, you are telling me that $\lambda\left(x^2+cy^2\right)=-2i \hbar xy(c-1)$ is satisfied for any integer value of $c$???