Introduction to Quantum Mechanics · Problem 8.28

The Variational Principle — David J. Griffiths

Physics — Quantum Mechanics · variational_yukawa_deuteron

Problem

In Yukawa's original theory (1934), which remains a useful approximation in nuclear physics, the "strong" force between protons and neutrons is mediated by the exchange of π\pi-mesons. The potential energy is V(r)=−r0V0e−r/r0r,(8.83)V(r) = -r_0 V_0 \frac{e^{-r/r_0}}{r}, \qquad (8.83) where rr is the distance between the nucleons, and the range r0r_0 is related to the mass of the meson: r0=ℏ/mπcr_0 = \hbar/m_\pi c. *Question:* Does this theory account for the existence of the deuteron (a bound state of the proton and the neutron)? The Schrödinger equation for the proton/neutron system is (see Problem 5.1): −ℏ22μ∇2ψ(r)+V(r)ψ(r)=Eψ(r),(8.84)-\frac{\hbar^2}{2\mu}\nabla^2\psi(\mathbf{r}) + V(r)\psi(\mathbf{r}) = E\psi(\mathbf{r}), \qquad (8.84) where μ\mu is the reduced mass (the proton and neutron have almost identical masses, so call them both mm), and r\mathbf{r} is the position of the neutron (say) relative to the proton: r=rn−rp\mathbf{r} = \mathbf{r}_n - \mathbf{r}_p. Your task is to show that there exists a solution with negative energy (a bound state), using a variational trial wave function of the form ψβ(r)=Ae−βr/r0.(8.85)\psi_\beta(\mathbf{r}) = Ae^{-\beta r/r_0}. \qquad (8.85) (a) Determine AA, by normalizing ψβ(r)\psi_\beta(\mathbf{r}). (b) Find the expectation value of the Hamiltonian (H=−ℏ22μ∇2+V)\left(H = -\frac{\hbar^2}{2\mu}\nabla^2 + V\right) in the state ψβ\psi_\beta. *Answer:* E(β)=ℏ22μr02β2[1−4γβ(1+2β)2],where γ≡2μr02ℏ2V0.(8.86)E(\beta) = \frac{\hbar^2}{2\mu r_0^2}\beta^2\left[1 - \frac{4\gamma\beta}{(1+2\beta)^2}\right], \quad \text{where } \gamma \equiv \frac{2\mu r_0^2}{\hbar^2}V_0. \qquad (8.86) (c) Optimize your trial wave function, by setting dE(β)/dβ=0dE(\beta)/d\beta = 0. This tells you β\beta as a function of γ\gamma (and hence—everything else being constant—of V0V_0), but let's use it instead to eliminate γ\gamma in favor of β\beta: Emin⁡=ℏ22μr02β2(1−2β)(3+2β).(8.87)E_{\min} = \frac{\hbar^2}{2\mu r_0^2}\frac{\beta^2(1-2\beta)}{(3+2\beta)}. \qquad (8.87) (d) Setting ℏ2/2μr02=1\hbar^2/2\mu r_0^2 = 1, plot Emin⁡E_{\min} as a function of β\beta, for 0≤β≤10 \le \beta \le 1. What does this tell you about the binding of the deuteron? What is the minimum value of V0V_0 for which you can be confident there is a bound state (look up the necessary masses)? The experimental value is 52 MeV.

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