Position and Momentum operators derived from annihilation and creation operators | Quantum Science Philippines

## Position and Momentum operators derived from annihilation and creation operators

By: Levine T. Poralan, MS Physics, MSU-IIT

Given the annihilation and creation operators below,

$a = \frac{1}{\sqrt{2}} (\hat{X} + i\hat{P})$

$a^{\dagger} = \frac{1}{\sqrt{2}} (\hat{X} - i\hat{P})$

Show that

(a) $\hat{X} = \frac{1}{\sqrt{2}} (a^{\dagger} + a)$

(b) $\hat{P} = \frac{i}{\sqrt{2}} (a^{\dagger} - a)$

(c) $\hat{X}$ and $\hat{P}$ are Hermitian but not $a$ and $a^{\dagger}.$

Solution:

(a) Adding the annihilation and creation operators

$\begin{array}{lll} a&=&\frac{1}{\sqrt{2}} (\hat{X} + i\hat{P})\\\\a^{\dagger}&=&\frac{1}{\sqrt{2}} (\hat{X} - i\hat{P}) \end{array}$

this results to

$a + a^{\dagger} = \frac{2}{\sqrt{2}} \hat{X}$

and this becomes

$\hat{X} = \frac{1}{\sqrt{2}} (a^{\dagger} + a)$.

(b) Now, finding the difference of the two operators

$\begin{array}{lll} a&=&\frac{1}{\sqrt{2}} (\hat{X} + i\hat{P})\\\\a^{\dagger}&=&\frac{1}{\sqrt{2}} (\hat{X} - i\hat{P}) \end{array}$

we can obtain

$a - a^{\dagger} = \frac{2}{\sqrt{2}} i\hat{P}$

multiplying the whole equation by (-i)

$( a - a^{\dagger} = \frac{2}{\sqrt{2}} i\hat{P}) -i$

$-i ( a - a^{\dagger} ) = \sqrt{2} \hat{P}$

$i ( a^{\dagger} - a ) = \sqrt{2} \hat{P}$

and this can be expressed as

$\hat{P} = \frac{i}{\sqrt{2}} (a^{\dagger} - a)$.

(c) Checking whether the operators are Hermitian,

$\begin{array}{lll} \hat{X}&=&\frac{1}{\sqrt{2}} (a^{\dagger} + a)\\\\(\hat{X})^{\dagger}&=&[\frac{1}{\sqrt{2}} (a^{\dagger} + a)]^{\dagger}\\\\&=&\frac{1}{\sqrt{2}} (a^{\dagger} + a)\\\\(\hat{X})^{\dagger}&=&\frac{1}{\sqrt{2}} (a^{\dagger} + a)\\\\(\hat{X})^{\dagger}&=&\hat{X}\end{array}$

Therefore,

$\hat{X}$ is Hermitian.

Now,

$\begin{array}{lll} \hat{P}&=&\frac{i}{\sqrt{2}} (a^{\dagger} - a)\\\\(\hat{P})^{\dagger}&=&[\frac{i}{\sqrt{2}} (a^{\dagger} - a)]^{\dagger}\\\\&=&-\frac{i}{\sqrt{2}} (a - a^{\dagger})\\\\(\hat{P})^{\dagger}&=&\frac{i}{\sqrt{2}} (a^{\dagger} - a)\\\\(\hat{P})^{\dagger}&=&\hat{P}\end{array}$

Therefore,

$\hat{P}$ is Hermitian.

Lastly,

$\begin{array}{lll} a&=&\frac{1}{\sqrt{2}}(\hat{X} + i\hat{P})\\\\a^{\dagger}&=&\frac{1}{\sqrt{2}}(\hat{X} - i\hat{P})\\\\(a^{\dagger})^{\dagger}&=&[\frac{1}{\sqrt{2}}(\hat{X} - i\hat{P})]^{\dagger}\\\\(a^{\dagger})^{\dagger}&=&\frac{1}{\sqrt{2}}(\hat{X} + i\hat{P})\end{array}$

From this, we can see that

$( a^{\dagger} )^{\dagger} = a$ and $( a^{\dagger} )^{\dagger} \neq a^{\dagger}$

Therefore,

$a$ and $a^{\dagger}$ are not Hermitian.