Principles of Quantum Mechanics · Problem 7.4.10

The Path Integral Formulation of Quantum Theory — R. Shankar

Physics — Quantum Mechanics · quantization-non-cartesian-coordinates

Problem

*(Important). Recall that we always quantize a system by promoting the Cartesian coordinates x1,…,xNx_1, \ldots, x_N and momenta p1,…,pNp_1, \ldots, p_N to operators obeying the canonical commutation rules. If non-Cartesian coordinates seem more natural in some cases, such as the eigenvalue problem of a Hamiltonian with spherical symmetry, we first set up the differential equation in Cartesian coordinates and then change to spherical coordinates (Section 4.2). In Section 4.2 it was pointed out that if H\mathscr{H} is written in terms of non-Cartesian but canonical coordinates q1…qNq_1 \ldots q_N; p1…pNp_1 \ldots p_N; H(qi→qi,pi→−iℏ ∂/∂qi)\mathscr{H}(q_i \to q_i, p_i \to -i\hbar\, \partial/\partial q_i) does not generate the correct Hamiltonian HH, even though the operator assignment satisfies the canonical commutation rules. In this section we revisit this problem in order to explain some of the subtleties arising in the direct quantization of non-Cartesian coordinates without the use of Cartesian coordinates in intermediate stages. (1) Consider a particle in two dimensions with H=px2+py22m+a(x2+y2)1/2\mathscr{H} = \frac{p_x^2 + p_y^2}{2m} + a(x^2+y^2)^{1/2} which leads to H→−ℏ22m(∂2∂x2+∂2∂y2)+a(x2+y2)1/2H \to \frac{-\hbar^2}{2m}\left(\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}\right) + a(x^2+y^2)^{1/2} in the coordinate basis. Since the problem has rotational symmetry we use polar coordinates ρ=(x2+y2)1/2\rho = (x^2+y^2)^{1/2}, ϕ=tan⁡−1(y/x)\phi = \tan^{-1}(y/x), in terms of which H→coordinate basis−ℏ22m(∂2∂ρ2+1ρ∂∂ρ+1ρ2∂2∂ϕ2)+aρ(7.4.41)H \xrightarrow{\text{coordinate basis}} \frac{-\hbar^2}{2m}\left(\frac{\partial^2}{\partial \rho^2} + \frac{1}{\rho}\frac{\partial}{\partial \rho} + \frac{1}{\rho^2}\frac{\partial^2}{\partial \phi^2}\right) + a\rho \tag{7.4.41} Verify that H=pρ22m+pϕ22mρ2+aρ\mathscr{H} = \frac{p_\rho^2}{2m} + \frac{p_\phi^2}{2m\rho^2} + a\rho (verify this) and directly promote all classical variables ρ,pρ,ϕ\rho, p_\rho, \phi, and pϕp_\phi to quantum operators obeying the canonical commutation rules. If we choose operators Pρ→−iℏ∂∂ρ,Pϕ→−iℏ∂∂ϕP_\rho \to -i\hbar\frac{\partial}{\partial \rho}, \quad P_\phi \to -i\hbar\frac{\partial}{\partial \phi} that obey the commutation rules, we end up with H→coordinate basis−ℏ22m(∂2∂ρ2+1ρ2∂2∂ϕ2)+aρ(7.4.42)H \xrightarrow{\text{coordinate basis}} \frac{-\hbar^2}{2m}\left(\frac{\partial^2}{\partial \rho^2} + \frac{1}{\rho^2}\frac{\partial^2}{\partial \phi^2}\right) + a\rho \tag{7.4.42} which disagrees with Eq. (7.4.41). (2) A way out is suggested by the fact that although the choice Pρ→−iℏ ∂/∂ρP_\rho \to -i\hbar\,\partial/\partial\rho leads to the correct commutation rule, it is not Hermitian! Verify that ⟨ψ1∣Pρ∣ψ2⟩=∫0∞∫02πψ1∗(−iℏ∂ψ2∂ρ)ρ dρ dϕ≠∫0∞∫02π(−iℏ∂ψ1∂ρ)∗ψ2ρ dρ dϕ=⟨Pρψ1∣ψ2⟩\langle \psi_1|P_\rho|\psi_2\rangle = \int_0^\infty\int_0^{2\pi} \psi_1^*\left(-i\hbar\frac{\partial\psi_2}{\partial\rho}\right)\rho\,d\rho\,d\phi \neq \int_0^\infty\int_0^{2\pi}\left(-i\hbar\frac{\partial\psi_1}{\partial\rho}\right)^*\psi_2\rho\,d\rho\,d\phi = \langle \mathbf{P}_\rho\psi_1|\psi_2\rangle (You may assume ρψ1∗ψ2→0\rho\psi_1^*\psi_2 \to 0 as ρ→0\rho\to 0 or ∞\infty. The problem comes from the fact that ρ dρ dϕ\rho\,d\rho\,d\phi and not dρ dϕd\rho\,d\phi is the measure for integration.) Show, however, that Pρ→−iℏ(∂∂ρ+12ρ)(7.4.43)P_\rho \to -i\hbar\left(\frac{\partial}{\partial\rho} + \frac{1}{2\rho}\right) \tag{7.4.43} is indeed Hermitian and also satisfies the canonical commutation rule. The angular momentum Pϕ→−iℏ ∂/∂ϕP_\phi \to -i\hbar\,\partial/\partial\phi is Hermitian, as it stands, on single-valued functions: ψ(ρ,ϕ)=ψ(ρ,ϕ+2π)\psi(\rho,\phi) = \psi(\rho,\phi+2\pi). (3) In the Cartesian case we saw that adding an arbitrary f(x)f(x) to −iℏ ∂/∂x-i\hbar\,\partial/\partial x didn't have any physical effect, whereas here the addition of a function of ρ\rho to −iℏ ∂/∂ρ-i\hbar\,\partial/\partial\rho seems important. Why? [Is f(x)f(x) completely arbitrary? Mustn't it be real? Why? Is the same true for the −iℏ/2ρ-i\hbar/2\rho piece?] (4) Feed in the new momentum operator PρP_\rho and show that H→coordinate basis−ℏ22m(∂2∂ρ2+1ρ∂∂ρ−14ρ2+1ρ2∂2∂ϕ2)+aρH \xrightarrow{\text{coordinate basis}} \frac{-\hbar^2}{2m}\left(\frac{\partial^2}{\partial\rho^2} + \frac{1}{\rho}\frac{\partial}{\partial\rho} - \frac{1}{4\rho^2} + \frac{1}{\rho^2}\frac{\partial^2}{\partial\phi^2}\right) + a\rho

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