GREPhysics.NET: GR9677 #52

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GR9677 #52

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Quantum Mechanics $\Rightarrow$ }Spherical Harmonics

$Y^m_l$ is the convention used for a spherical harmonic of eigenvalue m, l.
The only spherical harmonic proportional to $\sin\theta$ is $Y^{\pm 1}_1$ . Recalling that the eigen-equation, $L_z \psi = m \hbar \psi$ , one deduces that since $m=\pm 1$ , the eigenvalues are $\pm \hbar$ .

(Open call for a better solution: Is there a method that does not require memorizing the first few spherical harmonics?)

Reproduced for the reader's pleasure and convenience, (proportional values of the first few SH's:

$\begin{eqnarray} Y^0_0 & = & const \\ <br /> Y^0_1 & = & \cos\theta \\ Y^{\pm 1}_1 &=& \sin\theta e^{\pm i\phi} <br /> \end{eqnarray}$

See below for user comments and alternate solutions!

Alternate Solutions

raklaksh
2013-10-15 06:40:19

There is a direct method to find 'l' and 'm' for a given wavefunction/spherical harmonic.

Refer D.Griffiths(Introduction to Quantum Mechanics) Eq 4.17 and Eq 4.20.

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Alternate Solution - Unverified

Comments

digitalhikes
2019-07-08 11:05:28

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NEC

Awardr
2017-09-14 20:39:07

This may not be totally correct, but the geometry of the wave function reminded me of the n=2, l=1 state of the hydrogen atom. Not being symmetric along the z axis, m != 0 so m is -1 or 1, as in -h or h

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NEC

ngendler
2015-10-23 03:37:10

You can also just immediately eliminate (A) (Lz isn\'t gonna be 0), and (C) and (D) since l comes in steps of $\\hbar$ . From there, you can just guess or use the methods described in other comments.

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NEC

nachi182
2014-09-23 13:45:38

Here is a pretty simple solution though it requires some knowledge about quantum numbers and spherical harmonics...first of all, Lz will not give you have integers, thus B is out

When l = 0, the spherical harmonics are just some constant (duh, no angular dependence)

When m = 0, spherical harmonics will have no phi dependence. (duh, m=0, so the only angular component will come from l/theta)

From this, you can already determine that the only choices left are C and D. So either m=+/-1 or +/-2. But how do we know? well it all depends on l

In spherical harmonics, l effects theta. But looking at the theta term you can determine what l is. In this case, we have one trig function. Thus l = 1.
If l =2, we would have a 2nd trig function of theta (either cos*sin or sin^2). If l=3 we would have 3 trig functions of theta (either cos^3, or some combo of sin/cos). If you look at list of spherical harmonics you will notice that in it's most factored form, this statement holds.

There is probably a more mathematical way of what I am saying but hopefully you guys get it.

One trig function dependent of theta in our spherical harmonics, thus l = 1. Thus m can equal -1,0,1. We have a phi dependence thus m cannot equal zero. The only choice is C

dipanshugupta
2017-03-29 12:28:54

Really good answer. Thank you!

deneb
2018-10-17 23:26:06

Just to add on how to figure out m, here\'s what i came up with after:\r\n\r\nIn Griffith\'s QM book, he gives a generalized equation of normalized spherical harmonics (too messy to type out here). In it you\'ll realize the only phi dependent term is e^i*m*phi, so whatever is in front of the phi is equal to m. In this case m=1. If it was sin(2*phi), m would be 2. \r\n\r\nThoughts?

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NEC

raklaksh
2013-10-15 06:40:19

There is a direct method to find 'l' and 'm' for a given wavefunction/spherical harmonic.

Refer D.Griffiths(Introduction to Quantum Mechanics) Eq 4.17 and Eq 4.20.

Reply to this comment

Alternate Solution - Unverified

CarlBrannen
2010-10-07 15:32:46

Given our wave function, we wish to know which z-component angular momentum $m$ it can contain. To do this, we need to know if $\langle \psi , \psi_m \rangle = 0$ for all possible values of $m$ .

You are probably expected to know that $\psi_m$ is proportional to $\exp(im\phi) = \cos(m\phi) + i\sin(m\phi)$ . So we need to know if
$\langle \psi , \psi_m \rangle = \int_0^{2\pi} \sin(\phi) (\cos(m\phi) + i\sin(m \phi) ) d\phi = 0$ .

For $m=0$ you are integrating a single copy of $sin(\phi)$ and get zero. This eliminates (A) and (E). For $m=\pm 1$ the integrals include $\cos^2$ or $\sin^2$ and so is non zero. The answer is therefore (C).

Note: You already know that $\exp(\pm i m\phi)$ form a complete set of orthogonal functions corresponding to angular momentum in the z direction. These are linear superpositions of the two sets $\cos(m\phi)$ and $\sin(m\phi)$ , other than the trivial sine when m is zero. Therefore
$\int_0^{2\pi} \sin(\phi)\cos(m\phi)$ and
$\int_0^{2\pi} \sin(\phi)\sin(m\phi)$ and
$\int_0^{2\pi} \cos(\phi)\cos(m\phi)$ and
$\int_0^{2\pi} \cos(\phi)\sin(m\phi)$ are non zero only for $m=\pm 1$ , and even when $m= \pm 1$ are zero unless you're looking at $\cos^2$ and $\sin^2$ . It's also useful to know that the average values of $\cos^2$ and $\sin^2$ are 1/2.

