3.X Square Well - Negative Kinetic Energy

Consider a 1D finite square well that is 1 eV deep and L= 0.613 nm wide. For the ground state: $k=.595π/L,a = .27 nm, B/A=.594$ (a) From normalization calculate A and B in units of $1/(nm)^{1/2}$ Evanescent Left: $\psi(x)=Be^{-(x+L/2)/a}$ Evanescent Right: $\psi(x)=Be^{-(x-L/2)/a}$ Inside well: $\psi(x)=Acos(kx)$ For the ground state, we will assume that the left and right evanescent wave functions will be the same, therefore the normalization will look like: $$\int \mid \psi(x) \mid^2dx = 1 = \int_{L/2}^{\infty} 2B^2e^{-2(x-L/2)/a}dx + \int_{-L/2}^{L/2} A^2cos^2(kx)dx$$ $$=A^2[\int_{L/2}^{\infty} 2(0.594)^2e^{-2(x-L/2)/a}dx + \int_{-L/2}^{L/2} cos^2(kx)dx] $$ $$=A^2[\int_{0.613/2}^{\infty} 2(0.594)^2e^{-2(x-0.613nm/2)/0.27nm}dx + \int_{-0.613/2}^{0.613/2} cos^2(0.595\pi/0.613x)dx] $$ $$=A^2(0.0952657nm + 0.463221nm) =1$$ $$A=1.338115nm^{-1/2}$$ $$B/A=0.594$$ $$B=0.794841 nm^{-1/2}$$
 (b) What fraction of the normalization integral comes from outside of the well? $$(0.2699999nm)B^2 + (0.463221nm)A^2 = 1$$ $$(0.2699999nm)B^2 = (0.2699999nm)(0.794841 nm^{-1/2})^2=0.17$$  About $17$ percent of the normalization integral comes from the evanescent region outside the well. 
(c) Calculate the expectation value of the Kinetic Energy: $$\bar{T} = \int\psi(x)[-\hbar^2/(2m)\frac{d^2}{dx^2}]\psi(x)dx$$ $$= \int_{L/2}^{\infty} 2B^2e^{-(x-L/2)/a}[-\hbar^2/(2m)\frac{d^2}{dx^2}]e^{-(x-L/2)/a}dx$$ $$+ \int_{-L/2}^{L/2} A^2cos(kx)[-\hbar^2/(2m)\frac{d^2}{dx^2}]cos(kx)dx$$ Plugging in constants and using Wolfram Alpha to save time: $$\bar{T}=-0.177832eV + 0.293069eV = 0.115237eV=0.12eV$$ (d)What is the KE contribution from the region of space outside the well?  The kinetic energy contribution from outside of the well is $-0.177832eV$. 



The total kinetic energy expectation value is $$\bar{T}=0.12eV$$
(e)Show that the potential energy is the same as the fraction of the normalization integral inside the well times -1 eV.

Commentary added: The region outside the well is interesting, partly because there is not classical perspective from which to envision its meaning. Note, for example, that 17% of the normalization integral comes from outside the well. How would you interpret that? How would you describe that to a sibling, parent or friend?  What is going on? What does it mean?
      Note further that the contribution to the kinetic energy from that region is about -0.18 eV as shown by Lydia's calculations above. The total kinetic energy for this state is about .12 eV. How much less is this than what one would have for a infinite well of this width?
     The potential energy is, I believe, -0.87 eV. If this well had really tall sides (infinite) then the electron wave-function would be entirely inside the well and the potential energy would be -1 eV. So this is higher, right?
     Looking at the finite square well ground state provides a glimpse into the origin of regions of negative KE, which play a critical role in establishing bandwidth.