Expressing nth-eigenstate in terms of the zeroth using ladder operator | Quantum Science Philippines

## Expressing nth-eigenstate in terms of the zeroth using ladder operator

By: Argielou A. Flores, MSU-IIT

Show that $|\psi_n> = \frac{1}{\sqrt{n!}}(a^+)^n |\psi_0>$

Solution:

Given the following:

$|\psi_1> = a^+ |\psi_0>\\*|\psi_2> = \frac{1}{\sqrt{2}} ( a^+)^2 |\psi_0> = \frac{1}{\sqrt{2}} a^+ |\psi_1>$

So if normalize $|\psi_{n-1}>$ is given we can then obtain $|\psi_n>$ using ladder operator $a^+$ as shown below,

$|\psi_n> = c_n a^+ |\psi_{n-1}>$

Taking its inner product we have,

$<\psi_n|\psi_n> = |c_n|^2<\psi_{n-1}|\psi_{n-1}>$

Note that,

$[a, a^+] = aa^+ - a^+a = 1$

$\begin{array}{lll} aa^+ & = &a^+a +1\\&=& N+1\end{array}$

where $N = a^+a$ with eigenvalue $\nu = (n-1)$

Going on, we now have

$\begin{array}{lll} <\psi_n|\psi_n> & = &|c_n|^2<\psi_{n-1}|N+1|\psi_{n-1}>\\& = & |c_n|^2<\psi_{n-1}|(n-1)+1|\psi_{n-1}>\\& = & n|c_n|^2<\psi_{n-1}|\psi_{n-1}>\end{array}$

where $<\psi_n|\psi_n>=<\psi_{n-1}|\psi_{n-1}>=1$ for  normalized $\psi_n$ and $\psi_{n-1}$

By this we obtain the normalization constant $c_n$ which is

$c_n = \frac{1}{\sqrt{n}}$

Now, using $c_n$,

$\begin{array}{lll} |\psi_n> & = &\frac{1}{\sqrt{n}}a^+|\psi_{n-1}>\\& = &\frac{1}{\sqrt{n}} \frac{1}{\sqrt{n-1}}(a^+)^2 |\psi_{n-2}>\\& = &\frac{1}{\sqrt{n}} \frac{1}{\sqrt{n-1}} \frac{1}{\sqrt{n-2}}(a^+)^3 |\psi_{n-3}>\\& = & \cdots \end{array}$

From this, we can express (obtain) $|\psi_n>$ in terms of (from) $|\psi_0>$ as follows:

$|\psi_n>=\frac{1}{\sqrt{n!}}(a^+)^n |\psi_0>$