I'm teaching myself about QM, but there are something really puzzling me about the simple harmonic oscillator: $$H=\frac{p^2}{2m}+\frac{m\omega^2x^2}{2}.$$

I've learned how to use ladder operators to obtain the eigenvalues of this oscillator. Also, I'm able to write out the eigenstates in the form of position space wave functions: $$\langle x'|n\rangle=\left(\frac{1}{\pi^{1/4}\sqrt{2^n n!}}\right)\left(\frac{1}{x^{n+1/2}_0}\right)\left(x'-x^2_0\frac{d}{dx'}\right)^n\exp\left[-\frac{1}{2}\left(\frac{x'}{x_0}\right)^2\right],$$ where $$x_0\equiv \sqrt{\frac{\hbar}{m\omega}}.$$ So the initial wave function must be a superposition of these eigenfunctions.

However, given an arbitrary normalized wave function $\langle x'|\alpha\rangle$ which is not necessarily a proper superposition, I can use $|\alpha\rangle$ as an initial state and make it evolve according to the Schrödinger equation: $$\langle x'|i\hbar\frac{\partial}{\partial t}|\alpha;t\rangle = \langle x'|H|\alpha;t\rangle,$$ which seems to make sense.

So my question is:

  1. Can any normalized wave function be represented as a superposition of the eigenfunctions?
  2. If not, what would happen if I set the initial state to a wave function that is not a superposition of the eigenfunctions?

Also, there is another question which might be related:

  1. The numbers of eigenfunctions for $x$ and $p$ are obviously uncountably infinite. But how could it be that this number is countably infinite for $H$?
  1. The states $\{| n \rangle \}$ form a complete basis, so at any time you can expand any state $|\alpha(t)\rangle $ as a linear combination of $| n \rangle$,

$$ | \alpha(t) \rangle = \sum_n c_n(t)| n \rangle \tag{1} $$

Now is a matter of finding the coefficients $c_n$. To do that note that

\begin{eqnarray} i\hbar \frac{{\rm d}}{{\rm d}t}| \alpha(t) \rangle &=& i\hbar\sum_n \dot{c}_n(t) | n \rangle \\ H| \alpha(t) \rangle &=& \sum_n c_n(t) H | n \rangle = \sum_n c_n(t) E_n | n \rangle \\ \Rightarrow i\hbar\sum_n \dot{c}_n(t) | n \rangle &=& \sum_n c_n(t) E_n | n \rangle \end{eqnarray}

with $E_n = \hbar \omega(n + 1/2)$. Multiplying both sides by a state $| m\rangle$, and recalling that $\langle m | n \rangle = \delta_{nm}$

$$ i\hbar \dot{c}_m(t) = c_n(t)E_n $$

whose solution is

$$ c_n(t) = c_n(0)e^{-iE_n t/\hbar} \tag{2} $$

The coefficients $c_n(0)$ are easily obtained from Eq. (1):

$$ \langle m | \alpha(0)\rangle = \sum_n c_n(0)\langle m | n\rangle = c_m(0) $$

That is

$$ c_n(0) = \langle n | \alpha(0)\rangle \tag{3} $$

so, the evolution of any state $|\alpha(t)\rangle$ can be written as Eq. (1), where the coefficients $c_n(t)$ evolve according to Eq. (2) with initial conditions given by Eq. (3)

  1. Since the set $\{ | n \rangle \}$ you can always write any state as a linear combination of eigenstates of $H$

  2. Please follow this link, but intuitively speaking, the potential $V(x) = m\omega^2 x^2/2$ has infinitely many bound states, this means that no matter how large the energy of a particle is, you can always contain it with $V$. In this case the states are also countable, indeed, you can label them with a single integer $n$

  • 1
    $\begingroup$ Thanks. According to the link you gave, it seems that the space of any normalized wave function is $L^2(\mathbb R)$. So intuitively the eigenfunctions $\{ |n\rangle\}$ form a complete basis. But could you give more details about the completeness of this basis, e.g., a proof? $\endgroup$
    – OwUy
    May 25 '17 at 1:11
  • $\begingroup$ @OwUy Basically the reason is that $\{|n\rangle\$ are the eigenfunctions of a compact Hermitian operator with discrete eigenvalues. You could check this link for more details. Also, I think your last question goes beyond the scope of this post, please consider create a new one to address the issue of completness $\endgroup$
    – caverac
    May 25 '17 at 6:14

I'll address 3. The word "basis" is often defined to only allow linear combinations of finitely many basis elements. This is the definition in the famous theorem stating all bases have the same cardinality. Hilbert spaces are unusual in that any finite-norm linear combination of orthogonal elements will define a unique element of the space; we say it's metric-complete, which doesn't apply to arbitrary inner product spaces. (This is somewhat analogous to the fact that, although $\mathbb{R}$ contains the limits of all its Cauchy sequences, general Cauchy sequences in $\mathbb{Q}$ converge to arbitrary real numbers.) A "basis" of a Hilbert space is defined more generally. As a result, a Hilbert space's bases can vary in cardinality.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.