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I have been reading Sakurai's Modern Quantum Mechanics and I am confused by this:

A translation operator $\mathscr T(\delta x)$ is defined as $\mathscr T(\delta x) |x \rangle = | x + \delta x \rangle$ and it is used to derive the momentum operator $\hat p$ by the expression $\mathscr T(\delta x) = \exp(- \mathrm i \hat p \delta x/\hbar)$. And similarly to derive Hamiltionian $\hat H$ the time evolution operator $\mathscr U(t, t_0)$ is introduced.

Question:

Notice that

  1. $\mathscr T(\delta x) | 0 \rangle = | \delta x\rangle \neq | 0 \rangle$;
  2. $\mathscr T(\delta x) |\lambda x \rangle = |\lambda x + \delta x \rangle \neq \lambda\mathscr T(\delta x) |x \rangle$ ($\lambda \in \mathbb C$); ...

We know that the translations are not linear operators, but we still call them operators, and treat them as operators. We even find them Hermitian adjoints ($\mathscr T(\delta x)^\dagger \mathscr T(\delta x) \simeq \hat 1$, or "the infinitesimal translations are unitary"), which doesn't make any sense to me.

The problem also occurs in discussing the time evolution operators and Sakurai wrote: $\mathscr U(t, t_0)^\dagger \mathscr U(t, t_0) = \hat 1$, where the time evolution operators are not infinitesimal.

So when we call $\mathscr T(\delta x)$ and $\mathscr U(t, t_0)$ "operators" what do we want to say? If they are not operators in $\mathrm{End} (\mathbb H)$ (endomorphisms, i.e. linear operators), where do they belong to? Are they affine mappings in the affine group (I know little about this group)? And, how do we treat the Hermitian adjoints of them?

I would like a mathematically rigorous answer instead of an intuitive explaination, thank you!

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  • $\begingroup$ They are infinite-dimensional matrices, hence linear. $\endgroup$
    – Cosmas Zachos
    Commented Oct 3, 2020 at 2:54
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    $\begingroup$ You're confusing yourself with notation. $\lvert \lambda x \rangle$ means "The vector whose eigenvalue under the $\hat X$ operator is $\lambda x$. That is not the same thing as the vector $\lambda \lvert x \rangle$, which is "lambda times the vector whose eigenvalue under the $\hat{X}$ operator is $x$". Note that $\hat{X}(\lambda \lvert x \rangle) = x(\lambda \lvert x \rangle) \neq \lambda x (\lambda \lvert x \rangle)$. $\endgroup$
    – DanielSank
    Commented Oct 3, 2020 at 7:56

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(a) Not all operators are linear operators.

(b) Both spatial translations and time translations obey the property \begin{equation} \mathcal{O}(a |\psi_1\rangle + b | \psi_2 \rangle ) = a \mathcal{O} |\psi_1\rangle + b \mathcal{O}| \psi_2 \rangle \end{equation} and therefore are linear operators on Hilbert space.

Note that $|0\rangle \neq 0$ (ie the "0 ket" does not equal the "0 vector on Hilbert space"), this seems to be a confusion in your question.

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  • $\begingroup$ Thanks for your correction. However, could you please give an explicit demonstration (maybe take finite-dimensional vector space or 1-D world spaned by $| x \rangle$ as example?) for the linearity of translations? It is not clear to me yet. $\endgroup$
    – Hoyan Mok
    Commented Oct 3, 2020 at 4:06
  • $\begingroup$ I'm afraid you might be disappointed since I think this is just a definition... For example the translation operator $\mathcal{T}_a$ acts on position eigenstates as $\mathcal{T}_a | x \rangle = |x+a \rangle$. This property by itself doesn't tell you how the translation operator should act on a superposition like $|x_1\rangle + |x_2 \rangle$. But we can define the translation operator to be the linear operator, and then it is uniquely defined (up to an overall phase). $\endgroup$
    – Andrew
    Commented Oct 3, 2020 at 21:01
  • $\begingroup$ There is a nice relationship between the translation operator and the momentum operator, namely $\mathcal{T}_a = \exp\left(-\frac{i}{\hbar} a \hat{p}\right)$. In that form, you can see the translation operator is unitary, and you can also see that it is linear (since you can express the exponential as a sum of linear operators built out of $\hat{p}$). But while I do think this form gives a deeper understanding of the translation operator, I wouldn't say that it is a proof of the linearity of the translation operator. $\endgroup$
    – Andrew
    Commented Oct 3, 2020 at 21:04

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