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Heisenberg uncertainty principle derivation and canonically conjugate vairables? 
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#1
Nov910, 09:25 AM

P: 326

Hi,
I've just worked through a derivation of the H.U.P. that uses the Cauchy Schwarz inequality to come up with the expression [tex] (\Delta A)^2(\Delta B)^2 \geq \frac{1}{4}<[A,B]>^2 [/tex]. This much I am happy with, but then it seems that when dealing with two "canonically conjugate observables" you set [tex] [A,B] = i\hbar [/tex] to find the uncertainty principle [tex] (\Delta A)(\Delta B) \geq \frac{\hbar}{2} [/tex]. It clearly gives the result I was expecting, but I cannot seem to find out where this [tex][A,B] = AB  BA = i\hbar [/tex] comes from. Is this something that can be figured out? Or, is it just something that some quantum mechanic somewhere has found out by working out the commutators of loads of operators and discovered that the commutators of conjugate observables just happen to be equal to [tex] i\hbar [/tex]? 


#2
Nov910, 10:24 AM

P: 527

There is no derivation; by definition a pair of canonical conjugate variables satisfies this relation.
There is some motivation behind it though, which has to do with canonical coordinates in classical mechanics, and the rules of canonical quantization. A canonical pair of coordinates in classical mechanics satisfy the fundamental Poisson brackets, [tex]\{x,p\} = 1[/tex] Canonical quantization dictates that this is replaced by the commuator (along with the introduction of a Hilbert space and turning observables into operators). 


#3
Nov1010, 02:47 AM

P: 1,411

One way to see where it comes from is by considering momentum as a generator of space translations. If [tex]V(a)[/tex] is the unitary group of space translations:
[tex]V(a)\hat{x}V(a)^\dagger =\hat{x}+a[/tex] and if you write [tex]V(a)=\exp(i\hat{p}a/\hbar)[/tex] then, by differentiating at a=0 , you get the canonical commutation relations. Of course there is the question whym momentum is related to space translations? A partial answer is given by the fact that momentum is conserved when Hamiltonian is invariant under space translations (Noether's theorem). 


#4
Nov1110, 10:10 AM

P: 72

Heisenberg uncertainty principle derivation and canonically conjugate vairables?
To see why we use this in quantum mechanics imagine a differentiable function
[tex]\phi(x)[/tex] and define [tex]p = i\hbar\partial_x[/tex] Then [tex](xppx)\phi(x) = i\hbar\phi(x)[/tex] and so [tex] [x,p] = i\hbar I[/tex] 


#5
Nov1110, 03:28 PM

Sci Advisor
P: 5,443

Another way to see this is to look at de Broglie waves.
A plane wave carrying momentum p looks like exp(ipx). The derivative acting on this "wave function" produces the eigenvalue p. Of course this is no derivation, it simply shows that interpreting the derivative as momentum operator seems to be reasonable. 


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