1. ## Qm 1

Alright, got lots and lots of Quantum questions. A large variety.

Note everything in bold = operator

1.) These are all for a simple harmonic oscillator...

a.) Prove the expression of $\displaystyle \bold{a^\dagger}$ from $\displaystyle \bold{a}$ in the equation below (1.1) using $\displaystyle <\phi|\bold{A^\dagger}|\psi> = <\psi|\bold{A}|\phi>^*$

(1.1) $\displaystyle \bold{a} = \sqrt{m\omega/2\hbar}\bold{x} + i\bold{p}/\sqrt{2m\hbar\omega}$

(1.1 continued) $\displaystyle \bold{a^\dagger} = \sqrt{m\omega/2\hbar}\bold{x} - i\bold{p}/\sqrt{2m\hbar\omega}$

b.) Write $\displaystyle \bold{H}$ in terms of $\displaystyle \bold{x}$ and $\displaystyle \bold{p}$ and prove the expression of $\displaystyle H$ in terms of $\displaystyle \bold{a}$ and $\displaystyle \bold{a^\dagger}$.

c.) Find:

$\displaystyle [\bold{a^\dagger},\bold{a}] = ?$
$\displaystyle [\bold{H},\bold{a^\dagger}] = ?$
$\displaystyle [\bold{H},\bold{a}] = ?$

d.) If $\displaystyle |U_n>$ is the eigenstate of $\displaystyle \bold{H}$ w/ eigenvalue $\displaystyle E_n$, ie., $\displaystyle \bold{H}|U_n> = E_n|U_n>$, prove $\displaystyle |U_m> = \bold{a^\dagger} | U_n>$ is also an eigenstate of $\displaystyle H$ w/ a different eigenvalue. Determine the eigenvalue that corresponds to $\displaystyle |U_m>$.

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We know $\displaystyle \bold{H}|U_n = E_n|U_n>$ is the time indept. Schrod. equation, where $\displaystyle |U_n>$ is the eigenstate of $\displaystyle \bold{H}$.

We'll denote $\displaystyle \bold{a}|U_n>$ to be $\displaystyle U_m$.

I will prove that $\displaystyle \bold{H}|\bold{a}|U_N> = (E_n - \hbar\omega)\bold{a}|U_n>$. In order to do so, I conveniently have to know what $\displaystyle [\bold{H},\bold{a}] = ?$ (this is a question in part c...).

So my work:

$\displaystyle [\bold{H},\bold{a}] = [\hbar\omega\bold{a^\dagger}\bold{a} + \frac{1}{2}\hbar\omega, \bold{a}]$

We can use $\displaystyle [\bold{A},\bold{C}] + [\bold{B},\bold{C}] = [\bold{A} + \bold{B}, \bold{C}]$

$\displaystyle = [\hbar\omega\bold{a^\dagger},\bold{a}] + [\mbox{this ends up being 0 since we have a constant}]$

$\displaystyle = \hbar\omega[\bold{a^\dagger}\bold{a},\bold{a}]$

$\displaystyle = \hbar\omega\left(\bold{a^\dagger}[\bold{a},\bold{a}] + [\bold{a^\dagger},\bold{a}]\bold{a}\right)$

We know $\displaystyle [\bold{a},\bold{a}] = 0$, so we're left with $\displaystyle [\bold{a^\dagger},\bold{a}]$. Apparently this is equal to -1 but I'm not sure how, and I have to show why..

So any way, this would be equal to $\displaystyle -\hbar\omega\bold{a}$.

So we know $\displaystyle \bold{H}\bold{a} - \bold{a}\bold{H} = -\hbar\omega\bold{a}$

We have $\displaystyle \bold{a}\bold{H} - \hbar\omega\bold{a})|U_n>$

$\displaystyle = \bold{a}\bold{H}|U_n> - \hbar\omega\bold{a}|U_n>$

$\displaystyle = E_n\bold{a}|U_n> - \hbar\omega\bold{a}|U_n>$

$\displaystyle = (E_n - \hbar\omega)\bold{a}|U_n>$

$\displaystyle = H\bold{a}|U_n>$

And so we've shown $\displaystyle \bold{a}|U_n>$ is also an eigenstate.

So $\displaystyle \bold{H}|U_{n-1}> = E_{n-1}|U_{n-1}$.

So..

$\displaystyle |U_n> \to E_n$

$\displaystyle \bold{a}|U_n> \to E_n - \hbar\omega$

I believe $\displaystyle \bold{a}$ is referred to as a step-down operator.

$\displaystyle \bold{H}\bold{a}|U_{n-1}> = (E_{n-1} - \hbar\omega)\bold{a}|U_{n-1}>$

This is equivalent to: $\displaystyle \bold{H}\bold{a}\bold{a}|U_n> = (E_n - \hbar\omega - \hbar\omega)\bold{a}\bold{a}|U_n>$

$\displaystyle \bold{H}\bold{a^2}|U_n> = (E_n - 2\hbar\omega)\bold{a^2}|U_n>$

And so $\displaystyle |U_{n-2} = \bold{a^2}|U_n> \to E_n - 2\hbar\omega$.

We can generalize:

$\displaystyle \bold{H}\bold{a^n}|U_n> = (E_n - n\hbar\omega)\bold{a^n}|U_n>$

If $\displaystyle \bold{H}|U_n> = E_n|U_n>$, we consider

$\displaystyle \bold{a^\dagger}|U_n>$

$\displaystyle \bold{H}\bold{a^\dagger}|U_n> = (E_n + \hbar\omega)\bold{a^\dagger}|U_n>$

I think then I have to prove for $\displaystyle n+1, n+2, \ldots$..

$\displaystyle U_{n+1}> = \bold{a^\dagger}|U_n> \to E_n + \hbar\omega$

.. well this is where I'm at. Still many things unanswered.

