Question

Quantum Mechanics Question - How to normalize this wavefunction?

Answer 1

(the normalization constant is there; I was just wondering how they derived it)

Answer 2

I've seen this in ap biology but, I'm not sure how to do it. I wish I was more helpful.

Answer 3

A particle with mass mm is moving in one dimension. The wave function of the particle is

Ψ(x,t)=Axe−(km√/2ℏ)x2e−ik/m√(3/2)t
Ψ(x,t)=Axe−(km/2ℏ)x2e−ik/m(3/2)t
for −∞<x<∞−∞<x<∞, where kk and AA are constants. Normalize this wave function.

To normalize a wave function, we impose the condition
∫∞−∞|Ψ(x,t)|2dx=1
∫−∞∞|Ψ(x,t)|2dx=1
and solve for any constants (in this case, AA, and possibly kk). To make my life easier, I set D=km√ℏD=kmℏ, so Ψ(x,t)=Axe−(D/2)x2e−ik/m√(3/2)tΨ(x,t)=Axe−(D/2)x2e−ik/m(3/2)t, and |Ψ(x,t)|2=A2x2e−Dx2|Ψ(x,t)|2=A2x2e−Dx2. Thus, we are tasked with evaluating the integral
A2∫∞−∞x2e−Dx2dx
A2∫−∞∞x2e−Dx2dx
Using an integral table, I found that

∫∞−∞x2e−Dx2dx=14πD3−−−√erf(xD−−√)∣∣∣∞−∞−x2De−Dx2∣∣∣∞−∞
∫−∞∞x2e−Dx2dx=14πD3erf(xD)|−∞∞−x2De−Dx2|−∞∞
Supposedly, erf(∞)=1erf(∞)=1 and erf(−∞)=−1erf(−∞)=−1, so the first term evaluates to 12πD3−−−√12πD3. For the second term, since DD is a positive quantity, evaluating it at ∞∞ gives 00, but I can't seem to use similar reasoning and intuition for −∞−∞.

Crudely substituting −∞−∞ into the expression gives (−∞)(e∞)=(−∞)(∞)(−∞)(e∞)=(−∞)(∞). Is this an indeterminate form? And if so, how should L'Hospital's rule be used to help? I tried it a few different ways and didn't make any progress.

PS: It might be silly to want to do the last step of evaluating the integral at its limits, when I already used a table to get to that step, but I'm mostly just curious to see how it can be done analytically / algebraically.

I hope this is right but dont listen to me btw. Get confermation from someone else. I'm not to confadent about this answer >.<

Answer 4

This is Qu 10 in Boas' Calculus book, Ch 12 Sect 22, it comes with a few hints and certain "EXTRA"neous results that are necessary. They are "proved" separately in the book

\(\int\limits_{-\infty}^{\infty} dx ~ ~ \Psi^* \Psi \)


\(= \int\limits_{-\infty}^{\infty} dx ~ ~ N_n H_n e^{- \alpha x^2 /2}~ N_m H_m e^{- \alpha x^2 /2} \)


\(= N_n N_m \int\limits_{-\infty}^{\infty} dx ~e^{- \alpha x^2} ~ H_n( \sqrt \alpha x) ~ H_m ( \sqrt \alpha x)\)

with \(X = \sqrt \alpha x\)


\(= \frac{1}{\sqrt \alpha} N_n N_m \int\limits_{-\infty}^{\infty} dX ~e^{- X^2} ~ H_n(X) ~ H_m(X) \)

EXTRA 1: the Rodrigues formula: \(H_n = (-1)^n e^{X^2} \dfrac{d^n}{dX^n} e^{- X^2}\):

\(= \frac{1}{\sqrt \alpha} N_n N_m \int\limits_{-\infty}^{\infty} dX ~e^{- X^2} ~ (-1)^n e^{X^2} \dfrac{d^n}{dX^n} e^{- X^2} ~ H_m \)

By IBP:

\(= \frac{1}{\sqrt \alpha} N_n N_m (-1)^n \left( \left[\dfrac{d^{n-1}}{dX^{n-1}}e^{- X^2} H_m \right]_{- \infty}^{\infty} - \int\limits_{-\infty}^{\infty} dX ~ \dfrac{d^{n-1}}{dX^{n-1}} e^{- X^2} ~\dfrac{d}{dX} H_m \right)\)

EXTRA 2: the recursion result: \(H'_n = 2n H_{n-1}\), and a reverse-Rodriguez:

\(= \frac{1}{\sqrt \alpha} N_n N_m (-1)^n \left( \left[\dfrac{d^{n-1}}{dX^{n-1}}e^{- X^2} H_m \right]_{- \infty}^{\infty} - \int\limits_{-\infty}^{\infty} dX ~ \dfrac{1}{(-1)^{n-1}} H_{n-1} e^{- X^2} ~2m H_{m-1} \right)\)

\(= \frac{1}{\sqrt \alpha} N_n N_m (-1)^n \left( 0 - \dfrac{1}{(-1)^{n-1}} 2m\int\limits_{-\infty}^{\infty} dX ~ e^{- X^2} H_{n-1} H_{m-1} \right)\)

\(\implies I_{n,m} = \frac{2m}{\sqrt \alpha} N_n N_m \int\limits_{-\infty}^{\infty} dX ~ e^{- X^2} H_{n-1} H_{m-1} = 2m \dfrac{N_nN_m}{N_{n-1}N_{m-1} }I_{n-1, m-1}\)

EXTRA 3: Theeee bombshell, Hermite polynimials are orthogonal when weighted by \(e^{- X^2}\), so we can specify that n = m at any point for an actual wavefunction to exist

So \( I_{n,n} = 2n \dfrac{N_n^2}{N_{n-1}^2 }I_{n-1, n-1}\)

We can also drop all that normalisation palaver as it's a mess ***and we would get the same result by using actual values of the integrals*** ie really wish i had left those out and normalised later but there is soooo much Latex,...., so we'll ditch them mid stream.... and simplifying


\(I_{n,n} = 2n I_{n-1,n-1}\)

\(I_{0,0} = \int\limits_{-\infty}^{\infty}~ dx ~ e^{- \alpha x^2} = \sqrt {\dfrac{\pi}{\alpha}}\) as \(H_0 = 1\)

\(I_{1,1} = 2(1) \sqrt {\dfrac{\pi}{\alpha}}\)

\(I_{2,2} = 2\cdot 2 (1 \cdot 2) \sqrt {\dfrac{\pi}{\alpha}}\)

And generally: \(I_{n,n} = 2^n n! \sqrt {\dfrac{\pi}{\alpha}}\)

so if we NOW force normalisation \(N_n^2 I_{n,n} = 1 \implies N_n^2 = \dfrac{1}{2^n n! \sqrt {\dfrac{\pi}{\alpha}}}\)

which is your result.....

Answer 5

That's not really physics or chemistry though :)

Answer 6

Thank you so much, it must have taken a long time to do this ;;

Answer 7

Yes, Voca, it took ages 😃

but i learned to love the ladder operators more

this stuff doesn't even merit a mention in Eisberg/Resnick. hence Boas. and TUVM for the qu 😃