Proving an Eigenfunction of Momentum Operators

[Hart]

Homework Statement

Homework Equations

Stated in the question.

The Attempt at a Solution

It is a eigenfunction of L_z as it has no dependence 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 can't understand how to do it with these operators.

 

[Mark44]

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

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

 

[Hart]

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

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

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 dependence 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.

 

[gabbagabbahey]

Well, what is [tex]\hat{L}_z[/itex] in Cartesian coordinates?[/tex]

 

[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.

 

[Hart]

$$L_{z} = -ih(x\frac{\partial}{\partial y}-y\frac{\partial}{\partial z})$$

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

 

[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.

 

[gabbagabbahey]

$$L_{z} = -ih(x\frac{\partial}{\partial y}-y\frac{\partial}{\partial z})$$

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

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

 

[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!

 

[gabbagabbahey]

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!

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

 

[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 dependence 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?

 

[gabbagabbahey]

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?

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 $\hat{\mathbf{L}}$.

 

[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 $$\hat{L_{z}}$$ separately 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:

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

 

[gabbagabbahey]

I've just calculated the result:

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

Again, [tex]\mathbf{\hat{L}}=\hat{L}_x\mathbf{i}+\hat{L}_y\mathbf{j}+\hat{L}_z\mathbf{k}[/itex] is a <b>vector</b> operator. There is no such operator as $\hat{L}=\hat{L}_x+\hat{L}_y+\hat{L}_z$.<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> $$\hat{L_{x}} = 2ixz-2yz$$<br /> <br /> $$\hat{L_{y}} = 2xz+2iyz$$<br /> <br /> $$\hat{L_{z}} = 2ix^{2}-4xy-2iy^{2}$$ </div> </div> </blockquote><br /> You are being incredibly sloppy with your notation and your calculations here.<br /> <br /> $$\hat{L_{x}}=i\hbar\left(z\frac{\partial}{\partial y}-y\frac{\partial}{\partial z}\right)\neq 2ixz-2yz$$<br /> <br /> It is only when you operate $\hat{L}_x$ on the wavefunction you are given that you get something like $2ixz-2yz$. What you <i>should</i> say is that $\hat{L}_x\psi=-2\hbar z(x+iy)$<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> .. and then adding the three results together. Not forgetting the $$-i\hbar$$ factor of course. </div> </div> </blockquote><br /> I'll say this again just to drive home the point; you can't just add these together, they are components of a vector.<br /> <br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> * * Edit Post * * <br /> <br /> I think if I work them out one at a time fullly first that might be better.. </div> </div> </blockquote><br /> I agree, however...<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> From calculations:<br /> <br /> $$\hat{L_{x}^{2}} = \hat{L_{y}^{2}} = \hat{L_{z}^{2}} = -4 \hbar^{2} z^{2} \psi$$ </div> </div> </blockquote><br /> 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?). <br /> <br /> For starters, <br /> <br /> $$\hat{L}_{z}^2\psi=\hat{L}_z(\hat{L}_z\psi)=\hat{L}_z(2\hbar\psi)=2\hbar\hat{L}_z\psi=4\hbar^2\psi$$<br /> <br /> Second, $\hat{L}_x^2\psi\neq\hat{L}_y^2\psi$ and neither are equal to $-4\hbar^2z^2\psi$<br /> <br /> Third;<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Hence:<br /> <br /> $$\hat{L^{2}} = -12.\hbar^{2}.z^{2}.\psi$$<br /> <br /> Therefore:<br /> <br /> $$\lambda = -12 \hbar^{2} z^{2}$$ </div> </div> </blockquote><br /> 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?[/tex]

 

[Hart]

I've think realized 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:

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

 

[gabbagabbahey]

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

.. after some revised calculations..

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

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 $$\hat{L_z}=2\hbar$$.

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

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

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

hence:

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

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.

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

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

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].[/tex][/tex][/tex]