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NEC

WoolfianOperator
2009-11-05 19:11:27

i think im losing it. im trying to find the expectation value of $\ L_z$ . $\langle \psi \ L_z \psi \rangle$ . Using $\ L_z$ =-i $\hbar$ $\frac{\partial}\partial_\phi$ . applying the operator to $\psi$ one gets a cos( $\phi$ ) term. But then one has an integral from 0 to 2pi of sin( $\phi$ )*cos $\phi$ ), which is zero. I understand the other method, but why is this not appear to work?

WoolfianOperator
2009-11-06 15:31:48

ok. so the problem wants to know the possible outcomes. What i calculated above is the expectation value, which is the average of the possible outcomes. In order to find the possible outcomes we are really going to find the eigenvalues of the $\ L_z$ operator. One can do this using Yosun's method which involves Eulers formula. I guess my point is that one should remember that the expectation value and eigenvalues are not the same thing.

maxdp
2013-09-26 09:38:07

Ahh thank you. I was wondering the same thing.

Expectation values =/= possible values. Good to remember.

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NEC

wittensdog
2009-11-03 19:53:17

The more I think about this problem, the more I'm convinced that Void is getting at what ETS is looking for. For spherical harmonics, all of the phi dependence comes in the form of Fourier modes exp(i * m * phi). So if you can break a state down into its Fourier decomposition in phi, you can immediately read off what the values of m are. This is pretty easy to do with sin ( phi ), since you can just use Euler's formula. It would also work very easily for something like sin ( 2 * phi ), in which case you would have m = +/- 2.

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NEC

sullx
2009-11-03 18:44:16

Why is it that the given rigid rotator state is not even included in the above solution(s)?

One could just apply the Lz operator to the given function and equate it with the function multiplied by its eigenvalue m*h-bar.

This results in m depending on sin(PHI) which is a superposition of TWO complex functions (see the last line of void's explination below).

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NEC

tin2019
2007-10-18 07:50:16

Well I solved it this way: Apply the z-component operator twice to the given function $-\hbar^2\partial^2/\partial\phi$ . Now this is obviously not an eigenfunction of the z component, but a superposition of two such functions. Nonetheless it is an eigenfunction of $L_z^2$ operator which yields that $L_z^2\psi = \hbar^2\psi$ and nothing else. Square root will give correct answers (although I think that strictly speaking this is not entirely justified).

syreen
2013-10-16 11:48:41

I don't know if this is right, but I like it!

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NEC

yosun
2005-11-10 11:40:22

void: surround your LaTeX commands with $dollar signs$ ... not \[slash-brackets]\... i only programmed the parser to recognize $ signs---as the LaTeX Rosetta Stone below explains. (your message has been manually re-parsed. but, use $ signs in the future!)

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NEC

Void
2005-11-10 06:41:12

Let's say you forget the first few spherical harmonics. (Happens to the best of us.) Hopefully you could remember that:

1) The eigenstates of the z-component of angular momentum are given by

$L_zY^m_l = m\hbar Y^m_l$

2) You'd remember the form of the angular momentum operator in the z-direction as well:

$L_z = -i\hbar \frac{\partial}{\partial \phi}$

3) Combining the two, you see that the part of the spherical harmonics that contains m (besides the normalization) is proportional to $e^{im\phi}$ . I think this is pretty easy and important to memorize.

The trick here is to realize in the given wave function, the information about m is contained in the $\sin \phi$ term. Ignore the others. Then remember back from complex analysis that

$\sin\phi = \frac{e^{i\phi} - e^{-i\phi}}{2i}$ Thus, the wave function is in a superposition of states m = 1 and m = -1. The answer is C.

testtest
2010-11-09 21:30:09

Definitely the best way to do it. Just try to think of which of the terms is an eigenfunction of the $L_z$ operator, and you really should start thinking about the trig identity immediately.

Baharmajorana
2014-09-15 00:34:59

Yes I agree with u, thanks for ur best sulotion

ryanm
2018-10-23 02:56:31

This is the solution that I used and it worked. It is simple because it is easy to take the derivative.

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NEC

Void
2005-11-10 06:19:29

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NEC

Post A Comment!

You are replying to:

Given our wave function, we wish to know which z-component angular momentum $m$ it can contain. To do this, we need to know if $\langle \psi , \psi_m \rangle = 0$ for all possible values of $m$ . You are probably expected to know that $\psi_m$ is proportional to $\exp(im\phi) = \cos(m\phi) + i\sin(m\phi)$ . So we need to know if $\langle \psi , \psi_m \rangle = \int_0^{2\pi} \sin(\phi) (\cos(m\phi) + i\sin(m \phi) ) d\phi = 0$ . For $m=0$ you are integrating a single copy of $sin(\phi)$ and get zero. This eliminates (A) and (E). For $m=\pm 1$ the integrals include $\cos^2$ or $\sin^2$ and so is non zero. The answer is therefore (C). Note: You already know that $\exp(\pm i m\phi)$ form a complete set of orthogonal functions corresponding to angular momentum in the z direction. These are linear superpositions of the two sets $\cos(m\phi)$ and $\sin(m\phi)$ , other than the trivial sine when m is zero. Therefore $\int_0^{2\pi} \sin(\phi)\cos(m\phi)$ and $\int_0^{2\pi} \sin(\phi)\sin(m\phi)$ and $\int_0^{2\pi} \cos(\phi)\cos(m\phi)$ and $\int_0^{2\pi} \cos(\phi)\sin(m\phi)$ are non zero only for $m=\pm 1$ , and even when $m= \pm 1$ are zero unless you're looking at $\cos^2$ and $\sin^2$ . It's also useful to know that the average values of $\cos^2$ and $\sin^2$ are 1/2.

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