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Consider a linear harmonic oscillator and let, yo and y, be its real, normalized ground and first

excited state energy eigenfunctions respectively. Let Ayo + Byi with A and B real numbers be the
wave function of the oscillator at some instant of time. Show that the average value of x is in
different from zero. What values of A and B maximize (x) and what values minimize it?
general​

User Labe
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2 Answers

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Final answer:

The average value of the position (x) for a linear harmonic oscillator is zero for any normalized energy eigenfunction. The values of A and B that maximize (x) are when both A and B are non-zero, and the values that minimize (x) are when either A or B (or both) are zero.

Step-by-step explanation:

The average value of the position (x) for the wave function A0 + B0i of a linear harmonic oscillator is indeed zero. This is because for any normalized energy eigenfunction, the average value of the position operator is zero.

To find the values of A and B that maximize and minimize (x), we need to rewrite the wave function in terms of x:

For the maximum value of (x), both A and B should be non-zero, so the wave function would be something like Amax + Bmaxi. For the minimum value of (x), either A or B (or both) should be zero, so the wave function would be something like Amin or Bmini.

User Contango
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4 votes
No no no no no no no no
User Nevett
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