Even Wave Function in 1D Symmetric Potentials: Can We?

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SUMMARY

In one-dimensional symmetric potentials, it is impossible to construct an even wave function ψ(x) that equals zero at the origin. The discussion highlights two primary reasons: first, standing wave solutions like sin(kx) or cos(kx) cannot yield even wave functions due to their Taylor expansion starting with the x term. Second, while potentials can force wave functions to zero, only a divergent potential V(x) approaching infinity can achieve this, as demonstrated by the proposed combination of two infinite square wells with V=0 in specific regions and V=∞ outside.

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  • Understanding of quantum mechanics principles, particularly wave functions.
  • Familiarity with symmetric potentials in quantum systems.
  • Knowledge of standing wave solutions and their mathematical properties.
  • Concept of infinite square wells in quantum mechanics.
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  • Explore the mathematical properties of wave functions in quantum mechanics.
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  • Investigate the characteristics of infinite square wells and their applications.
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hokhani
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I like to know in one dimensional symmetric potentials, can we have any even wave functions which be zero in the origin?
 
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Do you want to construct an even potential V(x) such that ψ(x=0) = 0?

In an attractive potential having bound states there are two reasons for ψ(x) = 0; one is the generalization of standing waves like sin(kx) or cos(kx); of course you will never get an even wave function for such a standing wave b/c its Taylor expansion would start with the x1 term.

The second possibility is that the potential forces the wave function to zero; but b/c in QM you can have nonzero ψ even in classically forbidden regions (think about a double-well potential constructed like V(x) ~ (x+a)(x-a)) the only way to force the wave function ψ(x) to zero is a divergent potential V(x) → ∞.

So one simple possibility would be a combination of to infinite square wells with V=0 for x in [-b,-a] and x in [+a,+b], and V=∞ outside these two wells.
 
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