# Schrödinger equation for a harmonic oscillator

I have came across this equation for quantum harmonic oscillator

$$W \psi = - \frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} + \frac{1}{2} m \omega^2 x^2 \psi$$

which is often remodelled by defining a new variable $\varepsilon = \sqrt{m\omega/\hbar}\,x$. If i plug this in the equation above I know how to derive this equation (it is easy all i needed was a definition of differential):

$$\frac{d^2 \psi}{d x^2} + \left(\frac{W}{\left( \tfrac{\hbar \omega}{2} \right)} - \varepsilon^2 \right)\, \psi = 0$$

From this equation most of authors derive the energy equation:

$$W = \left(\tfrac{\hbar \omega}{2}\right)\cdot (\underbrace{2n+1}_{odd???})$$

QUESTION 1: I don't quite understand how this is done. Could anyone please explain this? I dont know why energy has to be odd function nor how this comes into play.

QUESTION 2: I am only guessing that if I would plug this in an equation above I would get I an equation below from which I could calculate possible $\psi$ functions for harmonic oscilator (please confirm).:

$$\frac{d^2 \psi}{d x^2} + \left( (2n +1)- \varepsilon^2 \right)\, \psi = 0$$

QUESTION 3: Where do Hermitean polinoms (which are often mentioned) come into play?

• The energy levels are spaced an energy $\hbar\omega$ apart, but the ground state has energy $\hbar\omega$/2. Your expression for $W$ gives these energies. The solutions for the QHO are Hermite polynomials, and their derivation is not particularly simple. See the Wikipedia article for the gory details: en.wikipedia.org/wiki/Quantum_harmonic_oscillator – John Rennie Apr 29 '13 at 7:42

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.

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.

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 $\psi$.
• Can i ask where did you use the commutator relation between $\hat{X}$ and $\hat{P}$? Does an $i$ means their order i need to multiply that by an $i$? – 71GA Apr 29 '13 at 18:22