Kashmir said:
Can anyone help me understand how for infinitesimal translation we've ##\hat{T}(d x)=\mathbb{1}-\frac{i}{\hbar} \hat{p}_{x} d x##?
1) Probability conservation: If the state |\psi \rangle is normalized to unity, the translated state |\psi^{\prime}\rangle = T(a)|\psi\rangle must also be normalized to unity. So, \langle \psi |\psi \rangle = \langle \psi |T^{\dagger}(a)T(a)|\psi \rangle = 1 . Thus, T(a) is a
unitary operator, T^{\dagger}(a) = T^{-1}(a) \ \ \ \ \ \ \ \ \ \ (1)
2) The set \{ T(a): \ a \in \mathbb{R} \} forms an infinite-dimensional unitary representation of the Abelian (Lie) group of translation. This means that the T(a) satisfies the following group axioms: T(a)T(b) = T(a + b) , \ \ \ \ \ \ \ \ \ (2)T^{-1}(a) = T(-a) , \ \ \ \ \ \ \ \ \ \ \ (3)T(0) = 1 , \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4)
3) Since T(0)|\psi \rangle = |\psi \rangle, then for an infinitesimal translation x \to x + \epsilon , \ \ |\epsilon | \ll 1 the operator T(\epsilon) must (by
continuity) equal to 1 plus term
linear in the infinitesimal parameter \epsilon. So, we may write T(\epsilon) = 1 - i \epsilon \ G . \ \ \ \ \ \ \ (5) Now it is a simple exercise to show that T(\epsilon) will satisfies the equations (1)-(4) if we take the operator G to be Hermitian: G^{\dagger} = G.
4) Notice that T(\epsilon)|x \rangle = |x + \epsilon \rangle , implies that \langle x |T(\epsilon) = \langle x - \epsilon|. Therefore \psi^{\prime}(x) = \left(T_{\epsilon}\psi\right)(x) \equiv \langle x |T(\epsilon)|\psi \rangle = \psi (x - \epsilon) . Now Taylor expansion of \psi (x - \epsilon) gives us \langle x|T(\epsilon)|\psi \rangle = \psi (x) - i \epsilon \ (- i \partial \psi)(x) + \mathcal{O}(\epsilon^{2}). Substituting Eq(5) in the LHS, we find (to first order in \epsilon) \psi (x) - i \epsilon \ (G\psi)(x) = \psi (x) - i \epsilon \ (- i \partial \psi )(x) .Thus, in the coordinate representation, the Hermitian operator G is given by G \psi (x) = - i \partial_{x}\psi (x) .
5) The relation between the
generator G and the
momentum operator P: We would like to obtain the commutation relation \big[Q , P \big] = i\hbar. So, consider Q|x \rangle = x \ |x \rangle and operate with T(\epsilon) from the left: T(\epsilon)Q|x \rangle = x T(\epsilon)|x \rangle = x |x + \epsilon \rangle . Next, consider T(\epsilon)|x \rangle = |x + \epsilon \rangle and operate with position operator Q: QT(\epsilon)|x \rangle = Q| x + \epsilon \rangle = (x + \epsilon) \ |x + \epsilon \rangle . Subtracting the results, we get \big[ Q , T(\epsilon) \big] \ | x \rangle = \epsilon \ |x + \epsilon \rangle . Then, from Eq(5), we obtain - i \epsilon \ \big[ Q , G \big] | x \rangle = \epsilon \ |x + \epsilon \rangle \approx \epsilon \ | x \rangle . Now, | x \rangle can be
any position eigen-ket, and we know that the position eigen-kets form a
complete set. Thus, we can write the
operator identity [Q , G] = i , and make the identification G = \frac{1}{\hbar} \ P . So, the infinitesimal transformation operator becomes T(\epsilon) = 1 - \frac{i}{\hbar} \epsilon \ P . From this infinitesimal form, a
finite translation T(a) can be obtained as T(a) = \mbox{lim}_{n \to \infty} \left( 1 - \frac{i}{\hbar}\left(\frac{a}{n}\right) P \right)^{n} = e^{- \frac{i}{\hbar}aP} .
