4
$\begingroup$

For a particle of mass m, confined to the xy plane and undergoing circular motion with fixed distance to the origin, r and polar angle, $\theta$ held constant at $90^{\circ}$, the Schrödinger equation in spherical coordinates simplifies to

$$\frac{-\hbar^2}{2I}\frac{\partial^2 \Psi}{\partial \varphi^2} = E \Psi$$

The normalized solutions to this equation are

$$\Psi_{m} = \frac{1}{\sqrt{2\pi}}e^{im\varphi}$$

where $m$ is an integer.

The eigenvalues returned by the $\hat{L_{z}}$ and $\hat{L^2}$ operators are

$$\hat{L_{z}}\Psi_{m} = -i\hbar\frac{\partial \Psi_{m}}{\partial \varphi} = m\hbar\Psi _{m}$$

$$\hat{L^2}\Psi_{m} = (\hat{L}_{x}^{2}+\hat{L}_{y}^{2}+\hat{L}_{z}^{2})\Psi_{m} = m^2\hbar^2\Psi _{m}$$

The last equation suggests that the magnitude of angular momentum is $m\hbar$, which, in turn, is equal to the magnitude of the component in the z direction, $m\hbar$. However, both values can only be the same if the components in the $x$ and $y$ direction are both equal to zero, in which case all threee components of the particle's angular momentum are completely defined. But this seems to violate the uncertainty principle for angular momentum, which states that two orthogonal components of angular momentum (for example $\hat{L_{x}}$ and $\hat{L_{y}}$) cannot be simultaneously known and measured with arbitrary precision. I would like to know if there is something wrong with my reasoning here, and if this example is even correctly stated.

Improve this question
$\endgroup$
5
  • 3
    $\begingroup$ "However, both values can only be the same if the components in the x and y direction are both equal to zero" - why? (Don't argue classically, show it in quantum mechanics!) $\endgroup$
    – ACuriousMind ♦
    Commented Feb 8, 2017 at 13:14
  • $\begingroup$ I am relatively new to this field of physics, so i would prefer if you forgave me for any mistakes that i made regarding the quantum mechanical formalism. $\endgroup$
    – Phillip
    Commented Feb 8, 2017 at 13:28
  • $\begingroup$ Moreover, if you Schrödinger equation only contains $\varphi$, you can't use momentum $L^2$. To put it differently, $L^2$ is the basically the Laplacian in spherical coordinates, but your functions $e^{im\varphi}$ are not eigenfunctions of the Laplacian. $\endgroup$
    – ZeroTheHero
    Commented Feb 8, 2017 at 13:34
  • $\begingroup$ @ZeroTheHero you can. It's just the equation in cylindrical coordinates after choosing $\psi(z)=\operatorname{const}.$ and removing the $\rho$-dependent factor. $\endgroup$
    – Ruslan
    Commented Feb 8, 2017 at 14:45
  • 1
    $\begingroup$ @ACuriousMind has given you the answer, but in case the helps to clarify it: The components of angular momentum in the $x$ and $y$ directions do not commute, so they cannot both be well-defined at the same moment. Therefore they certainly cannot both be zero. $\endgroup$
    – WillO
    Commented Mar 20, 2017 at 2:57

1 Answer 1

Reset to default
2
$\begingroup$

The magnitude of $L$ can't be $m \hbar $. To see this note that when you say $magnitude$ what you really mean is eigenvalue. After all in quantum mechanics what we measure in a specific experiment is an eigenvalue of an operator. Plus you have assumed that $\Psi_m$ is an eigenket of $L$. I know this because I can "prove" your statement by doing the following: $$ L^2 \Psi_m = L(L\Psi_m)= m\hbar L \Psi_m = m^2 \hbar^2 \Psi_m $$

But all this can't be true because it would mean that $L$ and $L_z$ both had the same eigenkets and therefore must commute. But this is not true, $[L,L_z] \neq 0$

Improve this answer
$\endgroup$
2
  • $\begingroup$ But isn't eigenvalue of the $\hat{L^2}$ operator is equal to the square of the magnitude of angular momentum of the particle? $\endgroup$
    – Phillip
    Commented Feb 8, 2017 at 14:08
  • $\begingroup$ First, magnitude of vectors do not any sense in quantum mechanics because magnitude requires you to know all three components of the operators simultaneously. Magnitude is a classical concept. Secondly, eigenvalue of $L^2$ is the square only if they have the same eigenkets but they do not. Don't believe me, write out what L is in terms of raising and lowering operators and $L_z$ and you will see. $\endgroup$
    – Amara
    Commented Feb 8, 2017 at 14:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.