Physics Asked on December 24, 2020
I know how to write the matrix for the operator $S_y$ in the ${|{↑}_zrangle , |{↓}_zrangle}$ basis, but I don’t understand how to write it in the ${|{↑}_xrangle , |{↓}_xrangle}$ basis.
Wouldn’t the calculation become very complex? Any help will be appreciated.
You know what $S_y$ looks like in the $z$ basis, and you should know what $alpha$ and $beta$ are in $$ |{uparrow}_xrangle =alpha |{uparrow}_zrangle+beta|{downarrow}_zrangle $$ so just plug into $$ langle {uparrow}_x|S_y| {uparrow}_xrangle= [alpha^*,beta^*] left[matrix{0&-icr i&0}right]left[matrix{alphacr beta}right]. $$ and similarly for the other three entries.
Correct answer by mike stone on December 24, 2020
The four matrix elements you are looking for are $$ langle {uparrow}_{x} |hat{S}_{y}| {uparrow}_{x} rangle, langle {uparrow}_{x} |hat{S}_{y}| {downarrow}_{x} rangle, langle {downarrow}_{x} |hat{S}_{y}| {uparrow}_{x} rangle, langle {downarrow}_{x} |hat{S}_{y}| {downarrow}_{x} rangle. $$
As you probably know, begin{align} | {uparrow}_{x} rangle=frac{1}{sqrt{2}}(| {uparrow}_{z} rangle+| {downarrow}_{z} rangle), | {downarrow}_{x} rangle=frac{1}{sqrt{2}}(| {uparrow}_{z} rangle-| {downarrow}_{z} rangle) end{align} and begin{align} | {uparrow}_{y} rangle=frac{1}{sqrt{2}}(| {uparrow}_{z} rangle+i| {downarrow}_{z} rangle), | {downarrow}_{y} rangle=frac{1}{sqrt{2}}(| {uparrow}_{z} rangle-i| {downarrow}_{z} rangle). end{align}
Play around with the equations above and knowing that $hat{S}_{y}| {uparrow}_{y} rangle=| {uparrow}_{y} rangle$ and $hat{S}_{y}| {downarrow}_{y} rangle=-| {downarrow}_{y} rangle$, you should get to your result.
Answered by Milarepa on December 24, 2020
Get help from others!
Recent Answers
Recent Questions
© 2024 TransWikia.com. All rights reserved. Sites we Love: PCI Database, UKBizDB, Menu Kuliner, Sharing RPP