Do electron spin like a ball?

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Do electron spin like a ball? you have heard that electron have spin and magnitude of that spin is $\dfrac{1}{2}$. well it's actually $\dfrac{1}{2}\hbar $ we just take $\hbar = 1$  and we all know that $\hbar$ is reduced plank constant. but is electron really spinning like ball and having angular momentum of $\dfrac{1}{2}\hbar$ ? hmmm... what is velocity of the electron then? let's try to calculate that , let's assume electron is solid sphere and it's radius is $r_e = 2.81×10^{-15}$ meters. (check this https://en.m.wikipedia.org/wiki/Classical_electron_radius) from high-school physics you can calculate moment of inertia of the solid sphere. and it's turn out to be $I = \dfrac{2}{5}mr^2$. and angular velocity $\omega = \dfrac{v}{r}$  so we have all the equipment. now we know that angular momentum $L = I\omega $  also $ L = \dfrac{1}{2}\hbar$  so we have $ I\omega = \dfrac{1}{2}\hbar$ putting Value of $I$ and $\omega$ $\dfrac{2}{5}mr^{2}\dfrac{v}{r} = \dfrac{1}{2}\hba...
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What is commutator of position and momentum operator?

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What is commutator of $ \hat{x} $ and $ \hat{p} $?

commutator of two operator $ \hat{A} $ and $ \hat{B} $ is defined as $ [ \hat{A}, \hat{B} ] \equiv \hat{A}\hat{B} - \hat{B}\hat{A} $

In quantum mechanics to even find the minimum uncertainty between two operator we need the commutator of two , so it's important to find such quantity.

here our position and momentum are also operators and we need to find the $ [ \hat{x}, \hat{p} ] $.

so how will we find it?? 

as the name suggests they are operator and they operate on the functions (function of the Hilbert space). 

for position operator we just multiply function by $x$ and for momentum operator we differentiate the function by $x$ and then multiple by $-i\hbar $. ( Note we are working only in 1 dimentions) 

generally working with operators we use test function $\phi(x)$. 

this $\phi(x)$ is arbitrary function.

so let's try to find 

$ [ \hat{x}, \hat{p} ]\phi(x) $ 

by using defination of commutator we can write it as 

$=[x(-i\hbar)\frac{d}{dx}-(-i\hbar)(\frac{d}{dx})x]\phi(x)$

$= [x(-i\hbar)\frac{d}{dx}\phi(x)-(-i\hbar)\frac{d}{dx}(x\phi(x)]$

$ = -i\hbar( x\frac{d\phi}{dx} - x\frac{d\phi}{dx}-\phi(x))$

$ = i\hbar\phi(x)$ 

now as $\phi(x)$ was arbitrary we'll Remove that and we get

$ [ \hat{x}, \hat{p} ] = i\hbar$

This equation is known as the cononical commutation relation. It's called canonical because position and momentum are related by Fourier transforms.

This is precisely the reason why minimum uncertainty in position and momentum is always great or equal to $\frac{\hbar}{2}$.

If for any 2 operators the commutator vanishes than uncertainty is zero that is you can measure both quantity simultaneously.( Basically $ \sigma_A\sigma_B = 0 $ ). Well if commutator is zero than they share simultaneous Eigen functions. But we'll discuss that into future posts.





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