I have came across this equation for quantum harmonic oscillator

$$ W \psi = - \frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} + \frac{1}{2} m \omega^2 x^2 \psi \tag{1}$$ which is often remodeled by defining a new variable $\varepsilon = \sqrt{m\omega/\hbar}\,x$. If I plug this in the equation above then we get

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

From this equation most of authors derive the energy equation

$$ W = \frac{\hbar \omega}{2} \cdot (\underbrace{2n+1}_{\text{odd}?}) \tag{3} $$

QUESTION 1: Where does the equation for $W$ come from? I don't know why energy has to be an odd function of $n$.

QUESTION 2: Is it correct that if we plug Eq. (3) into Eq. (2), then we get Eq. (4) from which we compute allowed $\psi$?

$$ \frac{d^2 \psi}{d x^2} + \left( (2n +1)- \varepsilon^2 \right)\, \psi = 0 \tag{4} $$

QUESTION 3: Where do Hermitean polynomials come into play?

  • 3
    $\begingroup$ 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 $\endgroup$ Commented Apr 29, 2013 at 7:42

1 Answer 1


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.

  • $\begingroup$ Thank you very much for all your effort! I will need some time to digest this as it is a big leap for me. I have been avoiding eigenvectors and eigenvalues untill now, but suprisingly i managed to understand every single thing until this one. Could you point me to any good video/book explaining eigenvalues or eigenvectors? I am allso unfamiliar with commutators... Am i in a big trouble? $\endgroup$
    – 71GA
    Commented Apr 29, 2013 at 17:39
  • $\begingroup$ It depends on your situation: if you are high school guy learning quantum mechanics for fun, then you will be fine :-) if you are second year undergrad having exams this week, then you definitely stuck. It is impossible to learn further quantum mechanics without these concepts. Depending you your learning style, I might recommend Schaum's Outlines series -- if you like learning stuff by solving on your own; Arfken of Riley -- if you prefer dense explanation with derivations. Actually, these are not mutually exclusive, so you can make use of both (three, actually). $\endgroup$
    – Eugene B
    Commented Apr 29, 2013 at 17:54
  • $\begingroup$ Also, you can spend some time googling for lecture notes (there are plenty of them as well as of examples) and select those you prefer. To be honest, I have never had difficulties at this point, therefore haven't mined literature, just solved problems from Schaum's. Really, it all depends on your learning style and on amount of spare time you have. $\endgroup$
    – Eugene B
    Commented Apr 29, 2013 at 17:57
  • $\begingroup$ 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 $i$? $\endgroup$
    – 71GA
    Commented Apr 29, 2013 at 18:22
  • $\begingroup$ It comes from the commutation relation between position and momentum operators. It is easy to derive and shall be your homework :-) $\endgroup$
    – Eugene B
    Commented Apr 29, 2013 at 18:25

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