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Suppose there exists a linear operator A that has an eigenvector |psi> with eigenvalue a,

A|psi> = a|psi>

Suppose that there also exists a linear operator B such that

[A, B] = B + 2BA^2

Show that the vector B|psi> is an eigenvector of A and find the associated eigenvalue.

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Solution Summary

The notation used in the statement of this problem is reviewed, and is then used in providing a complete, step-by-step solution of the problem, including a detailed explanation.

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[A, B] denotes the commutator of A and B:

[A, B] = AB - BA

Also, we are given that [A, B] = B + 2BA^2

Equating these two expressions for [A, B], we obtain:

AB - BA = B + 2BA^2

Now we need to show that B|psi> is an eigenvector of A,

that is, that there is some number b such that A(B|psi) = b(B|psi).

Using the ...

Solution provided by:
Georgia Martin, PhD
Education
  • AB, Hood College
  • PhD, The Catholic University of America
  • PhD, The University of Maryland at College Park
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