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omoplata
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From page 91 of "Modern Quantum Mechanics, revised edition", by J. J. Sakurai.
Some operators used below are,
<br /> a = \sqrt{\frac{m \omega}{2 \hbar}} \left(x + \frac{ip}{m \omega} \right)\\<br /> a^{\dagger} = \sqrt{\frac{m \omega}{2 \hbar}} \left(x - \frac{ip}{m \omega} \right)\\<br /> N = a^{\dagger} a<br />
In equation (2.3.16), he finds
<br /> a \mid n \rangle = \sqrt{n} \mid n - 1 \rangle<br />
Next he says it's similarly easy to show equation (2.3.17),
<br /> a^{\dagger} \mid n \rangle = \sqrt{n+1} \mid n + 1 \rangle<br />
I didn't find it to be so easy. Following the same procedure as equation (2.3.16), since from equation (2.3.12a),
<br /> Na^{\dagger} \mid n \rangle = (n+1) a ^{\dagger} \mid n \rangle<br />, I took
<br /> a^{\dagger} \mid n \rangle = c \mid n + 1 \rangle<br />, where c is some constant. Then,
<br /> \langle n \mid a a^{\dagger} \mid n \rangle = | c |^2<br />
To find a^{\dagger} a,
<br /> \begin{eqnarray}<br /> a a^{\dagger}<br /> & = & \left( \frac{m \omega}{2 \hbar} \right) \left( x + \frac{ip}{m \omega} \right) \left( x - \frac{ip}{m \omega} \right)\\<br /> & = & \frac{H}{\hbar \omega} + \frac{1}{2}\\<br /> \end{eqnarray}<br />
Since it is proven in equation (2.3.5) and (2.3.6) that H = \hbar \omega (N + \frac{1}{2}), I get,
<br /> a a^{\dagger} = N + 1<br />
Then,
<br /> \langle n \mid a a^{\dagger} \mid n \rangle = \langle n \mid N + 1 \mid n \rangle = n + 1<br />
Therefore, from above,
<br /> c = \sqrt{n+1}<br />, which gives,
<br /> a^{\dagger} \mid n \rangle = \sqrt{n+1} \mid n + 1 \rangle<br />, which is the desired result.
But is there an easier way to get this?
At the same time,
<br /> a a^{\dagger} = N^{\dagger}<br />
So,
<br /> \langle n \mid a a^{\dagger} \mid n \rangle = \langle n \mid N^{\dagger} \mid n \rangle = n^{*}<br />
Therefore,
<br /> n+1 = n^{*}<br />
How can this happen? I can't see this happening unless the real component of n tends to be infinity.
Some operators used below are,
<br /> a = \sqrt{\frac{m \omega}{2 \hbar}} \left(x + \frac{ip}{m \omega} \right)\\<br /> a^{\dagger} = \sqrt{\frac{m \omega}{2 \hbar}} \left(x - \frac{ip}{m \omega} \right)\\<br /> N = a^{\dagger} a<br />
In equation (2.3.16), he finds
<br /> a \mid n \rangle = \sqrt{n} \mid n - 1 \rangle<br />
Next he says it's similarly easy to show equation (2.3.17),
<br /> a^{\dagger} \mid n \rangle = \sqrt{n+1} \mid n + 1 \rangle<br />
I didn't find it to be so easy. Following the same procedure as equation (2.3.16), since from equation (2.3.12a),
<br /> Na^{\dagger} \mid n \rangle = (n+1) a ^{\dagger} \mid n \rangle<br />, I took
<br /> a^{\dagger} \mid n \rangle = c \mid n + 1 \rangle<br />, where c is some constant. Then,
<br /> \langle n \mid a a^{\dagger} \mid n \rangle = | c |^2<br />
To find a^{\dagger} a,
<br /> \begin{eqnarray}<br /> a a^{\dagger}<br /> & = & \left( \frac{m \omega}{2 \hbar} \right) \left( x + \frac{ip}{m \omega} \right) \left( x - \frac{ip}{m \omega} \right)\\<br /> & = & \frac{H}{\hbar \omega} + \frac{1}{2}\\<br /> \end{eqnarray}<br />
Since it is proven in equation (2.3.5) and (2.3.6) that H = \hbar \omega (N + \frac{1}{2}), I get,
<br /> a a^{\dagger} = N + 1<br />
Then,
<br /> \langle n \mid a a^{\dagger} \mid n \rangle = \langle n \mid N + 1 \mid n \rangle = n + 1<br />
Therefore, from above,
<br /> c = \sqrt{n+1}<br />, which gives,
<br /> a^{\dagger} \mid n \rangle = \sqrt{n+1} \mid n + 1 \rangle<br />, which is the desired result.
But is there an easier way to get this?
At the same time,
<br /> a a^{\dagger} = N^{\dagger}<br />
So,
<br /> \langle n \mid a a^{\dagger} \mid n \rangle = \langle n \mid N^{\dagger} \mid n \rangle = n^{*}<br />
Therefore,
<br /> n+1 = n^{*}<br />
How can this happen? I can't see this happening unless the real component of n tends to be infinity.
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