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Operators, normalised eigenstates and the generalised uncertainty relation

  • Thread starter humfri
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  • #1
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Homework Statement



Hi guys! Many time reader, first time poster... I've struggled big time with the following. Any advice at all would be great. I'm so muddled, it's just not funny any more... (plus i'm not really familiar with who to write the mathematic script so please be patient)

I have an operator, say A representing observable A with two normalised eigenstates with eigenvalues a1, a2 respectively (and the same with B in place of A)...
and if |A1> = 1/5(3|B1> +4|B2>) and |A2> = 1/5(4|B1> - 3|B2>). My question is how do i find the commutator of this, and if I do find the commutator of this, how can I then use this to determine the generalised uncertainty relation for operators Aand B?? So desperate!


Homework Equations


<(A)^2><(B)^2> > 1/4|<[A,B]>|^2

The Attempt at a Solution


I'm not sure why, but I felt that I could represent:
A: 1/5[3b1 4b1; 4b2 -3b2] ; and
B: 1/5[3a1 4a1; 4a2 -3a2]

for which I can then do matrix multiplication and subtraction (i.e. [A,B] = AB - BA (all operators))
and I get:
4/25 [4(b1a2-a1b2) -3(b1a2+a1b2); 3(b2a1-a2b1) 4(b2a1-a2b1)]

then not sure if i have anything remotely ok, although I do notice that the matrix looks kinda Hermitian or whatever?. really stuck from here. Basically have no idea how to get the 'expectation' value for this or whatever... any help would be great (and asap cos I've tried to work it out all week and now left my time seriously short - not that anyone HAS to help me of course!...) Cheers very much! humfri

(p.s. sorry if this post was annoying and confusing... still learning)
 

Answers and Replies

  • #2
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:( i think ill go eat some worms...... :(
 
  • #3
George Jones
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Can you write operator A in terms of its eigenvalues and eigenstates? Think projection operators.
 
  • #4
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thanks for your reply!
do you mean like operator A =
[<A1|A|> <A1|A|A2>; <A2|A|A1> <A2|A|A2>]
and
[<A1|B|> <A1|B|A2>; <A2|B|A1> <A2|B|A2>]


cos if I do that I get A = [a1 0; 0 a2]
and B= 1/25[9b1+16b2 12b1-12b2; 12b1-12b2 9b1 + 16b2]
but then I did do A on A and also A on B for some reason. does this make sense or have i made a boo boo.
 
  • #5
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LOL, apparently they operators commute anyway (something I found ages ago but thought I'd done something silly and so ignored my result), thus no uncertainty relation. Nevermind. Thanks for the help anyway, now I actually understand what I need to do next time!
 
  • #6
George Jones
Staff Emeritus
Science Advisor
Gold Member
7,261
790
thanks for your reply!
do you mean like operator A =
[<A1|A|> <A1|A|A2>; <A2|A|A1> <A2|A|A2>]
and
[<A1|B|> <A1|B|A2>; <A2|B|A1> <A2|B|A2>]


cos if I do that I get A = [a1 0; 0 a2]
and B= 1/25[9b1+16b2 12b1-12b2; 12b1-12b2 9b1 + 16b2]
but then I did do A on A and also A on B for some reason. does this make sense or have i made a boo boo.
Good.

In your first post, it seems that you calculated the matrices for A and B using two different bases. The same basis should be used to calculate both matrices.

In, this post you appear to have used the basis {|a1> , |a2>} to calculate the matrices for both A and B.

I might have made a mistake, but I get something a little different than you for the second component of the second row of the matrix for B.

LOL, apparently they operators commute anyway (something I found ages ago but thought I'd done something silly and so ignored my result), thus no uncertainty relation. Nevermind. Thanks for the help anyway, now I actually understand what I need to do next time!
A and B can't commute in general, because, in general, the eigenvectors for A and B are linearly independent.
 

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