Show that there is no acceptable solution to the (time-independent) Schrödinger equation for the infinite square well with E = 0 or E<0. (This is a special case of the general theorem in Problem 2.2, but this time do it by explicitly solving the Schrödinger equation, and showing that you cannot satisfy the boundary conditions.)
Question 2.3: Show that there is no acceptable solution to the (time-indep...
Equation 2.23 says
-\frac{\hbar^{2}}{2 m} \frac{d^{2} \psi}{d x^{2}}=E \psi (2.23).
\frac{d^{2} \psi}{d x^{2}}=-\frac{2 m E}{\hbar^{2}} \psi ; Eq. 2.26 says
ψ(0) = ψ(a) = 0 (2.26).
ψ(0) = ψ(a) = 0 , If E=0, d^{2} \psi / d x^{2}=0 so ψ (x) = A + Bx: ψ (0) = A = 0 ⇒ ψ = Bx; ψ(a) = Ba = 0 ⇒ B = 0, so ψ = 0. If E=0, d^{2} \psi / d x^{2}=κ^2 ψ , with \kappa \equiv \sqrt{-2 m E} / \hbar real, so \psi(x)=A e^{\kappa x}+B e^{-\kappa x} . This time ψ (0) = A + B = 0 ⇒ B = -A, so \psi=A\left(e^{\kappa x}-e^{-\kappa x}\right) , while \psi(a)=A\left(e^{\kappa a}-e^{-\kappa a}\right)=0 ⇒ either A = 0, so ψ = 0 , or else e^{\kappa a}=e^{-\kappa a} , so e^{2\kappa a}=1 , so 2κa = \ln(1) = 0 , so κ = 0, and again ψ = 0. In all cases, then, the boundary conditions force ψ = 0, which is unacceptable (non-normalizable).
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