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

  • #1
dark_matter_is_neat
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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.
 
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  • #2
Please edit your post and use $$ to bracket your LaTeX expressions to make them more legible.
 
  • #3
Thank you for fixing the LaTeX. How is ##L## defined here?
 
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  • #4
kuruman said:
Thank you for fixing the LaTeX. How is ##L## defined here?
##L=-i\hbar(\vec{r} \times \nabla)## is the angular momentum operator.
 
  • #5
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##.
 
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  • #6
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}}##
 
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  • #7
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.
 
  • #8
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).
 
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  • #9
dark_matter_is_neat said:
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.

dark_matter_is_neat said:
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!)
 
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  • #10
kuruman said:
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}|}##.
 
  • #11
Orodruin said:
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)##.
 
  • #12
dark_matter_is_neat said:
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 …
 
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