Is the Expectation Value of x Zero for an Even Potential Energy Function?

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SUMMARY

For a potential energy function V(x) that is even, the wave function for any stationary state can be classified as either even or odd. This symmetry leads to the conclusion that the expectation value of position x for any stationary state is zero. The probability distribution derived from the wave function is symmetric, which reinforces that the contributions to the expectation value from positive and negative x values cancel each other out.

PREREQUISITES
  • Understanding of quantum mechanics principles, specifically wave functions and potential energy functions.
  • Familiarity with the Time-Independent Schrödinger Equation (TISE).
  • Knowledge of symmetry properties in quantum systems.
  • Basic grasp of probability distributions in quantum mechanics.
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  • Study the Time-Independent Schrödinger Equation (TISE) in detail.
  • Explore the implications of symmetry in quantum mechanics, particularly in relation to wave functions.
  • Learn about the mathematical treatment of even and odd functions in quantum systems.
  • Investigate the concept of expectation values in quantum mechanics and their physical significance.
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Students of quantum mechanics, physicists studying wave functions, and anyone interested in the mathematical foundations of potential energy functions and their implications on expectation values.

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Homework Statement



Its a problem from a foreign book. It sounded simple to me but I am confused now.


If V(x), a potential energy function, is known to be an even function, what can you say about wave function for any stationary state? What shall be the expectation value of x for any stationary state ?

Homework Equations





The Attempt at a Solution



I grabbed a book by Griffith from my library and figured that if V(x) is even, the time independent wave function can be taken to be either even or odd. I know that expectation value of x for a stationary state has to be 0. But can some help me see this with the standpoint of V(x) being even ?


thanks in advance
 
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V(x) is even, so it is symmetric around the origin

the solutions to TISE will be either symmetric or anti-symmetric around the origin

note if [tex]\psi[/tex] is a solution, so will [tex]e^{i \phi}\psi[/tex] for any phi, ie the solution is only unique upto an overall phase

so in the anti-symmetric solution, there is no meaning in one side being negtive and the other positive positive, as it is equivalent to any solution with phase shifted by phi. the only point is that they are out of phase by p.

when you look at the probability distribution given by
[tex]P(r)dr = \psi \psi*[/tex]
the overall phase cancels and P(r)dr will be a non-negative symmetric function, thus the expectation position will be zero
 

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