The Schrödinger equation for the harmonic oscillator is
$$\left(-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}+\frac{m\omega^2}{2}x^2\right)\psi(x)=E\psi(x).$$
Then often $x$ gets substituted with $x=\xi a$, where $a=\sqrt{\hbar/m\omega}$, which leads to
$$\frac{\hbar\omega}{2}\left(-\frac{\partial^2}{\partial \xi^2}+\xi^2\right)\psi(\xi)=E\psi(\xi). $$
So my question is if now it should not be $\psi(a\xi)$, or do they define $\psi^\prime(\xi)= \psi(a\xi) $ and then rename it?
Quantum harmonic oscillator substitution
harmonic-oscillatornotationquantum mechanicsschroedinger equationwavefunction
Related Solutions
First of all you should recall that Schroedinger equation is an Eigenvalue equation. If you are unfamiliar with eigenvalue equations you should consult any math book or course as soon as possible.
Answer 1 (my apologies, I will use my own notation, as this is mainly copy-paste from my old notes):
First define constants \begin{equation} x_0 = \sqrt{\frac{\hbar}{m\omega}} , \end{equation} \begin{equation} p_0 = \frac{\hbar}{x_0} = \sqrt{\hbar m \omega} , \end{equation} and dimensionless operators \begin{equation} \hat{X} = \frac{1}{x_0} \hat{x} , \end{equation} and \begin{equation} \hat{P} = \frac{1}{p_0} \hat{p} . \end{equation}
Their commutation relation then is \begin{equation} \left[ \hat{X} , \hat{P} \right] = \left[ \frac{1}{x_0}\hat{x} , \frac{1}{p_0}\hat{p} \right] = \frac{1}{x_0 p_0} \left(\hat{x} \hat{p} - \hat{p} \hat{x} \right) = \frac{1}{x_0 p_0} \left[ \hat{x} , \hat{p} \right] = \frac{i\hbar}{x_0 p_0} = i , \end{equation} as \begin{equation} x_0 p_0 = \sqrt{\frac{\hbar}{m\omega}} \sqrt{\hbar m\omega} = \hbar . \end{equation}
Now write Hamiltonian in terms of $\hat{X}$ and $\hat{P}$. Start with \begin{equation} \hat{H} = \frac{p_0 ^2}{2m} \hat{P} ^2 + \frac{1}{2} m\omega^2 x_0 ^2 \hat{X}^2 . \end{equation}
Notice that \begin{equation} p_0 ^2 = \hbar m \omega \end{equation} and \begin{equation} x_0 ^2 = \frac{\hbar}{m \omega} , \end{equation} hence \begin{equation} \hat{H} = \frac{\hbar \omega}{2} \hat{P}^2 + \frac{\hbar \omega}{2} \hat{X}^2 = \frac{\hbar \omega}{2} \left(\hat{X}^2 + \hat{P}^2 \right). \end{equation}
Up to the commutation relation we can write \begin{equation} \left(X^2 + P^2 \right) = \left(X - iP \right) \left(X + iP \right) . \end{equation}
On the other hand, for operators this is not quite allowed, as \begin{align} \left(\hat{X} - i\hat{P} \right) \left(\hat{X} + i\hat{P} \right) &= \hat{X}^2 + i\hat{X}\hat{P} - i\hat{P}\hat{X} + \hat{P}^2 \nonumber \\ &= \hat{X}^2 +i \left(\hat{X}\hat{P} - \hat{P}\hat{X}\right) + \hat{P}^2 \\ &= \hat{X}^2 + i \left[ \hat{X} , \hat{P} \right] + \hat{P}^2 = \hat{X}^2 + \hat{P}^2 - 1 \nonumber , \end{align} so one has \begin{equation} \left(\hat{X}^2 + \hat{P}^2 \right) = \left(\hat{X} - i\hat{P} \right) \left(\hat{X} + i\hat{P} \right) + 1 . \end{equation}
Now we can define \begin{equation} \hat{a} = \frac{1}{\sqrt{2}} \left(\hat{X} + i\hat{P} \right) , \end{equation} and \begin{equation} \hat{a}^\dagger = \frac{1}{\sqrt{2}} \left(\hat{X} - i\hat{P} \right), \end{equation} calling this creation operator and $\hat{a}$ - annihilation operator. Notice that we can now express Hamiltonian in terms of creation and annihilation operators: \begin{equation} \hat{H} = \frac{\hbar\omega}{2} \left(\sqrt{2} \hat{a} ^\dagger \sqrt{2} \hat{a} + 1 \right) = \hbar \omega \left(\hat{a} ^\dagger \hat{a} + \frac{1}{2} \right) . \end{equation}
But we can also define the number operator, $\hat{N} = \hat{a} ^\dagger \hat{a}$, so finally get \begin{equation} \hat{H} = \hbar \omega \left(\hat{N} + \frac{1}{2} \right) . \end{equation}
Now go aside a bit and consider creation and annihilation operators. By definition,
\begin{equation} \hat{a}^\dagger \left| n \right\rangle = \sqrt{n + 1} \left| n + 1 \right\rangle , \end{equation} \begin{equation} \hat{a} \left| n \right\rangle = \sqrt{n} \left| n - 1 \right\rangle , \end{equation} where $\left| n \right\rangle$ is the eigenstate of creation and annihilation operators, as well as of the Hamiltonian (due to the fact that they commute - homework to prove).
