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How do I evaluate [x, SinPx] commutator

  • Thread starter sudipmaity
  • Start date
1. The problem statement, all variables and given/known data

Evaluate [x, SinPx] given [x, Px]=ih

2. Relevant equations

Px = h/I( d/dx)

3. The attempt at a solution
Let f (x) be a function of x.[ x, Sin Px ] f( x) ⇒ [ x sin{( h / i )d / dx} f ( x ) -Sin { (h/i) d / dx } x f(x)] . Does anybody concur .
 

DrClaude

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What you wrote is correct, but it does not answer the problem.
 
What is the answer?
 

DrClaude

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after a strenous ordeal Of 48 hr and heated debate with my pals and two sleepless nights, here is what I have come up with.we know in position space momentum oprtr is given -ih d/dx.what would be the opposite.what if I had acess to a position operator in momentum space.lets say x = ih d/dPx.then what happens.
This is what happens :[x , sinPx]f (Px)= [ ih (d / dPx) , sin Px ]f (Px) = ih [ d /dx ( fSin Px )-sin Px d/dx( f )] =
ih [ cos Px f(Px) + sinPx f' (Px)-sin Pxf' (Px) ] = ih cos Px f(Px) . Thrfr [ x, sin Px ] = ih Cos Px
does Anybody agree ??
 
Last edited:

DrClaude

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after a strenous ordeal Of 48 hr and heated debate with my pals and two sleepless nights, here is what I have come up with.we know in position space momentum oprtr is given -ih d/dx.what would be the opposite.what if I had acess to a position operator in momentum space.lets say x = ih d/dPx.then what happens.
This is what happens :[x , sinPx]f (x)= [ ih (d / dPx) , sin Px ]f (x) = ih [ d /dx ( fSin Px )-sin Px d/dx( f )] =
ih [ cos Px f(x) + sinPx f' (x)-sin Pxf' (x) ] = ih cos Px f(x) . Thrfr [ x, sin Px ] = ih Cos Px
does Anybody agree ??
There are a few typos along the way (##d/dx## instead of ##d/dP_x##), but it seems ok. You should also try using the Taylor series for the sine. You should find that
$$
\sin \left(\frac{\hbar}{i} \frac{d}{dx} \right) x = x \sin \left(\frac{\hbar}{i} \frac{d}{dx} \right) + \frac{\hbar}{i} \cos \left(\frac{\hbar}{i} \frac{d}{dx} \right)
$$
which leads you to the same answer.
 
I know.I did it that way earlier infact. But this method is more compact and satisfying as it doesnt involves any approximation.You can google out POSITION OPERATOR IN MOMENTUM SPACE.there are very good articles out there upon this.as i am a beginner today I am very happy when I came to know about this concept.
 

DrClaude

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I know.I did it that way earlier infact. But this method is more compact and satisfying as it doesnt involves any approximation.
The method I used is not approximate. You never truncate the Taylor series, you let it go to infinity, although you do not need to calculate all the terms, as the pattern is easily identifiable. It is more general, as the momentum space approach only worked because the differential operator wasn't in a function anymore. What if you had to calculate ##[ \cos(x), \sin(P_x)]##? You will also encounter things such as the exponential of operators, and the only correct way to express them is through their Taylor series.
 
Thanks for the insight.
 
You first need to prove [x , pn] = niħpn-1.
In order to do that take arbitrary values of n=1, 2, 3 and check.
You'll see for n=1 [x , p] = iħ result holds (as it is given and also we know)
for n=2 [x , p] = p[x , p] +[x , p]p = piħ +ipħ = 2ipħ = 2iħp2-1 again result holds
Similarly you can show for n=3 with the help of first two.
Now, assume for n=k this will hold i.e. [x , pk] = kiħpk-1
then show for n=k+1 and you will find [x , pk+1] = [x , pk]p + pk[x , p] = kihpk-1p + pkih
= kiħpk + kiħpk = (k+1)iħpk = (k+1)iħp(k+1)-1 again holds.
So, [x , pn] = niħpn-1 (PROVED)
Now, to evaluate [x , sin(p)], first expand sin(p) in taylor series.
You'll get sin(p) = p - (p3)/3! + (p5)/5! - .....
Put this in [x , sin(p)]
You'll get [x , sin(p)] = [x , p - (p3)/3! + (p5)/5! - .....]
= [x , p] - (1/3!)[x , p3] + (1/5!)[x , p5] - .....
= iħ - (1/3!)3iħp2 + (1/5!)5iħp4 - .....
= iħ [1 - p2/2! + p4/4! - .....]
= iħ cos(p)
∴ [x , sin(p)] = iħ cos(p)
And this is how you may do it.
 

PeroK

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You first need to prove [x , pn] = niħpn-1.
In order to do that take arbitrary values of n=1, 2, 3 and check.
You'll see for n=1 [x , p] = iħ result holds (as it is given and also we know)
for n=2 [x , p] = p[x , p] +[x , p]p = piħ +ipħ = 2ipħ = 2iħp2-1 again result holds
Similarly you can show for n=3 with the help of first two.
Now, assume for n=k this will hold i.e. [x , pk] = kiħpk-1
then show for n=k+1 and you will find [x , pk+1] = [x , pk]p + pk[x , p] = kihpk-1p + pkih
= kiħpk + kiħpk = (k+1)iħpk = (k+1)iħp(k+1)-1 again holds.
So, [x , pn] = niħpn-1 (PROVED)
Now, to evaluate [x , sin(p)], first expand sin(p) in taylor series.
You'll get sin(p) = p - (p3)/3! + (p5)/5! - .....
Put this in [x , sin(p)]
You'll get [x , sin(p)] = [x , p - (p3)/3! + (p5)/5! - .....]
= [x , p] - (1/3!)[x , p3] + (1/5!)[x , p5] - .....
= iħ - (1/3!)3iħp2 + (1/5!)5iħp4 - .....
= iħ [1 - p2/2! + p4/4! - .....]
= iħ cos(p)
∴ [x , sin(p)] = iħ cos(p)
And this is how you may do it.
Note that the previous posts are 6 years old.
 

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