Find the probability of a particle in the left half of an Infinite Square well

[Moolisa]

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,$\psi_n(x)=\psi_n(a-x)~\text{for}~0\leq x\leq a~\text{ You can use this to relate} \int_0^{{a/2}} \psi^*_m\psi_n\, dx~to\int_0^{{a}} \psi^*_m\psi_n\, dx~\text{when m, n are both odd.}$

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 $1/5\langle \Psi|(2\psi_1-i\psi_3)|\Psi\rangle$
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 I am 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

 

[berkeman]

At t = 0, a particle of mass m in an infinite square well with walls at
x = 0

Where is the 2nd wall? Or is it at infinity? Sorry, your statement is a bit confusing to me...

 

[Moolisa]

Where is the 2nd wall? Or is it at infinity? Sorry, your statement is a bit confusing to me...

Sorry, I should have been more careful!

Walls are at x=0 to x=a

 

[jtbell]

Set up the integral for the probability integral using $\psi_1$ and $\psi_3$ without substituting the actual forms of those wavefunctions. Multiply out the parentheses to get a sum of integrals.

I think the hint has a typographical error, and it should read "$\psi_n(x) = \psi_n(a-x)$". It's a statement about the symmetry of those wavefunctions. Using the complete hint, you should be able to change the limits on the integrals and put them in a form that you can write down the values immediately without having to plug in the wavefunctions and grind them out. Hint: the wavefunctions are normalized!