Diagonalizabilty versus spread for uncertainty (discrete)

In summary, the uncertainty principle states that two non-commuting Hermitian operators cannot be simultaneously diagonalized.
  • #1
nomadreid
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I have seen two characterizations of the problem in measuring a discrete variable of a state ψ exactly with each of two non-commuting Hermitian operators A and B:
(1) that the product of the standard deviations ( = √(<ψ|A2|ψ>-<ψ|A|ψ>2), & ditto for B) ≥ 1
(2) that one cannot simultaneously diagonalize the matrix representations of A and B
(i.e., if A = UCU and B = VDV, for unitary U and V and diagonal C and D, with denoting the adjoint, then U≠V.
Where is the link between these two?
 
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  • #2
nomadreid said:
I have seen two characterizations of the problem in measuring a discrete variable of a state ψ exactly with each of two non-commuting Hermitian operators A and B:
(1) that the product of the standard deviations ( = √(<ψ|A2|ψ>-<ψ|A|ψ>2), & ditto for B) ≥ 1
(2) that one cannot simultaneously diagonalize the matrix representations of A and B
(i.e., if A = UCU and B = VDV, for unitary U and V and diagonal C and D, with denoting the adjoint, then U≠V.
Where is the link between these two?

As explained here: http://en.wikipedia.org/wiki/Uncert...2.80.93Schr.C3.B6dinger_uncertainty_relations, the link is that

[itex]\sigma_A \sigma_B \geq \frac{1}{2} | \langle [A,B] \rangle|[/itex]
 
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  • #3
Thanks, stevendaryl. Light is starting to seep in, as I read in the link you cited "... in wave mechanics, the uncertainty relation between position and momentum arises because the expressions of the wavefunction in the two corresponding orthonormal bases in Hilbert space are Fourier transforms of one another ... A nonzero function and its Fourier transform cannot both be sharply localized." So I apparently have to delve into the theory of Fourier transforms next.
Forgive me for appearing a bit dense (a poor imitation of the Tortoise in Lewis Carroll's "What the Tortoise said to Achilles") , but I do not see that the link (between the HUP and the non-simultaneous-diagonalizability complementarity) is the more general case (than my use of "1" in units of ħ/2) of the HUP which you stated , since in looking at the derivation it appears to me that all that is required to show that |<[ A,B]>| is non-zero is the non-commutativity, and not the non-simultaneous-diagonalizability.
 

Related to Diagonalizabilty versus spread for uncertainty (discrete)

1. What is diagonalizability and how does it relate to uncertainty in discrete systems?

Diagonalizability is a mathematical property of a matrix where it can be transformed into a diagonal matrix using a similarity transformation. In the context of uncertainty in discrete systems, it means that the system can be represented in a simpler form, making it easier to analyze and understand.

2. Can a non-diagonalizable matrix represent a discrete system with uncertainty?

Yes, a non-diagonalizable matrix can still represent a discrete system with uncertainty. However, it may be more complex and difficult to analyze compared to a diagonalizable matrix.

3. Are there any advantages to having a diagonalizable matrix in representing a discrete system with uncertainty?

Yes, there are several advantages to having a diagonalizable matrix. It allows for easier analysis and understanding of the system, as well as the possibility of finding simpler solutions to the system. Additionally, diagonalizable matrices have well-defined eigenvalues and eigenvectors, which can provide useful information about the system.

4. How does the spread of uncertainty affect the diagonalizability of a matrix?

The spread of uncertainty refers to the range of possible values for a variable in a system. In general, a larger spread of uncertainty can make a matrix less diagonalizable. This is because a larger spread can result in a more complex system, making it more difficult to find a diagonalizable representation.

5. Can a matrix be partially diagonalizable?

Yes, a matrix can be partially diagonalizable. This means that some of the eigenvalues and eigenvectors of the matrix can be found, but not all of them. In this case, the matrix may still provide some useful information about the system, but it may not be fully diagonalizable.

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