I quote Sakurai$^1$
\begin{equation*} \mathscr{T}(dx^\prime)~| x^\prime \rangle = ~ | x^\prime+dx^\prime \rangle \tag{1.194} \end{equation*} where a possible arbitrary phase factor is set to unity by convention.
In (1.194), the operator $ \mathscr{T}(dx^\prime) $ is an infinitesimal translation operator ( a spatial displacement operator ), and the $| x^\prime \rangle$ are position eigenkets, which are expressible in terms of the Dirac delta function.
Sakurai$^1$ says that the operator $ \mathscr{T}(dx^\prime)$, may be applied to arbitrary kets.
My question is: Where does the arbitrary phase factor of the infinitesimal translation operator, that appears in (1.194), on pg.40 of Sakurai, come from?
Other Information
Displacing a ket in various ways, is a very big idea for Sakurai’s book. It is crucial for getting anywhere using the ‘Active View’ of transformations, that is used in the book. The other view being the ‘Passive View’, from which transformations are looked upon as changes to the coordinate system.
See Sakurai$^1$, sections 1.6.3, 2.1.1 and 3.1.2, for material on the various types of displacement considered by him.
Reference:
1, J.J.Sakurai and Jim Napolitano, Modern Quantum Mechanics Third Edition, Cambridge University Press & Assessment (2021).
We should, eventually, be able to see where the arbitrary phase factor comes from, not by looking directly, at how a ket is displaced, but by looking at how the physical system represented by the ket, is displaced.
Let’s now explain the above, in detail.
I think it’s always possible to consider a normalizable ket from some Hilbert space as a function of some set of variables. This is not taking into account any ket normalized in terms of the Dirac delta function.
It is possible to use a function based notation for kets.
Let our Hilbert Space of kets ( function space of functions ) be occupied by normalizable functions of a variable ‘$x$’. A value of the variable ‘$x$’ is taken to mean a value of a particle's position coordinate. If we choose the function, $\psi(x)$, from the Hilbert space, then we may say that we have chosen the ket $| \psi(x)\rangle$.
I don’t think the operator $ \mathscr{T}(dx^\prime)$ in (1.194) of Sakurai$^1$ can be derived by just considering the “displacement”of a definite initial ket. This is because the displacement of a function, by $dx^\prime$, is a definite thing, see the following.
An operator for the displacement of any of our kets, say the function $| \psi(x)\rangle$, can be derived as follows
If you displace the function $|\psi(x)\rangle$ by $dx^\prime$, it becomes a new function $|\phi(x)\rangle$, that must satisfy the condition \begin{equation*} |\phi(x+ dx^\prime )\rangle~=~| \psi(x)\rangle \end{equation*} Changing the arguments of the functions $\phi$ and $\psi$, in the same way, gives another true equation ( assuming both functions are defined at the new values of their arguments ). Hence the rule for the new function ket is \begin{equation*} |\phi(x)\rangle= | \psi(x-dx^\prime)\rangle \end{equation*}
We would now be able to define an infinitesimal displacement operator in our Hilbert space by \begin{equation*} T(dx^\prime) ~ |\psi(x)\rangle~= ~|\psi(x-dx^\prime)\rangle \end{equation*}
So using the idea, that we are merely displacing a function, does not lead to any arbitrary phase factor.
Dirac$^2$ discusses ‘Displacement Operators’ on pages 99-103, he says that displacing a state is a definite thing, but that the displacement of a ket is not such a definite thing though. He also talks about various arbitrary phase factors.
So, let’s analyse the displacement of our one dimensional physical system.
Take our physical system to be in the state represented by the ket $| \psi(x)\rangle $, this ket is normalised such that $\|\psi(x)\| ~= ~1$.
Displace the physical system it represents, by $dx^\prime$.
