Determine unit vector n such that (L dot n)psi(r) = (m*hbar)psi(r)

[dark_matter_is_neat]

Homework Statement

For $\psi(\vec{r}) = (x+y+3z)f(r)$ where $r = \sqrt{x^{2}+y^{2}+z^{2}}$
Determine $\hat{n}$ and m such that $(\hat{n} \cdot L) \psi(r) = m\hbar\psi(r)$ Where $L=-i\hbar(\vec{r} \times \nabla)$ is the angular momentum operator.

Relevant Equations

$\psi(\vec{r}) = (x+y+3z)f(r)$

First I calculated $(\vec{n} \cdot L) \psi(r) = -i\hbar(n_{x}(3y-z)+n_{y}(z-3x)+n_{z}(x-y))f(r)$ and then tried to solve for $n_{i}$ such that I get (x+y+3z)f(r), and then divide $n_{i}$ by the magnitude of $\vec{n}$ to get the unit vector and m, but when I try doing this, I get the system of equations: $3n_{x}-n_{z} = 1$, $n_{y}-n_{x} = 3$ and $n_{z}-3n_{y} = 1$, but there is no solution to this system of equations.

I'm not sure how else I would determine $\hat{n}$ and m.

 

[kuruman]

Please edit your post and use $$ to bracket your LaTeX expressions to make them more legible.

 

[kuruman]

Thank you for fixing the LaTeX. How is $L$ defined here?

 

[dark_matter_is_neat]

Thank you for fixing the LaTeX. How is $L$ defined here?

$L=-i\hbar(\vec{r} \times \nabla)$ is the angular momentum operator.

 

[kuruman]

I would rewrite the given wavefunction as a linear combination of spherical harmonics $Y_1^m$ and then operate on it with $n_xL_x+n_yL_y+n_zL_z.$ In other words, treat it like a hydrogen atom wavefunction in the form $$\psi(\vec r)=f(r)\sum_{m} A_{m}Y_1^m.$$Note that in spherical coordinates $x=r\sin\theta\cos\phi \implies x \propto (Y_1^1+Y_1^{- 1} )~$ and similarly for $y$ and $z$.

 

[dark_matter_is_neat]

So rewriting $\psi(\vec{r})$, $\psi(\vec{r}) = rf(r)\sqrt{\frac{2\pi}{3}}[(i-1)Y^{1}_{1} + \frac{6}{\sqrt{2}}Y^{0}_{1} + (1+i)Y^{-1}_{1}]$
Applying $n_{x}L_{x}$ to this I get:
$n_{x}rf(r)\hbar\sqrt{\frac{2\pi}{3}}(\sqrt{2}iY^{0}_{1} + \frac{6}{\sqrt{2}}(Y^{1}_{1}+Y^{-1}_{1}))$
Applying $n_{y}L_{y}$ to this I get:
$n_{y}rf(r)\hbar\sqrt{\frac{2\pi}{3}}(-\sqrt{2}iY^{0}_{1} + \frac{6}{\sqrt{2}}i(Y^{1}_{1}-Y^{-1}_{1}))$
Applying $n_{z}L_{z}$ to this I get:
$n_{z}rf(r)\hbar\sqrt{\frac{2\pi}{3}}((i-1)Y^{1}_{1}-(i+1)Y^{-1}_{1})$

Matching factors in front of each spherical harmonic in the original expression and the new expression (and ignoring $\hbar$, since the final answers will just be divided by it), I get this system of equations:
$\sqrt{2}in_{x}-\sqrt{2}in_{y} = \frac{6}{\sqrt{2}}$
$3n_{x}+3in_{y}+(i-1)n_{z} = (i-1)$
$3n_{x}-3in_{y}-(i+1)n_{z} = (i+1)$

The solution to this system of equations is:
$n_{x} = -\frac{7i}{6}$
$n_{y} = \frac{11i}{6}$
$n_{z} = -\frac{9i}{2}$

To get $\hat{n}$ and m
I pull out the factor of $i$ in each $n_{i}$ to get:
$n_{x} = -\frac{7}{6}$
$n_{y} = \frac{11}{6}$
$n_{z} = -\frac{9}{2}$
Then just divide $n_{i}$ by $\sqrt{n_{x}^{2}+n_{y}^{2}+n_{z}^{2}}$ .
Then m = $i\sqrt{n_{x}^{2}+n_{y}^{2}+n_{z}^{2}}$

