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

and:

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

so:

<br /> <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 /> <br />

?

 

[gabbagabbahey]

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

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]

 

[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 \hat{L}_z[/itex], what equation must be true?

 

[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

\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?

 

[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 /> <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?