Difference between normalized wave function and orthogonal wave function? I am confused about when we say that a wave function is NORMALIZED so that we say that the indefinite integral of the wave function squared = 1, vs. when we say that the wave function represents ORTHOGONAL spaces so that the indefinite integral of the wave function squared = 0.
I have been watching a helpful video about finding the average expectation values of x and p, here:
In this video, the particular part that is confusing is attached as an image to this question. The thing that is confusing is that the indefinite integral of ([psi(x)])^2 becomes 1. I know that the magnitude of the integral of ([psi(x,t)])^2=1 but thought that that integral just in terms of x is ([psi(x)])^2=0, due to orthogonal vector spaces.
Would someone please explain any missing steps? Thank you.

 A: When your integral (over all space) is of the product if two different (orthogonal) wavefunctions, it will equal zero. This is the orthogonality condition. When your integral (over all space) is the product of a wavefunction with itself, i.e. the squared magnitude of a wavefunction, it will equal 1. This is the normality condition. To determine whether an integral (over all space) of a product of wavefunctions will equal 0 or 1, you need simply consider whether the two wavefunctions are the same or not.
A: OK, I just checked the youtube video. I guess you are considering a quantum harmonic oscillator whose initial state is a superposition of the ground and first excited states, namely |psi(0)>=A(3|0>+4|1>) where psi_0(x) = <x|0> and psi_1(x) = <x|1>. You may check the wikipedia page on quantum harmonic oscillator. It seems that the eigenstates psi_k(x) of a quantum harmonic oscillator can be described by a Gaussian function multiplied by Hermite functions.
(a) The initial state will be normalised when <psi(0)|psi(0)> = 1. Since the eigenstates |0> and |1> are normalised, <psi(0)|psi(0)> = |A|^2 (9<0|0>+16<1|1>) = 25 |A|^2 due to the orthogonality between eigenstates, <0|1> = 0. Therefore, A = 1/5 when A is real-valued.
(b) The unitary dynamics of a quantum state can be described by applying a unitary operator U(t) = \sum_{n=0}^{infinity} |n><n| exp(-i omega (n+0.5) t) to the initial state |psi(0)> where omega is the frequency of the quantum harmonic oscillator and |n> is the n-th excited state. Therefore, the state at time t is U(t)|psi(0)> = (1/5) (3 U(t)|0> + 4 U(t)|1>) = (3/5) e^(-i 0.5 omega t) |0> + (4/5) e^(-i 1.5 omega t) |1>.
(c) Now a position operator is described by x = c(a+a^dagger) where c is a scaling factor and a and a^dagger are annihilation and creation operators, respectively, of the harmonic oscillator. Therefore, the average position at time t is given by <x(t)> = <psi(t)|x|psi(t)> = c <psi(t)|(a+a^dagger)|psi(t)> = c (1/25) (3 <0| + 4 exp(i omega t) <1|)(a+a^dagger)(3 |0> + 4 exp(-i omega t) |1>). Based on the properties of a and a^dagger operators, namely a^dagger|0> = |1> and a|1> = |0>, you can check that the average position is given by <x(t)> = (1/25) (3*4) (exp(i omega t) + exp(-i omega t)) = (24/25) cos(omega t), which oscillates with the frequency omega of the quantum harmonic oscillator. This simply means that even if the average position of each eigenstate |0> and |1> is zero, a superposition of the eigenstates |0> and |1> can result in non-zero average position, meaning that the statistical properties of a superposed quantum state (nonzero <x(t)> for |0>+|1>) can be different from the classical average of the statistical properties of individual states (<0|x|0> = <1|x|1> = 0). This means that the coherence between eigenstates, namely <0|x|1>, can induce time-dependence of the average position.
I hope this will be helpful! :)