 

[kuruman]

According to the statement of the problem $\hat n$ is a unit vector and cannot have a magnitude greater than one which is what your solution indicates. I think that the source of the problem is that the angular part of the wavefunction is not normalized. Note that $$\begin{align} & \left[(i-1)Y^{1}_{1} + \frac{6}{\sqrt{2}}Y^{0}_{1} + (1+i)Y^{-1}_{1}\right]=\left[-\sqrt{2}~e^{(-i\pi/4)}Y^{1}_{1} + \frac{6}{\sqrt{2}}Y^{0}_{1} + \sqrt{2}~e^{(i\pi/4)}Y^{-1}_{1}\right] \nonumber \\ & =N \left[-e^{(-i\pi/4)}Y^{1}_{1} +3Y^{0}_{1} + e^{(i\pi/4)}Y^{-1}_{1}\right] .\nonumber
\end{align}$$ Finding the normalization constant $N$ such that $$1=N^2\int \left[-e^{(-i\pi/4)}Y^{1}_{1} +3Y^{0}_{1} + e^{(i\pi/4)}Y^{-1}_{1}\right]^*\left[-e^{(-i\pi/4)}Y^{1}_{1} +3Y^{0}_{1} + e^{(i\pi/4)}Y^{-1}_{1}\right]d\Omega$$is easy. Using the orthonormality of the spherical harmonics, it follows that $N=\sqrt{\frac{1}{11}}.$ Finally, use the normalized expression $$\varphi=\sqrt{\frac{1}{11}}\left[-e^{(-i\pi/4)}Y^{1}_{1} +3Y^{0}_{1} + e^{(i\pi/4)}Y^{-1}_{1}\right]$$ to find $(\hat{n} \cdot\vec L) \varphi$ and proceed as before.

 

[Orodruin]

You can, of course, brute force it. However, you will get a long way here by considering $\psi(\vec r) = (\vec v \cdot \vec r) f(r)$ and applying symmetry arguments (or simply some basic vector computations).

 

[Orodruin]

First I calculated $(\vec{n} \cdot L) \psi(r) = -i\hbar(n_{x}(3y-z)+n_{y}(z-3x)+n_{z}(x-y))f(r)$ and then tried to solve for $n_{i}$ such that I get (x+y+3z)f(r),

But this is not what you should get, you should get $-i\hbar m(x+y+3z)f(r)$, for all combinations of $x,y,z$. You cannot try to solve the above first and then normalise to get $m$, the $m$ needs to be in there from the start.

and then divide $n_{i}$ by the magnitude of $\vec{n}$ to get the unit vector and m, but when I try doing this, I get the system of equations: $3n_{x}-n_{z} = 1$, $n_{y}-n_{x} = 3$ and $n_{z}-3n_{y} = 1$, but there is no solution to this system of equations.

So here you would instead get $3n_x - n_z = m$, $n_y - n_x = 3m$ and $n_z - 3n_y = m$. Does this system have any solutions? (Note that $m$ should also be determined!)

 

[dark_matter_is_neat]

According to the statement of the problem $\hat n$ is a unit vector and cannot have a magnitude greater than one which is what your solution indicates. I think that the source of the problem is that the angular part of the wavefunction is not normalized. Note that $$\begin{align} & \left[(i-1)Y^{1}_{1} + \frac{6}{\sqrt{2}}Y^{0}_{1} + (1+i)Y^{-1}_{1}\right]=\left[-\sqrt{2}~e^{(-i\pi/4)}Y^{1}_{1} + \frac{6}{\sqrt{2}}Y^{0}_{1} + \sqrt{2}~e^{(i\pi/4)}Y^{-1}_{1}\right] \nonumber \\ & =N \left[-e^{(-i\pi/4)}Y^{1}_{1} +3Y^{0}_{1} + e^{(i\pi/4)}Y^{-1}_{1}\right] .\nonumber
\end{align}$$ Finding the normalization constant $N$ such that $$1=N^2\int \left[-e^{(-i\pi/4)}Y^{1}_{1} +3Y^{0}_{1} + e^{(i\pi/4)}Y^{-1}_{1}\right]^*\left[-e^{(-i\pi/4)}Y^{1}_{1} +3Y^{0}_{1} + e^{(i\pi/4)}Y^{-1}_{1}\right]d\Omega$$is easy. Using the orthonormality of the spherical harmonics, it follows that $N=\sqrt{\frac{1}{11}}.$ Finally, use the normalized expression $$\varphi=\sqrt{\frac{1}{11}}\left[-e^{(-i\pi/4)}Y^{1}_{1} +3Y^{0}_{1} + e^{(i\pi/4)}Y^{-1}_{1}\right]$$ to find $(\hat{n} \cdot\vec L) \varphi$ and proceed as before.

