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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.

enter image description here

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  • $\begingroup$ Thanks, @Jakob . I’ve just added to my question to make it clearer! $\endgroup$
    – Yelena
    Apr 18 at 20:10
  • $\begingroup$ I guess so. Or maybe you could say it is just in terms of x, I thought that meant orthogonal eigenspaces vs. In terms of x AND t, normalisation? Earlier in the video, we see that the [psi(x)]^2 = 0, because of orthogonal eigenspaces. So why is it different here? $\endgroup$
    – Yelena
    Apr 18 at 20:15
  • $\begingroup$ Please note that question here should be as self-contained as possible. That is, you should provide all equations etc., such that one does not have to watch a video. Further, please try to typeset your equations with MathJax. You can find a tutorial here. $\endgroup$
    – Jakob
    Apr 18 at 20:17
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    $\begingroup$ These integrals are definite integrals, not indefinite integrals. $\endgroup$
    – G. Smith
    Apr 18 at 20:21
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    $\begingroup$ Normalization is a property of one wave function. Orthogonality is a relationship between two different wave functions. Just like for vectors, because a wavefunction is a vector, in an infinite-dimensional vector space of functions. $\endgroup$
    – G. Smith
    Apr 18 at 20:23
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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.

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  • $\begingroup$ Thanks very much! @electronpusher so this is true but you have to have the order of operations correct (I.e., if you have the wave function squared with itself multiplied by x, then you have to integrate the whole thing, not just x by saying that equals one because of product rule), yes, is this correct? $\endgroup$
    – Yelena
    Apr 19 at 22:50
  • $\begingroup$ Normality and orthogonality apply when you only have two wavefunctions in the integral, no other functions. If there is an operator between the wavefunctions, such as x for position, then the integral will not generally equal 0 or 1; it will equal the expected/average value of the operator. This gives the definition of the expected value for an operator in quantum mechanics; it works only when the wavefunctions in the integral are the same, representing one state (the average value of an operator for a system in two different states isn't something that makes sense at the introductory level). $\endgroup$ Apr 21 at 1:19
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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! :)

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    $\begingroup$ You need to fix all your equations, etc, by using MathJax. $\endgroup$
    – Bill N
    Apr 18 at 21:20

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