# Translation operator and position operator

I am a little bit confused about the translation and position operator and hope for some clarification.

Let $$\hat{x}$$ be the position operator, which satisfies $$\hat{x} \vert x \rangle = x \vert x \rangle$$ and $$\hat{T}_{a}$$ the translation operator, which satisfies $$\hat{T}_{a} \vert x \rangle = \vert x+a \rangle$$. With that being said, it is known, that $$[\hat{x},\hat{T}_a] \neq 0$$, since $$\hat{T}_a \hat{x} \vert x \rangle = x \vert x+a \rangle$$, whereas $$\hat{x} \hat{T}_a \vert x \rangle = (x+a) \vert x+a \rangle$$.

With that being said I am confused about the same operators in the position representation. Let $$\psi(x)$$ be any wavefunction, that solves the Schrödinger equation.

To reproduce the same results in the position representation

$$\hat{T}_a \hat{x} \psi(x) = \hat{T}_a x \psi(x) = x \hat{T}_a \psi(x) = x\psi(x+a)$$

and

$$\hat{x} \hat{T}_a \psi(x) = \hat{x} \psi(x+a) = (x+a) \psi(x+a)$$

must be true.

However I previously thought, that the postion operator is always "just x" in the position representation. In this case, we would get

$$\hat{T}_a \hat{x} \psi(x) = \hat{T}_a x \psi(x) = (x+a) \psi(x+a)$$

and

$$\hat{x} \hat{T}_a \psi(x) = x \psi(x+a) = x \psi(x+a)$$

which does not fit to the generalized statements mentioned at the beginning.

To conclude everything: which - if any - result is the correct one and how exactly is the position operator defined for the position representation?

First, a quick clarification. If you define the action of $$T_a$$ on the position ket $$|x\rangle$$ as $$T_a|x\rangle = |x+a\rangle$$, then one would have that $$T_a|\psi\rangle = \int \mathrm dx \ \psi(x) T_a |x\rangle = \int \mathrm dx \ \psi(x) |x+a\rangle = \int \mathrm dx \ \psi(x-a) |x\rangle$$ Similarly, $$T_a X|\psi\rangle = \int \mathrm dx \ \psi(x) T_a X |x\rangle = \int \mathrm dx\ \psi(x) x T_a |x\rangle = \int \mathrm dx \ \psi(x) x |x+a\rangle$$ $$=\int \mathrm dx \ (x-a) \psi(x-a)|x\rangle$$ whereas $$X T_a\psi\rangle = \int \mathrm dx \ \psi(x-a) X|x\rangle = \int \mathrm dx \ x \psi(x-a) |x\rangle$$

Avoiding bra-ket notation and defining the action of these operators directly at the level of the wavefunction would yield $$\big[(T_a X)\psi\big](x) = (x-a) \psi(x-a) \qquad \big[(XT_a)\psi\big](x) = x \psi(x-a)$$

The first equality could be understood by letting $$X\psi \equiv \phi$$, so $$\phi(x) = x \psi(x)$$. From there, application of $$T_a$$ would yield $$\big(T_a \phi\big)(x)=\phi(x-a) = (x-a)\psi(x-a)$$.

Altogether, I think the important thing is to remember that $$\psi$$ is a function while $$\psi(x)$$ is a number, and operators act on functions (in the position representation). Case in point,

$$\hat{T}_a \hat{x} \psi(x) = \hat{T}_a x \psi(x) = x \hat{T}_a \psi(x) = x\psi(x+a)$$

is incorrect. It should be understood as acting on $$\psi$$ with $$\hat x$$, then acting on the result with $$T_a$$, and only then evaluating the result at $$x$$. In other words, as $$\big(\hat T_a \hat x \psi\big)(x)$$. This makes it much easier to understand that $$T_a$$ shifts the argument of $$\hat x \psi$$: $$\big(\hat T_a \hat x \psi\big)(x) = \big(\hat x \psi\big)(x-a) = (x-a)\psi(x-a)$$ because $$\big(\hat x\psi\big)(u) = u \psi(u)$$.

In the position representation, we need to (or can) evaluate the action of $$T_a$$ on the bra $$\langle x|$$. To this end, note that $$\langle x|T_a = \int \mathrm dy \, \langle x|T_a|y\rangle \langle y| = \int \mathrm dy \, \langle x|y+a\rangle \langle y| = \langle x-a| \quad ,$$ where we've used the completeness relation $$\displaystyle \int \mathrm d x\, |x\rangle \langle x| = \mathbb I$$ and $$\langle x|y\rangle = \delta(x-y)$$.

Now let us define $$\psi (x) \equiv \langle x|\psi\rangle$$

and $$(O \psi)\,(x) \equiv \langle x|O|\psi\rangle$$ for some operator $$O$$ and state $$|\psi\rangle$$. Then $$(T_ax \psi)\, (x) = \langle x|T_a x|\psi\rangle = \langle x-a |x|\psi\rangle = (x-a) \langle x-a|\psi\rangle =(x-a)\, \psi(x-a) \quad ,$$ while $$(xT_a\psi)\,(x) = \langle x| x T_a|\psi\rangle =x \langle x|T_a|\psi\rangle = x \langle x-a|\psi\rangle = x\, \psi(x-a)\quad .$$

Eventually, this shows that $$\langle x| [T_a,x] |\psi\rangle =(x-a)\psi(x-a) - x \psi(x-a) = -a \psi(x-a) = -a \langle x-a|\psi\rangle = -a \langle x|T_a|\psi\rangle \quad .$$ Since this should hold for all $$|\psi\rangle$$ (in a specified domain) and $$\langle x|$$, we conclude $$[T_a,x] = -a \,T_a \quad .$$