In the original physical system, the probability of finding the particle, at ‘$x$’ within the interval $[x,x+dx]$, is given by
\begin{equation*} |~|\psi(x)\rangle~|^2~dx \end{equation*} or dropping the ‘$| \rangle$’, by \begin{equation*} |~\psi(x)~|^2~dx \end{equation*}
The physical system, has each of it’s points, ‘$x$’, displaced to ‘$x+dx^\prime$’. The probabilities, at each ‘$x$’, must be reproduced at the corresponding points ‘$x+dx^\prime$’ from the new system ket, $|\phi\rangle$. In symbols \begin{equation*} |~|\phi(x+dx^\prime)\rangle~|^2~dx =|~ |\psi(x)\rangle~|^2~dx \end{equation*} Dividing through by $dx$ and again dropping the ‘$| \rangle$’, we have \begin{equation*} |~\phi(x+dx^\prime)~|^2 = |~\psi(x)~|^2 \tag{1} \end{equation*}
Compare ($\mathbf{1}$) to \begin{equation*} |~ z_1~|^2 = |~ z_2~|^2 \end{equation*} Which can be solved, for $z_1$ in terms of $z_2$ as $z_1 = e^{i\alpha}~z_2$, where $\alpha$ is an arbitrary real number. Solving ($\mathbf{1}$) in like manner, for $\phi$, gives \begin{equation*} \phi(x+dx^\prime)= e^{i\alpha} ~\psi(x) \end{equation*}
Changing the arguments of the functions $\phi$ and $\psi$, in the same way, gives another true equation ( assuming both functions are defined at the new values of their arguments ). Hence the rule for the new function ket is \begin{equation*} \phi(x)= e^{i\alpha} ~\psi(x-dx^\prime) \tag{2} \end{equation*}
We can now specify an operator in our Hilbert space, to within an arbitrary phase factor $e^{i\alpha}$, by \begin{equation*} T(dx^\prime)~\psi(x)= e^{i\alpha} ~\psi(x-dx^\prime) \end{equation*} Or, in ket notation \begin{equation*} T(dx^\prime)~|\psi(x)\rangle= e^{i\alpha} ~|\psi(x-dx^\prime) \rangle \tag{3} \end{equation*}
Now, we see that the possible arbitrary phase factor for $\mathscr{T}$ has been brought to our attention. If we put the arbitrary phase factor in ($\mathbf{3}$) equal to one, we obtain ($\mathbf{4}$), an equation that is consistent with (1.194), this is shown in the Other Information.
\begin{equation*} T(dx^\prime)~|\psi(x)\rangle= |\psi(x-dx^\prime) \rangle \tag{4} \end{equation*}
It has been shown how an arbitrary phase factor appears when we displace a ket, see how ($\mathbf{3}$) was derived.
In the derivation of ($\mathbf{3}$) we did not look directly, at how a ket is displaced, but looked at how the physical system represented by the ket, was displaced.
Other Information
Defining Ket displacement
We end up with (4), defining how to infinitesimally displace (spatially translate) a ket, this is after choosing the arbitrary phase factor to be equal to one. Our operator of choice for ket displacement, is exactly what the operator would have been, if we just thought of ket displacement as the displacement of a function!
To show that ($\mathbf{4}$), is consistent with (1.194).
Note, \begin{equation*} |x^\prime\rangle= | \delta(x-x^\prime) \rangle \end{equation*} and \begin{equation*} ~~~|x^\prime+dx^\prime\rangle= | \delta(x-(x^\prime+dx^\prime) )\rangle \end{equation*}
The $|x^\prime\rangle$ are not really function kets from a Hilbert space, but allowing our operator $T(dx^\prime)$ to operate on these kets |$x^\prime\rangle$ and putting \begin{equation*} |\psi(x) \rangle= |x^\prime\rangle= | \delta(x-x^\prime) \rangle \end{equation*}
in ($\mathbf{4}$), we have \begin{equation*} T(dx^\prime)~| \delta(x-x^\prime) \rangle= ~ | \delta(x-(x^\prime+dx^\prime) )\rangle \end{equation*}
Which shows directly a Dirac delta function way of expressing things, but, we may also write, based on the above, \begin{equation*} T(dx^\prime)~| x^\prime \rangle= ~ | x^\prime+dx^\prime \rangle \tag{5} \end{equation*}
Compare ($\mathbf{5}$) with Sakurai$^1$ (1.194) \begin{equation*} \mathscr{T}(dx^\prime)~| x^\prime \rangle= ~ | x^\prime+dx^\prime \rangle \tag{1.194} \end{equation*}
Our operator $T(dx^\prime)$ has the same effect on the position eigenkets as $\mathscr{T}(dx^\prime)$, as the position eigenkets will form a basis of our space, this indicates, to me, that the two operators may well be equal.
References
1, J.J.Sakurai and Jim Napolitano, Modern Quantum Mechanics Third Edition, Cambridge University Press & Assessment (2021).
2, P.A.M.Dirac, The Principles Of Quantum Mechanics, Fourth Ed.(Revised), Clarendon Press, Oxford (1958).