Next: Bibliography
Up: Algebraic solution
Previous: Algebraic solution
A more canonical approach to the previous section is as follows. We write
the Hamiltonian as
![$\displaystyle H = \frac{1}{2m}P^2 + \frac{m\omega^2}{2}Q^2$](img145.png) |
(4.6) |
This can be written dimensionlessly with the introduction of substitute operators,
In parallel with the previous section, we consider the operators,
From basic quantum mechanics we know that
, thus
,
and so
, whereby
![$\displaystyle H = \frac{1}{2}\hbar \omega(aa^{\dagger} + a^{\dagger}a)=\hbar\omega(aa^{\dagger} - \frac{1}{2}) = \hbar \omega (a^{\dagger}a + \frac{1}{2})$](img155.png) |
(4.13) |
The Hamiltonian thus shares its eigenvalue spectrum with
, and
![$\displaystyle [N,a] = a^{\dagger}aa - aa^{\dagger}a=[a^{\dagger},a]a=-a$](img157.png) |
(4.14) |
![$\displaystyle [N,a^{\dagger}]= a^{\dagger}aa^{\dagger} - a^{\dagger}a^{\dagger}a=a^{\dagger}[a,a^{\dagger}]=a^{\dagger}$](img158.png) |
(4.15) |
Thus we find,
![$\displaystyle N\vert n\rangle=n\vert n\rangle \ \Longrightarrow Na\vert n\rangle =a(N-1)\vert n\rangle = (n-1)a\vert n\rangle$](img159.png) |
|
|
(4.16) |
![$\displaystyle Na^b\vert n\rangle=(n-b)a^b\vert n\rangle$](img160.png) |
|
|
(4.17) |
Thus
is an eigenvector of N with eigenvalues
.
We require these eigenvalues be positive for the following reason. An operator
is required to have a real expectation value since the expectation value
is what we would measure, i.e. we require that for an operator R,
This implies R is Hermitian, and given
then we have,
Since the eigenvalues are real, then the squared norm of these eigenvectors
follow the form:
![$\displaystyle (\langle n\vert a^{\dagger})(a\vert n\rangle)=\langle n \vert N \vert n \rangle = n\langle n \vert n \rangle \geq 0$](img166.png) |
(4.18) |
and so we conclude that there must be a
cutoff
As well,
![$\displaystyle Na^{\dagger}\vert n\rangle = a^{\dagger}(N+1)\vert n\rangle = (n+1)a^{\dagger}\vert n\rangle$](img168.png) |
|
|
(4.19) |
![$\displaystyle (\langle n \vert a)(a^{\dagger}\vert n\rangle)=(n+1)\langle n \vert n \rangle$](img169.png) |
|
|
(4.20) |
We compare equation 22 with the fact that
and
conclude that
![% latex2html id marker 1694
$\displaystyle a^{\dagger}\vert n\rangle = \beth \v...
...le)=(n+1) \ \therefore \ a^{\dagger}\vert n\rangle = \sqrt{n+1}\vert n+1\rangle$](img171.png) |
(4.21) |
A similar argument yields
.
It follows that
![$\displaystyle \vert n\rangle=\frac{a^{\dagger}^n\vert\rangle}{\sqrt{n!}}$](img173.png) |
(4.22) |
These are the so-called Fock states.
Next: Bibliography
Up: Algebraic solution
Previous: Algebraic solution
Timothy D. Jones
2007-01-29