This article is the Part 4 of the article series I wrote about Quantum Harmonic Oscillator. If you have not read the Part-1: Introduction to Quantum Harmonic Oscillator and Part2: Schrödinger’s Equation with Dimensionless Terms and Part-3: The Asymptotic solution, then you cannot understand what I am going to explain in this article, so reading those articles is a must.

Okay…without any further delay let’s just start…

The objective of this article is to find a general solution for the harmonic oscillator by finding the series solution.

From the equation 7 of my last article, we obtained an expression, and to solve Schrodinger’s equation for this problem, we wish to explicitly build in knowledge of the exponential asymptotic behaviour of ψ that was established in previous article.

So, there is a one way to do so, that is by assuming that can be expressed as the product of two functions, one which will have the asymptotic behaviour of wavefunction and other one is an unknown function, lets say it as H(ξ). We can express what we said as follows,

ψ(ξ) = H(ξ) e^(-ξ²/2)          (1)

This is just an assumption for the solution taken by only trial and error method. We have to see whether it satisfies our conditions or not.

From elementary calculus we know that any function of x ( here, ξ) can be expressed as an infinite polynomial using Taylor or McLaurin Series, so,

H(ξ) = a+ aξ + aξ² + … = n an ξⁿ  [n ranges from 0 to ∞]     (2)

Here, n ranges from 0 because we cannot have negative powers for ξ  as it would lead to infinities, which in this case will lead to not normalizable.

Now, we have to solve Schrodinger’s equation for harmonic oscillator that is equation 11 of part-2 article this this series of QHO.

d^2 ψ / d ξ^2  + ( λ ξ^2) ψ = 0       (3)

Differentiating equation 1 we get, we get,

d² ψ / d ξ² = [ d²H/d ξ² - 2ξ dH/dξ + ( ξ² - 1)H ] e^(-ξ²/2)      (4)

Note that we have dropped the explicit dependence of H on ξ for sake of clarity.

Now, substituting equation 4 and equation 1 in equation 3, we get,

d²H/d ξ² - 2ξ dH/dξ + (λ - 1) H = 0         (5)

Now, we have everything in our hands to attempt a series solution for H(ξ). So, by differentiating equation 2, we get,

dH/dξ = n nan ξⁿ⁻¹   [n ranges from 0 to ∞]   (6)

again differentiating,

d²H/d ξ² = n n(n-1)an ξⁿ⁻² [n ranges from 0 to ∞]   (7)

Substituting equation 2, 6 and 7 in equation 5, we get,

n n(n-1) an ξⁿ⁻² +n (λ -1-2n) an ξ= 0      [n ranges from 0 to ∞]    (8)

Dealing with equation 8 will be hard as expanding out the sums would give contributions of the form ξ from both terms for i ≥ 2 and thus resulting into not all terms having same power in ξ collected together. This can be solved. Let’s see how…

Firstly, see that the first sum will be equal to 0 if n=0 or 1, so, we can write as follows,

n n(n-1) an ξⁿ⁻² = 0 + 0 + n n(n-1) an ξⁿ⁻²       (9)

Where, last term of RHS having n ranges from n=2 to ∞ and in the term of LHS, n ranges from n=0 to ∞.

What we see in equation 9 is that, here, n is a dummy index, which means that because the sum runs over all possible integral values which are greater than 0 up to ∞, it is clearly irrelevant what label is actually assigned to index of summation. In short, we can change or transform the index as we wish we solve the equation easily. So, let’s consider i = n 2, to solve the problem we discussed two paragraphs above. Now equation 9 converts into,

n n(n-1) an ξⁿ⁻² = ∑ᵢ (i+2) (i+1) ai+2 ξ      (10)

Where, in the term of LHS, n ranges from n=0 to ∞ and in the term of RHS, i ranges from i=0 to ∞

Now, in equation 10 both n and i are dummy indexes, so, we can define n=i,

n n(n-1) an ξⁿ⁻² = n (n+2) (n+1) an+2 ξ      (11)

If you have a doubt on equation 11, then just expand the equation on both sides, you will find they are absolutely identical.

