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

 

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