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

 

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