How do we know (BΨ) is not equal to zero?

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Discussion Overview

The discussion revolves around the properties of operators in quantum mechanics, specifically addressing the relationship between eigenfunctions and the implications of the operator B acting on an eigenfunction Ψ of operator A. The participants explore whether the result of BΨ can be zero and the conditions under which this might occur.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant posits that if operators A and B commute and Ψ is an eigenfunction of A, then BΨ should also be an eigenfunction of A.
  • Another participant argues that since Ψ is in the domain of B, BΨ can only equal zero if Ψ itself is zero, which they consider trivial and thus excluded from analysis.
  • A different viewpoint suggests that while Ψ and BΨ are both eigenfunctions of A with the same eigenvalue, BΨ could potentially be zero if the corresponding eigenvalue is zero.
  • One participant states that if BΨ equals zero, it is acceptable as the zero vector is a trivial eigenvector for any linear operator.

Areas of Agreement / Disagreement

Participants express differing views on whether BΨ can equal zero, with some asserting it cannot under certain conditions while others argue that it is possible. The discussion remains unresolved regarding the implications of BΨ being zero.

Contextual Notes

There are assumptions about the nature of the operators and the eigenfunctions that are not fully explored, particularly regarding the implications of linear dependence and the treatment of the zero vector in this context.

PFfan01
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Suppose that operators A and B have complete sets of eigenfunctions, [A, B] = 0, and Ψ is an eigenfunction of A with the eigenvalue a, namely AΨ=aΨ. Then we have A(BΨ) = BAΨ = a (BΨ). They say (BΨ) is also the eigenfunction of A. Why? How do we know (BΨ) is not equal to zero?
 
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Well Psi is the domain of B, so that the only way that B psi =0 is that psi =0, which is trivial, the null vector is excluded in any analysis.
 
dextercioby said:
Well Psi is the domain of B, so that the only way that B psi =0 is that psi =0, which is trivial, the null vector is excluded in any analysis.

Thanks, but I am thinking in a different way.
Ψ and (BΨ) are both eigenfunctions of A with the same non-degenerate eigenvalue a, and they must be linearly dependent, namely BΨ=b*Ψ. Ψ is a common eigenfunction of A and B. But the eigenvalue b could be zero (b=0), so that BΨ=0. Right?
 
If Bpsi =0, then it's no problem, the 0 vector is a trivial eigenvector for any linear operator.
 

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