Find the relationship between $a_+\psi_n$ and $\psi_{n+1}$

My attempt:

I was able to prove that

$\int{(a_+\psi)^*(a_+\psi)dx} = \int{\psi^*({a_-a_+\psi})dx}\qquad\qquad (1)$


$(a_-a_+-\frac{\hbar\omega}{2})\psi = E\psi \qquad\qquad \qquad\qquad\qquad(2)$

Therefore from $(1)$ and $(2)$

$\int{(a_+\psi)^*(a_+\psi)dx}$ $ = \int{\psi^*}(E+\frac{\hbar\omega}{2})\psi dx$ $=(E+\frac{\hbar\omega}{2})\int{\psi^*}\psi dx $ $= (E+\frac{\hbar\omega}{2}) $

$a_+\psi_n$ is proportional to $\psi_{n+1}$

So, $a_+\psi_n = c_n\psi_{n+1}$

Normalising on both sides

$\int{|a_+\psi_n|^2 dx} = \int{|c_n\psi_{n+1}|^2}dx$

$E_n+\frac{\hbar\omega}{2} = |c_n|^2$

but, $E_n = (n + \frac{1}{2})\hbar\omega$


$c_n = \sqrt{(n+1)\hbar\omega} $

But the answer given in book is

$c_n = i\sqrt{(n+1)\hbar\omega} $

A similar result I got for $a_-\psi_n$ and $\psi_{n-1}$ is

$c_n = \sqrt{n\hbar\omega}$

But the answer given in book is

$c_n = -i\sqrt{n\hbar\omega}$

And the reason given for these answers is "$i$'s are there to make the wavefunction real"

Please help me understand this statement or point out where I've gone wrong in proving the given equation


2 Answers 2


Your answer is perfectly fine. As you can see one can choose an abritrary phase $\exp(i\phi)$ for $c_n$ in the equation $$E_n + \frac{\hbar \omega}{2} = |c_n|^2$$ and it will still hold. This relates to the fact that you can always choose an arbitrary phase for the eigenfunctions $\psi_n$. All physical observables (e.g. $A_{nn} =\langle \psi_n|\hat{A}|\psi_n\rangle $) are invariant under the choice of the phase. In your book they just chose it to their convenience.


Recall that states or wave-functions are only defined up to an overall phase, i.e. $\psi(x)$ and $e^{i \alpha(x)} \psi(x)$ are both wave-functions that describe the same state. The wave-function generically is a complex function of the form $\psi = f(x) e^{i h(x)}$ where $f(x)$ and $h(x)$ are real functions. It is then often convenient to make a choice of the phase $\alpha(x) = - h(x)$ so that $e^{i \alpha(x)} \psi(x) = f(x)$ is real. This is an completely equivalent and good definition of the quantum state since all we have done is changed the wave function by a phase.

This is what your book is doing. They are making a choice of the phase so that all wave-functions $\psi_n$ turn out to be real.


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