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9.2 The operators $ \hat{a}$ and $ \hat{a}^{\dagger}_{}$.

In a previous chapter I have discussed a solution by a power series expansion. Here I shall look at a different technique, and define two operators $ \hat{a}$ and $ \hat{a}^{\dagger}_{}$,

$\displaystyle \hat{a}$ = $\displaystyle {\frac{1}{\sqrt 2}}$$\displaystyle \left(\vphantom{ y + \frac{d}{dy} }\right.$y + $\displaystyle {\frac{d}{dy}}$ $\displaystyle \left.\vphantom{ y + \frac{d}{dy} }\right)$,    $\displaystyle \hat{a}^{\dagger}_{}$ = $\displaystyle {\frac{1}{\sqrt 2}}$$\displaystyle \left(\vphantom{ y - \frac{d}{dy} }\right.$y - $\displaystyle {\frac{d}{dy}}$ $\displaystyle \left.\vphantom{ y - \frac{d}{dy} }\right)$. (9.4)

Since

$\displaystyle {\frac{d}{dy}}$(yf (y)) = y$\displaystyle {\frac{d}{dy}}$f (y) + f (y), (9.5)

or in operator notation

$\displaystyle {\frac{d}{dy}}$$\displaystyle \hat{y}$ = $\displaystyle \hat{y}$$\displaystyle {\frac{d}{dy}}$ + $\displaystyle \hat{1}$, (9.6)

(the last term is usually written as just 1) we find
$\displaystyle \hat{a}$$\displaystyle \hat{a}^{\dagger}_{}$ = $ {\frac{1}{2}}$$\displaystyle \left(\vphantom{\hat
y^{2}-\frac{d^{2}}{dy^{2}} +\hat 1}\right.$$\displaystyle \hat{y}^{2}_{}$ - $\displaystyle {\frac{d^{2}}{dy^{2}}}$ + $\displaystyle \hat{1}$ $\displaystyle \left.\vphantom{\hat
y^{2}-\frac{d^{2}}{dy^{2}} +\hat 1}\right)$,  
$\displaystyle \hat{a}^{\dagger}$$\displaystyle \hat{a}$ = $ {\frac{1}{2}}$$\displaystyle \left(\vphantom{\hat
y^{2}-\frac{d^{2}}{dy^{2}} -\hat 1}\right.$$\displaystyle \hat{y}^{2}_{}$ - $\displaystyle {\frac{d^{2}}{dy^{2}}}$ - $\displaystyle \hat{1}$ $\displaystyle \left.\vphantom{\hat
y^{2}-\frac{d^{2}}{dy^{2}} -\hat 1}\right)$. (9.7)

If we define the commutator

[$\displaystyle \hat{f}$,$\displaystyle \hat{g}$] = $\displaystyle \hat{f}$$\displaystyle \hat{g}$ - $\displaystyle \hat{g}$$\displaystyle \hat{f}$, (9.8)

we have

[$\displaystyle \hat{a}$,$\displaystyle \hat{a}^{\dagger}_{}$] = $\displaystyle \hat{1}$. (9.9)

Now we see that we can replace the eigenvalue problem for the scaled Hamiltonian by either of

$\displaystyle \left(\vphantom{{\hat a}^{\dagger} \hat a +{\textstyle{\frac{1}{2}}}}\right.$$\displaystyle \hat{a}^{\dagger}$$\displaystyle \hat{a}$ + $ {\frac{1}{2}}$$\displaystyle \left.\vphantom{{\hat a}^{\dagger} \hat a +{\textstyle{\frac{1}{2}}}}\right)$u(y) = $\displaystyle \epsilon$u(y),  
$\displaystyle \left(\vphantom{\hat a {\hat a}^{\dagger} -{\textstyle{\frac{1}{2}}}}\right.$$\displaystyle \hat{a}$$\displaystyle \hat{a}^{\dagger}$ - $ {\frac{1}{2}}$$\displaystyle \left.\vphantom{\hat a {\hat a}^{\dagger} -{\textstyle{\frac{1}{2}}}}\right)$u(y) = $\displaystyle \epsilon$u(y). (9.10)

By multiplying the first of these equations by $ \hat{a}$ we get

$\displaystyle \left(\vphantom{\hat a {\hat a}^{\dagger} \hat a +{\textstyle{\frac{1}{2}}}\hat a}\right.$$\displaystyle \hat{a}$$\displaystyle \hat{a}^{\dagger}$$\displaystyle \hat{a}$ + $ {\frac{1}{2}}$$\displaystyle \hat{a}$ $\displaystyle \left.\vphantom{\hat a {\hat a}^{\dagger} \hat a +{\textstyle{\frac{1}{2}}}\hat a}\right)$u(y) = $\displaystyle \epsilon$$\displaystyle \hat{a}$u(y). (9.11)

If we just rearrange some brackets, we find

$\displaystyle \left(\vphantom{\hat a {\hat a}^{\dagger} +{\textstyle{\frac{1}{2}}}}\right.$$\displaystyle \hat{a}$$\displaystyle \hat{a}^{\dagger}$ + $ {\frac{1}{2}}$$\displaystyle \left.\vphantom{\hat a {\hat a}^{\dagger} +{\textstyle{\frac{1}{2}}}}\right)$$\displaystyle \hat{a}$u(y) = $\displaystyle \epsilon$$\displaystyle \hat{a}$u(y). (9.12)

If we now use

$\displaystyle \hat{a}$$\displaystyle \hat{a}^{\dagger}$ = $\displaystyle \hat{a}^{\dagger}_{}$$\displaystyle \hat{a}$ - $\displaystyle \hat{1}$, (9.13)

we see that

$\displaystyle \left(\vphantom{{\hat a}^{\dagger} \hat a +{\textstyle{\frac{1}{2}}}}\right.$$\displaystyle \hat{a}^{\dagger}$$\displaystyle \hat{a}$ + $ {\frac{1}{2}}$$\displaystyle \left.\vphantom{{\hat a}^{\dagger} \hat a +{\textstyle{\frac{1}{2}}}}\right)$$\displaystyle \hat{a}$u(y) = ($\displaystyle \epsilon$ - 1)$\displaystyle \hat{a}$u(y). (9.14)

Question: Show that

 
$\displaystyle \left(\vphantom{{\hat a}^{\dagger} \hat a +{\textstyle{\frac{1}{2}}}}\right.$$\displaystyle \hat{a}^{\dagger}$$\displaystyle \hat{a}$ + $ {\frac{1}{2}}$$\displaystyle \left.\vphantom{{\hat a}^{\dagger} \hat a +{\textstyle{\frac{1}{2}}}}\right)$$\displaystyle \hat{a}^{\dagger}_{}$u(y) = ($\displaystyle \epsilon$ + 1)$\displaystyle \hat{a}^{\dagger}$u(y). (9.15)

We thus conclude that (we use the notation un(y) for the eigenfunction corresponding to the eigenvalue $ \epsilon_{n}^{}$)

$\displaystyle \hat{a}$un(y) $\displaystyle \propto$ un - 1(y),  
$\displaystyle \hat{a}^{\dagger}_{}$un(y) $\displaystyle \propto$ un + 1(y). (9.16)

So using $ \hat{a}$ we can go down in eigenvalues, using a$\scriptstyle \dagger$ we can go up. This leads to the name lowering and raising operators (guess which is which?).

We also see from (9.15) that the eigenvalues differ by integers only!


next up [*]
Next: 9.3 Eigenfunctions of through Up: 9. Ladder operators Previous: 9.1 Harmonic oscillators

© 1998 Niels Walet, UMIST
Email Niels.Walet@umist.ac.uk