Harmonic Oscillator Finding <x^2>

**bugatti79** · November 13th 2010, 01:06 PM

Dear Experts,

I am having difficulty undersanding the pattern when working with operators which is part of the work to evaluate <x^2>. See attached jpg.

Likewise the same with evaluating $a_{+}a_{-}\psi_{n}(x)$

Thanks

**mr fantastic** · November 13th 2010, 02:23 PM

Originally Posted by bugatti79

Dear Experts,

I am having difficulty undersanding the pattern when working with operators which is part of the work to evaluate <x^2>. See attached jpg.

Likewise the same with evaluating $a_{+}a_{-}\psi_{n}(x)$

Thanks

$a_{+} \sqrt{n+1} \, \psi_{n+1} = \sqrt{n+1} \, a_{+} \psi_{n+1}$ and $a_{+} \psi_{n+1} = \sqrt{n+2} \, \psi_{n+2}$ by definition of the raising operator.

**bugatti79** · November 14th 2010, 01:56 AM

On that basis,

I calculate $a_{+}a_{-} \psi_{n}=\sqrt{n} a_{+} \psi_{n-1}=\sqrt{n} \sqrt{n+1} \psi_{n}$

$a_{-} a_{+}\psi_{n}=\sqrt{n+1} a_{-} \psi_{n+1}=\sqrt{n+1} \sqrt{n} \psi_{n}$

$a_{-}^2 \psi_{n}=\sqrt{n} a_{-} \psi_{n-1}=\sqrt{n} \sqrt{n-1} \psi_{n-2}$

The first 2 calculations are wrong by my lectures notes, so obviously dont understand the definition.

For example, it there is an adding operator, I will sum it with n. Ie, adding a_+ operator to n+1 gives n+2, subtracting a_- to n gives n-1 etc.....

**mr fantastic** · November 14th 2010, 09:58 AM

Originally Posted by bugatti79

On that basis,

I calculate $a_{+}a_{-} \psi_{n}=\sqrt{n} a_{+} \psi_{n-1}=\sqrt{n} \sqrt{n+1} \psi_{n}$

$a_{-} a_{+}\psi_{n}=\sqrt{n+1} a_{-} \psi_{n+1}=\sqrt{n+1} \sqrt{n} \psi_{n}$

$a_{-}^2 \psi_{n}=\sqrt{n} a_{-} \psi_{n-1}=\sqrt{n} \sqrt{n-1} \psi_{n-2}$

The first 2 calculations are wrong by my lectures notes, so obviously dont understand the definition.

For example, it there is an adding operator, I will sum it with n. Ie, adding a_+ operator to n+1 gives n+2, subtracting a_- to n gives n-1 etc.....

Use the definitions more carefully. eg. $a_{+} \psi_{n-1} \neq \sqrt{n+1} \psi_{n}$ etc.

**bugatti79** · November 14th 2010, 11:25 AM

Originally Posted by mr fantastic

Use the definitions more carefully. eg. $a_{+} \psi_{n-1} \neq \sqrt{n+1} \psi_{n}$ etc.

I have it now, when I compare
$a_{+}\psi_{n}=\sqrt{n+1} \psi_{n+1}$ and $a_{+} \psi_{n-1}$ together I know hows it derived.

Cheers

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