Schrodinger equation

• logearav
logearav
ψ and its derivatives occur only linearly in the Schrodinger equation, that is, second or higher powers of these quantities do not appear in the equation.
Schrodinger equation for a free particle is
i$\hbar$∂ψ(x,t)/∂t = (-$\hbar$2/2m)(∂2ψ(x,t)/∂x2)
Here (∂2ψ(x,t)/∂x2) is second power of ψ. Then how can we justify the statement "second or higher powers of ψ do not appear"?

Gold Member
That's the second derivative. The statement "linear" means no terms like (dψ/dx)^2 or ψ^2 appear.

Gold Member
To follow that up, the key issue with the equation being linear is that if two wave functions are possible solutions (like two plane waves in your example), then any linear superposition of them is also a solution. This gives us the concept of "wave packets", and is important in a lot of quantum mechanics.

logearav
Thanks for your replies Matterwave and Ken G

juanrga
ψ and its derivatives occur only linearly in the Schrodinger equation, that is, second or higher powers of these quantities do not appear in the equation.
Schrodinger equation for a free particle is
i$\hbar$∂ψ(x,t)/∂t = (-$\hbar$2/2m)(∂2ψ(x,t)/∂x2)
Here (∂2ψ(x,t)/∂x2) is second power of ψ. Then how can we justify the statement "second or higher powers of ψ do not appear"?

The Schrödinger equation is a differential linear equation in ψ.

For a free particle it is

i$\hbar$∂/∂t ψ = K ψ

Since it is linear this means that if ψ1 and ψ2 are solutions, then any linear combination ψT = c1ψ1 + c2ψ2 is a solution as well.