The Orthogonality Theorem: Mathematical Corner-stone for Superposition theorem and Perturbation Theorem!!!

Today’s the article will be a little bit more mathematical as this article will deal with the mathematical architecture and the building blocks of the theories like Superposition theorem and Perturbation Theorem.

So, without any further delay, let’s dive in…

As always we will start by considerations as we all know that physics is full of that!!!

So, consider there are two wave functions 𝝍ₙ and 𝝍ₖ. Both satisfy the Schrodinger’s equation for some potential V(x).

Now, if their energies are Eₙ and Eₖ respectively then Orthogonality theorem states that

∫ 𝝍ₖ*(x) 𝝍ₙ(x) dx =0 (Eₙ ≠ Eₖ)          (1)     

Here, the limits of the integral is the limit of the system and 𝝍ₖ* is the imaginary part of 𝝍ₖ.

Well, that’s it… its Orthogonality theorem’s main statement. But we are here to derive it also…so let’s finish this task….

As I said earlier, the above-mentioned wave functions obey the Schrodinger’s equations so,

- (ħ²/2m)(d²𝝍ₙ/dx²) + V(x) 𝝍ₙ = Eₙ 𝝍ₙ     (2)

And,

- (ħ²/2m)(d²𝝍ₖ/dx²) + V(x) 𝝍ₖ = Eₖ 𝝍ₖ     (3)

Now, if we multiply 𝝍ₖ* and 𝝍ₙ with eq. (2) and the complex conjugate of eq.(3) and then separate V(x) by subtracting eq.(3) from eq.(2). After all these jumble operations, we get this expression…

- (ħ²/2m)[ 𝝍ₖ* (d²𝝍ₙ/dx²) - 𝝍ₙ( d²𝝍ₖ/dx²) ] = (Eₙ - Eₖ) 𝝍ₖ* 𝝍ₙ       (4)

Now, just integrate this beautiful expression. The limits should be relevant to the problem or situation. After integrating we get….

- (ħ²/2m) ∫ [ 𝝍ₖ* (d²𝝍ₙ/dx²) - 𝝍ₙ( d² 𝝍ₖ/dx²) ] dx= (Eₙ - Eₖ) ∫ 𝝍ₖ* 𝝍ₙ    (5)

Well, we are almost there and just a few steps away from deriving the ‘pillar’ condition for the theorem!!!

Now, just take the left side of Eq. (5)…

By using some basic calculus we can write the left side of eq.(5) as …

- (ħ²/2m) ∫ (d/dx)[ 𝝍ₖ* (d𝝍ₙ/dx) - 𝝍ₙ(d𝝍ₖ/dx) ] dx       (6)

Thus eq.(6) transforms into…

- (ħ²/2m) )[ 𝝍ₖ* (d𝝍ₙ/dx) -𝝍ₙ( d𝝍ₖ/dx) ]        (7)

Now, if we make correct assumption then the wave functions vanish asymptotically and eq.(7) will be equal to 0. This will mainly occur in boundaries.

After all these mathematical drama our lovely eq.(5)  finally converts into this expression….

(Eₙ - Eₖ) ∫ 𝝍ₖ* 𝝍ₙ = 0          (8)

 If Eₙ ≠ Eₖ, then we finally proved eq.(1).

So, from eq.(8) we can say that, any set of functions 𝜓ⱼ (x) such that any two numbers of the set obey an integral constraint of this form is said to constitute an orthogonal set of functions.

If each individual member of the set is normalised, then they are called an orthonormal set of functions and the expression for Orthogonality in this case can be expressed by Dirac Notation as follows:

∫ 𝝍ₖ* 𝝍ₙ = ⟨ 𝝍ₖ* | 𝝍ₙ ⟩ = ẟₖⁿ           (9)

Here, ẟₖⁿ represents the Kronecker Delta and defined by:

ẟₖⁿ = 1   (k=n)

ẟₖⁿ = 0   (otherwise)

Orthogonality helps us to answers an important question, i.e. why two wave functions don’t overlap each other? But, now with the help of the above derivation, we found out that only and only if parameter k and n are same then only the integral of eq.(9) will have some value. Otherwise, it will be zero. Also from eq.(8), we can say that Eₙ ≠ Eₖ and it proves our statement.

So, that’s it…that’s all you need to know about Orthogonality and Orthonormality to start understanding the mathematical structure of Quantum Superposition and Perturbation theorem. Till now, on this website I only explored the theoretical side of the Superposition, but now onwards I will try to explain the mathematical architecture of it. So, be with me and we will explore this together.

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                                                                 -Ratnadeep Das Choudhury

                                                  Founder and writer of The Dynamic Frequency