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At $t=0$, the wave function of a particle with Hamiltonian $$\mathcal{H}=\mu B L_y \equiv \omega L_y$$ is given by $$\left \langle \mathbf{r}|\alpha \right \rangle \equiv \psi\left ( \mathbf{r} \right )= f\left ( r \right )\left ( r+z \right ).$$ Write the time evolution of the state.

I tried to write everything in Dirac notation; so, since $$r+z=r\left ( 1+\frac{z}{r} \right )=r\sqrt{4\pi}\left ( Y_0^0 + \frac{1}{\sqrt{3}}Y_1^0\right ),$$ putting $R(r)\equiv rf(r)$, I wrote the normalized ket in the $\left \{ \left|L,L_z\right\rangle \right \}$ basis as $$\left|\psi \right \rangle = \frac{\sqrt{3}}{2}R\left ( r \right )\left [ \left|00\right \rangle + \frac{1}{\sqrt{3}}\left|10\right \rangle \right ].$$ I've got trouble in writing $\left|\psi (t) \right\rangle$.

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    $\begingroup$ It would be better to re-express $\vert 00\rangle$ and $\vert 1,0\rangle$ in a basis of eigenstates of $L_y$; in this basis the evolution is trivial. You can convert back to the original basis after you have evolved your system. $\endgroup$ – ZeroTheHero Jan 8 '18 at 18:52
  • $\begingroup$ @ZeroTheHero, really sorry to bother you again. How can i re-express $\left| 10 \right\rangle$ and $\left| 00 \right\rangle$ in the basis of eigenstates of $L_y$? $\endgroup$ – Vincenzo Ventriglia Jan 8 '18 at 19:44
  • $\begingroup$ @ZeroTheHero, I tried to write $L_z$ eigenkets in terms of $L_y$ eigenkets. If I'm right, it's $$\left|10_z\right \rangle = -i\frac{\sqrt{3}}{2}\left [ \left|11_y \right\rangle - \left|1-1_y\right\rangle \right ].$$ Now, I don't know how to write $\left|00_z\right \rangle$: is it correct to put $$\left|00_z\right \rangle = \left|00_y\right \rangle?$$ $\endgroup$ – Vincenzo Ventriglia Jan 8 '18 at 21:46
  • $\begingroup$ Looks about right although I did not check explicitly. The $\vert 10\rangle$ probably has $1/\sqrt{2}$ rather than the factor you have, which gives a non-normalized state. $\endgroup$ – ZeroTheHero Jan 8 '18 at 23:31
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This is but a classical rotation around the y-axis masquerading as a quantum problem. It amounts to rotation of z to x and back.

The evolution operator is $$\exp (-itH/\hbar)= \exp (-it\omega L_y/\hbar)=\exp(\theta K)\equiv R(\theta),$$ an orthogonal 3d rotation matrix, where $\theta= t\omega/\hbar$ and the Hermitean spin (-one) matrix generator $L_y$ is proportional to the Cartesian antisymmetric rotation generator $K_y$, $$ L_y= i \left[\begin{array}{ccc} 0 & 0 & 1 \\ 0 & 0 & 0\\ -1 & 0 & 0 \end{array}\right] =iK_y. $$

By the Rodrigues rotation formula, by inspection, $$ R(\theta)= I + (\sin\theta) K_y + (1-\cos\theta) K_y ^2= \left[\begin{array}{ccc} \cos \theta & 0 & \sin \theta \\ 0 & 1 & 0\\ -\sin \theta & 0 & \cos \theta \end{array}\right]. $$

This is just a busy way of promoting (x,z) to $(x\cos \theta +z\sin\theta, z\cos\theta -x\sin\theta )$, and of course leaving the scalar r alone. Thus, the future wave function is $f(r)(r+z \cos\theta -x\sin \theta)$, with θ as defined above.

  • To check unitarity, indeed, orthogonality, of R, simply note $K_y^3=-K_y$, and $K_y^4=-K_y^2$, so $R R^T=\mathbb{1}$.

  • NB. The reason I use this classical-friendly Cartesian basis for the Ls and not the strictly equivalent more conventional QM spherical tensor one, T , is because the eigenvectors of the Ls are far more evident, instantly specified by the components of (x,y,z), just as they are in sophomore physics.

  • If you however insist on sticking to the inferior spherical basis, your evolution matrix turns into a monster, $$ \left[\begin{array}{ccc} (1+\cos\theta)/2 & -\sin \theta /\sqrt{2} & (1-\cos\theta)/2 \\ \sin \theta /\sqrt{2} & \cos\theta & -\sin \theta /\sqrt{2} \\ (1-\cos\theta)/2 & \sin \theta /\sqrt{2} & (1+\cos\theta)/2 \end{array}\right], $$ orthogonal and collapsing to the identity at t=0, naturally!

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  • $\begingroup$ I followed your idea. So $$\exp{\left\{-\frac{i}{\hbar}L_y\omega t \right\}}= 1\!\!1 -\left (\frac{L_y}{\hbar}\right)^2\left (1-\cos\omega t \right) -i\left (\frac{L_y}{\hbar} \right )\sin\omega t;$$ using $$L_y =\frac{i\hbar}{\sqrt2}\begin{bmatrix} 0 & -1 &0 \\ 1 &0 &-1 \\ 0& 1 &0 \end{bmatrix},$$ I get $$\exp{\left\{-\frac{i}{\hbar} L_y\omega t \right\}}=-\frac{1}{\sqrt2}\left|11\right\rangle\sin\omega t +\left|10\right\rangle\cos\omega t+\frac{1}{\sqrt2}\left|1-1\right\rangle\sin\omega t.$$ The problem, now, is that $\left|\psi\right\rangle$ is no longer normalized. $\endgroup$ – Vincenzo Ventriglia Jan 12 '18 at 10:59
  • $\begingroup$ How can a unitary rotation send a normalized state to an unnormalized one? I think your problem is in our translation to kets. I gave you the expression in terms of the wavefunction above the bullet item. $\endgroup$ – Cosmas Zachos Jan 12 '18 at 13:02
  • $\begingroup$ Wait wait! the $|\psi\rangle$ the evolution operator acts on also contains the singlet $|00\rangle$ which evolves by multiplication by 1--it stays put. What you wrote down is just the evolute of $|1,0\rangle$ which evolves unitarily, in its subspace. So what you wrote, with the initial t=0 normalization is very normalized... it has to be, as the evolution operator is an orthogonal/unitary one! $\endgroup$ – Cosmas Zachos Jan 12 '18 at 17:49
  • $\begingroup$ ?? you get the second, parenthesis, term to be $\frac{1}{3} ( \sin^2 /2 + \cos^2 + \sin^2/2)=1/3$, so what's inside the square bracket is 4/3 .... Your can't cheat unitarity! $\endgroup$ – Cosmas Zachos Jan 12 '18 at 19:27

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