2. Originally Posted by DiscreteW
1.) These are all for a simple harmonic oscillator...

a.) Prove the expression of $\displaystyle \bold{a^\dagger}$ from $\displaystyle \bold{a}$ in the equation below (1.1) using $\displaystyle <\phi|\bold{A^\dagger}|\psi> = <\psi|\bold{A}|\phi>^*$

(1.1) $\displaystyle \bold{a} = \sqrt{m\omega/2\hbar}\bold{x} + i\bold{p}/\sqrt{2m\hbar\omega}$

(1.1 continued) $\displaystyle \bold{a^\dagger} = \sqrt{m\omega/2\hbar}\bold{x} - i\bold{p}/\sqrt{2m\hbar\omega}$
$\displaystyle < \phi | \bold{a} | \psi >$

$\displaystyle = < \phi | \sqrt{m\omega/2\hbar}\bold{x} + i\bold{p}/\sqrt{2m\hbar\omega} | \psi >$

$\displaystyle = < \phi | \sqrt{m\omega/2\hbar}\bold{x} | \psi > + < \phi | i\bold{p}/\sqrt{2m\hbar\omega} | \psi >$

$\displaystyle = \sqrt{\frac{m\omega}{2\hbar}} < \phi | \bold{x} | \psi > + \frac{i}{\sqrt{2m\hbar\omega}} < \phi | \bold{p} | \psi >$

So
$\displaystyle < \phi | \bold{a} | \psi > ^* = \sqrt{\frac{m\omega}{2\hbar}} < \phi | \bold{x} | \psi >^* - \frac{i}{\sqrt{2m\hbar\omega}} < \phi | \bold{p} | \psi >^*$

etc.

-Dan

3. Originally Posted by DiscreteW
b.) Write $\displaystyle \bold{H}$ in terms of $\displaystyle \bold{x}$ and $\displaystyle \bold{p}$ and prove the expression of $\displaystyle H$ in terms of $\displaystyle \bold{a}$ and $\displaystyle \bold{a^\dagger}$.
This one should be in your text somewhere.
$\displaystyle \bold{H} = \bold{T} + \bold{U}$

$\displaystyle \bold{H} = \frac{1}{2m} \bold{p^2} + \frac{1}{2}m \omega ^2 \bold{x^2}$

Probably the best way to continue is to redefine this in terms of unitless operators:
$\displaystyle \bold{X} = \sqrt{\frac{m \omega}{\hbar}}\bold{x}$
and
$\displaystyle \bold{P} = \frac{1}{\sqrt{m \hbar \omega}}\bold{p}$

So that
$\displaystyle \bold{H} = \frac{\hbar \omega}{2}(\bold{X^2} + \bold{P^2})$

We may easily derive that
$\displaystyle [ \bold{X}, \bold{P} ] = i$

Now set
$\displaystyle \bold{a} = \frac{1}{\sqrt{2}}(\bold{X} + i \bold{P})$
and
$\displaystyle \bold{a}^{\dagger} = \frac{1}{\sqrt{2}}(\bold{X} - i \bold{P})$

Then we have that
$\displaystyle \bold{X} = \frac{1}{\sqrt{2}}(\bold{a}^{\dagger} + \bold{a})$

$\displaystyle \bold{P} = \frac{1}{\sqrt{2}}(\bold{a}^{\dagger} - \bold{a})$

And using the commutation relation we can verify that
$\displaystyle [ \bold{a}, \bold{a}^{\dagger} ] = 1$
so that

$\displaystyle \bold{H} = \hbar \omega \left ( \bold{a} \bold{a}^{\dagger} - \frac{1}{2} \right )$

-Dan

4. Originally Posted by DiscreteW
d.) If $\displaystyle |U_n>$ is the eigenstate of $\displaystyle \bold{H}$ w/ eigenvalue $\displaystyle E_n$, ie., $\displaystyle \bold{H}|U_n> = E_n|U_n>$, prove $\displaystyle |U_m> = \bold{a^\dagger} | U_n>$ is also an eigenstate of $\displaystyle H$ w/ a different eigenvalue. Determine the eigenvalue that corresponds to $\displaystyle |U_m>$.
We know that
$\displaystyle [ \bold{N}, \bold{a}^{\dagger} ] = \bold{a}^{\dagger}$
where $\displaystyle \bold{N} = \bold{a}^{\dagger} \bold{a}$

Since $\displaystyle [ \bold{H}, \bold{N} ] = 0$, any eigenstate of H is an eigenstate of N. So basically we are to prove that $\displaystyle \bold{a}^{\dagger}|U_n >$ is an eigenstate of N.

So:
$\displaystyle [ \bold{N}, \bold{a}^{\dagger} ] |U_n> = \bold{a}^{\dagger} |U_n>$

So
$\displaystyle \bold{N}\bold{a}^{\dagger}|U_n> = \bold{a}^{\dagger}\bold{N}|U_n> + \bold{a}^{\dagger}|U_n>$

$\displaystyle = n\bold{a}^{\dagger}|U_n> + \bold{a}^{\dagger}|U_n>$
(where n is the eigenvalue of $\displaystyle |U_n>$ with respect to the operator N.)

$\displaystyle = (n + 1) \bold{a}^{\dagger}|U_n>$

In summation:
$\displaystyle \bold{N} \left ( \bold{a}^{\dagger}|U_n> \right ) = (n + 1) \left ( \bold{a}^{\dagger}|U_n> \right )$

Thus $\displaystyle \bold{a}^{\dagger}|U_n>$ is an eigenvalue of N with eigenvalue n + 1. I leave it to you to translate this eigenvalue in terms of the corresponding energy eigenvalue.

-Dan