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

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


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

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

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


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

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 $$ [T_a,x] = -a \,T_a \quad .$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.