I ran through the math and you're just going to end up with essentially the same system of equations as before, since $\varphi$ is the same thing as $\psi$ other than for some constant which doesn't affect how $\hat{n} \cdot L$ acts on it.

The magnitude of $|\vec{n}|$ doesn't equal one, but you can easily get a vector that does have a magnitude of 1 and does what you want when dotted with L by just calculating $\frac{\vec{n}}{|\vec{n}|}$.

 

[dark_matter_is_neat]

So here you would instead get $3n_x - n_z = m$, $n_y - n_x = 3m$ and $n_z - 3n_y = m$. Does this system have any solutions? (Note that $m$ should also be determined!)

By looking adding equation 1 to equation 3 you get:
$3n_{x}-3n_{y} = 2m$ which multiplying each side by -1 is $3n_{y}-3n_{x} = -2m$. The only way for this equation to be consistent with equation 2 is for m=0.

So setting m = 0, I get that $n_{x} = n_{y}$ and $n_{z} = 3n_{x} = 3n_{y}$. Choosing $n_{z} = 3$, I get that the vector is (1, 1, 3) which equals $\frac{1}{\sqrt{11}}$(1,1,3) when normalized. So then m = 0 and $\hat{n} = \frac{1}{\sqrt{11}}(1,1,3)$.

 

[Orodruin]

By looking adding equation 1 to equation 3 you get:
$3n_{x}-3n_{y} = 2m$ which multiplying each side by -1 is $3n_{y}-3n_{x} = -2m$. The only way for this equation to be consistent with equation 2 is for m=0.

So setting m = 0, I get that $n_{x} = n_{y}$ and $n_{z} = 3n_{x} = 3n_{y}$. Choosing $n_{z} = 3$, I get that the vector is (1, 1, 3) which equals $\frac{1}{\sqrt{11}}$(1,1,3) when normalized. So then m = 0 and $\hat{n} = \frac{1}{\sqrt{11}}(1,1,3)$.

Yes, this a valid solution. More generally and easier to see is the following:

Assuming that $\psi = (\vec v \cdot \vec r) f(r)$ for some constant $\vec v$ implies that
$$
\nabla\psi = f(r) \nabla(\vec v\cdot \vec r) + (\vec v\cdot \vec r) f'(r) \nabla r = f(r) \vec v + (\vec v\cdot \vec r) f'(r) \vec e_r
$$
We directly find that
$$
\hat L \psi = -i\hbar (\vec r \times \nabla\psi) = -i\hbar f(r) \vec r \times \vec v
$$
since $\vec r \times \vec e_r = 0$. Consequently,
$$
(\vec n \cdot \hat L)\psi = -i\hbar f(r) \vec n \cdot (\vec r \times \vec v)
= - i\hbar f(r) \vec v \cdot (\vec n \times \vec r)
$$
This is proportional to $\psi$ with $m \neq 0$ only if $\vec n \times \vec r \propto \vec r$ which is false. We conclude that $m = 0$.

For $m = 0$ we must have $\vec v \cdot(\vec n \times \vec r) = \vec r \cdot (\vec v \times \vec n) = 0$ for arbitrary $\vec r$, ie, if $\vec v \times \vec n = 0$. With both $\vec v$ and $\vec n$ being non-zero, this occurs if and only if $\vec n \propto \vec v$ and, since $\vec n$ is a unit vector,
$$
\vec n = \vec v/v
$$

In your case $\vec v = \vec e_x + \vec e_y + 3 \vec e_z$

Edit: I totally did not have this post prepared since 10 hours ago just waiting for the OP to solve the problem. Totally …