Question

Suppose a particle has a wavefunction image. Sketch the form of this wavefunction. Where is the...

Suppose a particle has a wavefunction image. Sketch the form of this wavefunction. Where is the particle most likely to be found? At what values of x is the probability of finding the particle reduced by 50 per cent from its maximum value?

I believe I understand the first two parts of the question, but I am confused on how to set up an equation to solve the last part. Please explain/show steps. Thanks!

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Answer: V(x) = inf V(x)-0 : V(x)-inf (a,0,0) I will be doing time-independent quantum mechanics for particle confined betweenWe have two solutions to the Schrodingers equation ()-Aand Note: Ai and A2 are complex numbers A more general solution will be ψ(z) = Aie+,wz+Age-ur-Ai (cos(wx) + 1 sin(wx))+A2(cos(wx)-1 sin(wx))-((A1+Ag) cos(wr)+(A1-1A2) sin(wx)) C cos(wr) Dsin(wx) Using Boundary Condition (a) 0 ψ(a) = C cos(wa) + D sin(wa) = 0 Using Boundary Condition (b)-0 C +Dtan(wb0 tan(wa) = tan(wb) So, b+a where n is a positive integer and C--Dtan(wa)tan (wa) +tan(wa)- Now, the probability density function P(x) DD(sin(wx) - tan(wa) cos(wx(sin(wx) - tan(wa) cos(wx)) P(z) (sin2 (wx) tan (wa) cos(wx) - tan(wa) sin(2wx)) For maximum probabilty dP(x) d(sina (wa)cos(ztan daxNow 2i+ tan(a)(7 Here. b-a =- n =ー-, where ω = So for a particular Energy state En we have a particular quantum number n and1sin2(wa)+tan2(a) sin2(wa) sec(wa) 1 +sin (wa) (1 + tan2(wa)) Tsec(wa) (1 +tan2(wa)) + a Now we need to find a point c such thatP(c) =-P P(c) = + tan2(wa)sin(we)+ tanr(w)()tan 1+tan2(wa) 2w (cos(wa))[(sin(we)-tan(wa) cos(wc)12 =- (cos(wa))[sin(wc)-tan(wa) cos(wc)] = → (cos(wa)

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