As we known, to express the position operator $x$ in terms of the creation and annihilation operator $a^{+}$ and $a$, one way is:
$$x= \sqrt{\frac{\hbar}{2\mu\omega}}(a^++a);$$
$$p= i\sqrt{\frac{\mu\hbar\omega}{2}}(a^+-a).$$
But how about
$$x= -i\sqrt{\frac{\hbar}{2\mu\omega}}(a^+-a);$$
$$p= \sqrt{\frac{\mu\hbar\omega}{2}}(a^++a)? $$
I want to know whether the two expressions are fine.
PS: It seems that I can use them to calculate something like fluctuation $\langle\Delta_x\rangle^2$ at coherent state $|\alpha\rangle$.
Both expressions can give the right answer.
By the 1st expression,
$\langle\alpha|x^2|\alpha\rangle=\frac{\hbar}{2\mu\omega}[(\alpha+\alpha^*)^2+1]$,
$\langle\alpha|x|\alpha\rangle^2=\frac{\hbar}{2\mu\omega}[(\alpha+\alpha^*)^2]$
then,
$\langle\Delta_x\rangle^2=\langle\alpha|x^2|\alpha\rangle – \langle\alpha|x|\alpha\rangle^2 =\frac{\hbar}{2\mu\omega}$.
By the 2nd way,
$\langle\alpha|x^2|\alpha\rangle=-\frac{\hbar}{2\mu\omega}[(\alpha-\alpha^*)^2-1]$,
$\langle\alpha|x|\alpha\rangle^2=-\frac{\hbar}{2\mu\omega}[(\alpha-\alpha^*)^2]$
therefore,
$\langle\Delta_x\rangle^2=\langle\alpha|x^2|\alpha\rangle – \langle\alpha|x|\alpha\rangle^2 =\frac{\hbar}{2\mu\omega}$.
The result is the same.
And if we check the dimension, this formula is also OK.
Best Answer
You are just dealing with the unitary transformation $$U : L^2(\mathbb R) \to L^2(\mathbb R)\:.$$ defined by the unique linear continuous extension of $$U|n\rangle := i^n |n\rangle\:,$$ which implies $$ U ^\dagger a U = i a\:,\qquad U ^\dagger a^\dagger U = -i a^\dagger\:,$$ The two pairs of operators $x,p$ are related by means of the same unitary transformations. As is well-known, unitary transformations preserve the structure of quantum mechanics.