Exploring the Kinetic Energy Operator: Why is it Differentiated Twice?

[Mayan Fung]

When I learned about operators, I learned <x> = ∫ Ψ* x Ψ dx, <p> = ∫ Ψ* (ħ/i ∂/∂x) Ψ dx. The book then told me the kinetic energy operator

T = p²/2m = -ħ²/2m (∂²/∂x²)
I am just think that why isn't it -ħ²/2m (∂/∂x)²
Put in other words, why isn't it the square of the derivative, but differentiating it twice?

 

[strangerep]

When I learned about operators, I learned <x> = ∫ Ψ* x Ψ dx, <p> = ∫ Ψ* (ħ/i ∂/∂x) Ψ dx. The book then told me the kinetic energy operator

T = p²/2m = -ħ²/2m (∂²/∂x²)
I am just think that why isn't it -ħ²/2m (∂/∂x)²
Put in other words, why isn't it the square of the derivative, but differentiating it twice?

They're just different notations for the same thing: the 2nd derivative is the square of the 1st derivative.
Cf. 2nd derivative.

 

[vanhees71]

As the name suggests operators operate on the Hilbert space vectors. They are linear operators mapping $|\psi \rangle \in \mathcal{H}$ to another vector $\hat{A} |\psi \rangle \in \mathcal{H}$. By definition the product of two operators is the composition of these operations, i.e., $\hat{B} \hat{A}|\psi \rangle$ first maps $|\psi \rangle$ to $\hat{A} |\psi \rangle$ and then this vector to $\hat{B}(\hat{A} |\psi \rangle)=\hat{B} \hat{A} |\psi \rangle$. Thus $\partial_x^2=\partial_x \partial_x$ applied to a position-space wave function means. You first take the derivative of the wave function and then take the derivative of the result again, i.e., you take the 2nd derivative of the wave function. Then the usual notation is
$$\left (\frac{\partial}{\partial x} \right)^2=\frac{\partial^2}{\partial x^2}.$$

 

[Mayan Fung]

Thanks all!