# Are there two different ways to express the position operator $x$ in terms of the creation and annihilation operator?

As we known, to express the position operator $x$ in terms of the creation and annihilation operator $a^{+}$ and $a$, one way is:
$$x= \sqrt{\frac{\hbar}{2\mu\omega}}(a^++a);$$

$$p= i\sqrt{\frac{\mu\hbar\omega}{2}}(a^+-a).$$

$$x= -i\sqrt{\frac{\hbar}{2\mu\omega}}(a^+-a);$$

$$p= \sqrt{\frac{\mu\hbar\omega}{2}}(a^++a)?$$

I want to know whether the two expression 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.

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Fundamental is the canonical commutation relation (CCR) between x and p, [x,p]=i. Any pair satisfying it is acceptable. If there is a finite set of coordinates and momenta different choices are unitarily equivalent. This needs no longer be true in the infinite case (quantised fields). – Urgje Apr 30 '14 at 9:22
@Urgje Thanks for comments! So we may get different expectation value for the same operator. For example, the expectation values of the position operator in the two cases are different. How can we explain this? – jiadong Apr 30 '14 at 9:32
If you perform a unitary transformation U, the operators transform as A→UAU^(-1) and the states as f→Uf. Thus expectation values are invariant. – Urgje Apr 30 '14 at 9:46
@Urgje Thanks a lot! I see now. – jiadong Apr 30 '14 at 9:49
@Jiadong Both set of x and p satisfy the commutation relation and give the same expression for Harmonic oscillator hamiltonian. Expectation value of position in both set is same, zero. – Rahul kumar walia Apr 30 '14 at 10:01

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.