So, now we have solved our problem which I discussed a few paragraphs above and thus now from equation 8 and 11,

n {(n+2) (n+1) an+2 + (λ -1-2n) an} ξ= 0      [n ranges from 0 to ∞]    (12)

From, equation 12 we can see that the terms inside the curly braces are pure numbers and don’t have any dependence on ξ. Also, the whole sum results in 0, and from our knowledge we know that in general, ξ is not equal to 0, but still, the sum is 0. The only way this can happen if every term inside curly braces vanishes, so this can be achieved by,

an+2 = {(2n+1- λ)/(n+2) (n+1)} an+2       (13)

Now, from our knowledge from series solution theory, we know that equation 13 is called as a recursion relation as it specifies a given expansion coefficient recursively in terms of the preceding coefficient in the series. So, we need to find a and a terms only and then we can find the even and odd terms on the basis of aand arespectively.

Before we progress further, I want to remind you guys that we have a boundary condition that is the wavefunction must be finite as ξ tends to ±∞.  So, in the equation 1, we know that e^(-ξ²/2) is convergent as we derived that in this way in our part-3 article on asymptotic solution. Thus, as long as H(ξ) is convergent or at worst case, if it diverges no more strongly than the exp(+ξ²/2), then the overall solution will remain convergent.

But, unfortunately, if you consider a general series expansion of then you will find it behaves asymptotically like exp(2ξ²). General series expansion of exp(ηξ),

exp(ηξ) = 1 + ηξ + η²ξ²ⁱ/2! + η³ξ³ⁱ/3! + …         (14)

Now, from equation 2 and 13,

Term(n+1) / Term(n) = (an+2 / an) ξ² = {(2n+1- λ)/(n+2) (n+1)}   (15)

And the ratio of successive terms in the exponential function of equation 14,

Term(n+1) / Term(n) = ηξ/(η+1)         (16)

Equation 15 and 16 at n tends to ∞,

Term(n+1) / Term(n) ~ 2 ξ²/n      (17)


Term(n+1) / Term(n) ~ ηξ/n      (18)

On comparing equation 17 and 18, we get, η=2 and i=2. Thus, we conclude that harmonic oscillator solution behaves as an exponential of the form exp(ηξ) = exp(2ξ²). This means that we can represent wavefunction as,

ψ(ξ) = H(ξ) e^(-ξ²/2) ~ e^(2ξ²) e^(-ξ²/2) ~ e^(3ξ²/2)       (19)

Equation 19 is clearly divergent and non-normalizable. The only way to solve this problem and prevent this catastrophe is to prevent the series for H(ξ) from going to an infinite number of terms. Thus, terminating the series of H(ξ) after a finite number of terms.

So, if the highest power of ξ appearing in H(ξ) is n, and if all a= 0 for i>n, then from the recursion relation (equation 13),

λ = 2n+1               

So, from above equation and equation 2 and 10 from Part-2 article that is Schrödinger’s Equation with Dimensionless Terms, we can find the energy corresponding to quantum number n:

λ = 2n+1 =  2E / ħω          


E = (n+1/2) ħω = (n+1/2) ħ(√k/m),    n=1,2,3,4,…          (20)

From equation 20, we can see that E is proportional to n, which indicates that the harmonic potential energy levels are equally spaced. These equally spaced energy levels are known as vibrational levels as the harmonic potential mimic the attractive forces between the molecules.

Apart from these, if you observe a spectrum then you will find that molecular spectra exhibit lines between the equally spaced vibrational transitions. These are called rotational levels. I will someday write a brief article on this topic.

So, that’s it, if you survived till the last then congrats as not as many people can do so. In my next article, which will be part-5 in this series, I will write on Hermite Polynomials and Quantum Harmonic Oscillator Wavefunctions. So, stay tuned for that. Once it is uploaded, the link will be available here.

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