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I am going through Griffiths' Intro to QM, and in his solution the quantum harmonic oscillator, he just derived the recursion formula:

$$a_{j+2}=\frac{-2(n-j)}{(j+1)(j+2)}$$

Using this, we can find the Hermite polynomials (although with different constants in front), so I can find now the solutions (with $A$ some constant: $$\psi(\xi)=AH(\xi)e^{-\xi^2/2}$$ However, Griffiths now says that the solution is: $$\psi_n(x)=\left(\frac{m\omega}{\hbar\pi}\right)^{1/4}\frac{1}{\sqrt{2^nn!}}H(\xi)e^{-\xi^2/2}$$

But I just don't know where that value for A comes from. He does say that the $2^n$ is for convenience, but that is all that I see spoken about it. How could I prove where this comes from?

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  • $\begingroup$ In the last equation you changed from $\psi(\xi)$ to $\psi_n(x)$, while on the right hand side there is no $x$ (at least explicitly). $\endgroup$ Commented Jan 26, 2021 at 21:21
  • $\begingroup$ I know, io found it strange too, but that is the way it is written on griffiths 3rd edition. Is this a mistake? $\endgroup$ Commented Jan 26, 2021 at 21:22
  • $\begingroup$ I don't have the book, but I could imagine that $\xi$ is some dimensionless quantity which depends on $x$? Edit: Comparing with the equation given at Wikipedia yields the desired relation: en.wikipedia.org/wiki/Quantum_harmonic_oscillator . $\endgroup$ Commented Jan 26, 2021 at 21:23

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Constants for a wave function always come from normalisation. Recall that the wave function is in fact a probability distribution, so the following holds (since we have a one dimensional QHO):

$$\int_{-\infty}^{\infty}\psi^*(\zeta)\psi(\zeta) d\zeta = 1$$

If you evaluate this integral, you will find it checks out with the constants as given in Griffiths.

P.S. This is actually a nice enough integral to do using integration by parts. You can try it for the first few Hermite polynomials.

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  • $\begingroup$ Is there a way to normalize the general "n" state? Or do I have to normalize each individual states and then look for a pattern? $\endgroup$ Commented Jan 26, 2021 at 21:24
  • $\begingroup$ @NickHeumann As you can see in the expression you gave in the post, the normalization constant 'depends on $n$'. Hence for each $\psi_n$ there is a normalization constant $A_n$. But you can do the integral for an arbitrary $n$ to find $A_n$ explicitly. $\endgroup$ Commented Jan 26, 2021 at 21:27
  • $\begingroup$ @NickHeumann Which definition of Hermite polynomials are you familiar with? Take a look at the physicists' Hermite polynmials on the wiki page here :en.wikipedia.org/wiki/Hermite_polynomials It gives a non-recirsive formula which is helpful $\endgroup$
    – Meep
    Commented Jan 26, 2021 at 21:31

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