Indeed, you need to brush up your calculus.
Note that the wavefunction of the free particle is
ψ(x,t)=cexp[ℏi(p⋅x−Et)] and when we expand the dot product of momentum and position vector, we have
ψ(x,t)=cexp[ℏi(pxx+pyy+pzz−Et)] Let consider only the partial derivative with respect to
x only
−2mℏ2∂x2∂2ψ(x,t)=−2mℏ2∂x∂[∂x∂ψ(x,t)] I would let you continue and you would get
−2mℏ2∂x2∂2ψ(x,t)=−2mℏ2(ℏipx)2cexp[ℏi(pxx+pyy+pzz−Et)] Do the partial derivative with respect to
y and
z, we would obtain a similar result.
−2mℏ2∂y2∂2ψ(x,t)=−2mℏ2(ℏipy)2cexp[ℏi(pxx+pyy+pzz−Et)]−2mℏ2∂z2∂2ψ(x,t)=−2mℏ2(ℏipz)2cexp[ℏi(pxx+pyy+pzz−Et)] Combine the result, we would get
−2mℏ2(∂x2∂2+∂y2∂2+∂z2∂2)ψ(x,t)=−2mℏ2((ℏipx)2+(ℏipy)2+(ℏipz)2)ψ(x,t) After some simplification, you should get the kinetic energy of the free particle as
EK=2mpx2+py2+pz2 There is some incomplete info in post #5, the energy expression that I wrote in the post is kinetic energy NOT total energy.
[−2mℏ2(∂x2∂2+∂y2∂2+∂z2∂2)+V0]ψ(x,t)=(2mpx2+py2+pz2+V0)ψ(x,t) The total energy is
E=2mpx2+py2+pz2+V0