Now \begin{equation} \hat{a}^\dagger\hat{a} \left| n \right\rangle = \hat{a}^\dagger \sqrt{n} \left| n - 1 \right\rangle = \sqrt{n} \sqrt{n} \left| n \right\rangle = n \left| n \right\rangle , \end{equation} so conclude that the eigenvalue of a number operator, $\hat{N}$, is just $n$, so if we now apply Hamiltonian in the Schroedinger equation, get
\begin{equation} \hat{H} \psi = E \psi , \end{equation} \begin{equation} E_n = \hbar \omega \left(n + \frac{1}{2} \right) , \end{equation} which is exactly the result you were looking for.
Answer 2:
First of all you should remember that the general aim of solving an eigenvalue problem is to find a set of eigenvectors, but not a single eigenvector. In your case, equation should be modified to
\begin{equation} \frac{d^2 \psi_n}{dx^2} + \left[(2n + 1) - \varepsilon^2\right]\psi_n = 0 , \end{equation} where $\psi_n$ are eigenvectors (eigenfunctions) that correspond to eigenvalues $E_n$. Try to think a little bit and explain physical meaning of having many energy eigenvalues in quantum mechanics.
Now return to the general theory of eigenvalue equations. Although I have never met the equation you wrote, I cannot find any place it can be wrong apart from the one just pointed out. Though, I don't see how far can you go from it.
Answer 3:
Hermite polynomials are usually beyond standard quantum mechanics courses. If you know Legendre, Chebyshev and/or other polynomials, you may guess that Hermite polynomials are derived as solution to some differential equation, and this does not contradict to the definition of $\psi$.
As I've already mentioned, Hermite polynomials are usually beyond standard quantum mechanics courses. Usually you are not supposed to derive them at this level. However, if you are still interested, you may want to consult with google or ask another question here.
Hope your questions have now been answered in full. However, should you need any further comment - you are welcome.
Indeed, as suggested by phase-space quantization, most of these equations are reducible to generalized Laguerre's, the cousins of Hermite. As universally customary, I absorb $\hbar$, M and ω into r,E. Note your E is twice the energy.
Since $r\geq 0$ you don't lose negative values, and you may may redefine $r^2\equiv x$, so that $$ r\partial_r = 2x \partial_x \qquad \Longrightarrow r\partial_r (r\partial_r)= r^2\partial_r^2+ r\partial_r=4(x^2\partial_x^2+x\partial_x), $$ hence your radial equation reduces to $$ \left ( \partial_x^2+ \frac{1}{x}\partial_x +\frac{E-x}{4x} -\frac{m^2}{4x^2} \right ) R(m,E)=0 ~. $$
Now, further define $$ R(m,E)\equiv x^{|m|/2} e^{-x/2} ~ \rho(m,E), $$ to get $$ \partial_x R(m,E)= x^{|m|/2} e^{-x/2} \left (-1/2 +\frac{|m|}{2x} + \partial_x \right )~ \rho(m,E) \\ \partial_x^2 R(m,E)= x^{|m|/2} e^{-x/2} \left (-1/2 +\frac{|m|}{2x} + \partial_x \right )^2~ \rho(m,E), $$ whence the generalized Laguerre equation for non-negative m=|m|, $$ x \partial_x^2\rho(m,E) +\left({m+1} -x\right )\partial_x \rho(m,E)+\frac{1}{2}(E/2-m-1) \rho(m,E)=0~. $$ This equation has well-behaved solutions for non-negative integer $$k=(E/2-m-1)/2\geq 0 ~,$$ to wit, generalized Laguerre (Sonine) polynomials $L^{(m)}_k (x)=x^{-m}(\partial_x -1)^k x^{k+m}/k!$.
Plugging into the factorized solution and the above substitutions nets your eigen-wavefunctions. The ground state is $k=0=m$, ($E=2$ in your conventions), so a radially symmetric Gaussian, $e^{-r^2/2}$.
Again, in your idiosyncratic convention, the degeneracy is E/2.
So, degeneracy 2 for $E=4$ : $m=1$, $k=0$; you may check this is just $r e^{-r^2/2 +i\phi} $. You may choose the $\cos \phi$ and $\sin \phi$ solutions, if you wish, constituting a doublet of the underlying degeneracy group SU(2).
Best Answer
Yes, you are right, one just redefines the $\psi(x)=\psi_{new}(\xi)=\psi(a\xi)$. The Schrödinger equation is thus $$\frac{\hbar\omega}{2}\left(-\frac{\partial^2}{\partial \xi^2}+\xi^2\right)\psi_{new}(\xi)=E\psi_{new}(\xi) $$ The basic motivation is to non dimensionalize the independent variable. In order to recover the original $\psi(x)$, we have $$\psi(x)=\psi_{new}(\frac{x}{a})$$