Determining Eigenfunction of Operator

[Hart]

Homework Statement

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

Homework Equations

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

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. :|

 

[gabbagabbahey]

Homework Statement

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

Homework Equations

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

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

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

You need to double check this.

 

[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.

 

[gabbagabbahey]

I was meant to state this:

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

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

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

Recheck your calculation.

 

[Hart]

I have:

$$<br /> (x\frac{\partial \psi}{\partial y}) = x(2cy) = 2cxy<br />$$

and:

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

so:

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

?

 

[gabbagabbahey]

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

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

[tex]\frac{\partial\psi}{\partial x}=\frac{\partial}{\partial x}\left(x^2+cy^2\right)=2x[/itex][/tex]

 

[Hart]

It should be:

$$(y\frac{\partial \psi}{\partial x}) = y(2x + 0) = 2xy$$

then.

Ok, so:

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

 

[gabbagabbahey]

Ok, so:

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

Yup, so if $\psi$ is an eigenfunction of [tex]\hat{L}_z[/itex], what equation must be true?[/tex]

 

[Hart]

$$\lambda = \frac{-2i \hbar xy(c-1)}{x^{2}+cy^{2}}?$$

 

[gabbagabbahey]

$$\lambda = \frac{-2i \hbar xy(c-1)}{x^{2}+cy^{2}}?$$

Sure, but I'd write this as

[tex]\lambda\left(x^2+cy^2\right)=-2i \hbar xy(c-1)[/itex]<br /> <br /> 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 $\lambda=0$?[/tex]

 

[Hart]

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

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

 

[gabbagabbahey]

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

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

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

 

[Hart]

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

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

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

$$c = 1<br />$$

 

[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 $y$?

 

[Hart]

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

 

[gabbagabbahey]

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

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 $c$?