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 5
Homework Statement:

Image of problem in Attempt
At t = 0, a particle of mass m in an infinite square well with walls at
x = 0, x=a, is in a superposition of the (normalized) first and third stationary states
##\Psi(x,0)=A(2\psi_1(x)i\psi_3(x))##
(a) What must A be in order for Ψ to be a normalized wavefunction?
##A=1/\sqrt5##
(b) What is the probability that the particle is in the left half of the box
(i.e.,between x = 0 and x = a/2) at t = 0?
Hint: Instead of trying to do a complicated integral directly, notice that for odd n,##\psi_n(x)=\psi_n(ax)~\text{for}~0\leq x\leq a~\text{ You can use this to relate} \int_0^{{a/2}} \psi^*_m\psi_n\, dx~to\int_0^{{a}} \psi^*_m\psi_n\, dx~\text{when m, n are both odd.}##
Question 1: Is it possible to do this in a simpler way? Perhaps using dirac notation?
Question 2: How do I use the hint
Relevant Equations:

Stationary state of an infinite square well
##\psi_n(x)=\sqrt{\frac {2} {a}} ~~ sin(\frac {n\pi x} {a})##
Attempt: I'm sure I know how to do this the long way using the definition of stationary states(##\psi_n(x)=\sqrt{\frac {2} {a}} ~~ sin(\frac {n\pi x} {a})## and ##\int_0^{{a/2}} {\frac {2} {a}}(1/5)\left[~ \left(2sin(\frac {\pi x} {a})+i~ sin(\frac {3\pi x} {a})\right)\left( 2sin(\frac {\pi x} {a})i ~sin(\frac {3\pi x} {a})\right)\right]dx ## then combining and using trig identities then integrate
But
Assuming my A is correct, I actually have 2 questions about part (b). One is about how to use the hint, the other if it is possible to find this using another method
Question 1: Is it possible to do this in a simpler way? Perhaps using dirac notation?
I tried ##1/5\langle \Psi(2\psi_1i\psi_3)\Psi\rangle##
If this is possible and the correct set up, how do I continue? Do I try to separate them? We just learned this week, so Im still pretty confused about it
If not possible to use, why?
Question 2: How can I use the hint
I want to understand how to do it according to the hint, if I use the hint does this mean
##\int_0^{{a/2}} {\frac {2} {a}}(1/5)\left[~ \left(2sin(\frac {\pi (x(a/2))} {a})+i~ sin(\frac {3\pi (x(a/2))} {a})\right)\left( 2sin(\frac {\pi (x(a/2))} {a})i ~sin(\frac {3\pi (x(a/2))} {a})\right)\right]dx ##
I'm assuming this is wrong because I don't see how this makes it any simpler unless I'm missing something crucial
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