Is \(x^k p_x^m\) Hermitian?

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SUMMARY

The operator \(x^k p_x^m\) is definitively not Hermitian for positive integers \(k\) and \(m\). In contrast, the symmetrized operator \(\frac{x^k p_x^m + p_x^m x^k}{2}\) is Hermitian. The discussion highlights the importance of correctly applying conjugation to integrals and derivatives, emphasizing that the conjugation of an integral involves conjugating the entire integrand and switching the bra and ket in the inner product notation.

PREREQUISITES
  • Understanding of Hermitian operators in quantum mechanics
  • Familiarity with the notation of bra-ket formalism
  • Knowledge of integration and differentiation in the context of quantum wave functions
  • Basic principles of operator algebra
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  • Study the properties of Hermitian operators in quantum mechanics
  • Learn about the implications of operator symmetrization
  • Explore the mathematical foundations of bra-ket notation
  • Investigate the role of conjugation in quantum mechanical operators
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Students and professionals in quantum mechanics, particularly those studying operator theory and its applications in physics. This discussion is beneficial for anyone seeking to deepen their understanding of Hermitian operators and their significance in quantum systems.

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Homework Statement


Show that the operator x^kp_x^m is not hermitian, whereas \frac{x^kp_x^m+p_x^mx^k}{2} is, where k and m are positive integers.




The Attempt at a Solution



Is this valid?

<x^kp_x^m>^*=\left(\int_{-\infty}^\infty\Psi^*x^k(-i\hbar)^m\frac{\partial^m\Psi}{\partial x^m} dx\right)^*
=\int_{-\infty}^\infty\Psi^*(i\hbar)^m\frac{\partial^m(x^k\Psi^*)}{\partial x^m} dx \neq <x^kp_x^m>

That is, can you conjugate an integral by conjugating its integrand? Can you conjugate a derivative by conjugating the function you are differentiating?

And assuming that you can, did I carry out the conjugation correctly?
 
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You can conjugate the integrand, however, you have to conjugate all of the terms. Basically what happens is that the bra and the ket switch, and the operator is conjugated.

\langle \phi | A | \psi \rangle^* = \langle \psi | A^* | \phi \rangle
 

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