Suppose, I have a wavefunction given by $\psi(x,t)$. This wavefunction, over time, becomes $\psi(\alpha x,t)$. I've been asked to compute the final kinetic energy of this new wavefunction, in terms of the initial kinetic energy.
We know, $$\langle T_i\rangle=\langle \psi(x)|(-\frac{\hbar}{2m} \nabla_x^2)|\psi(x)\rangle$$
This is the initial kinetic energy, in Bra-Ket notation. We can write the final kinetic energy as :
$$\langle T_f\rangle=\langle \psi(\alpha x)|(-\frac{\hbar}{2m} \nabla_x^2)|\psi(\alpha x)\rangle$$
However, changing variable to $u$ such that $u=\alpha x$, we can see :
$$\frac{\partial}{\partial u} = \frac{\partial}{\partial x}\frac{\partial x}{\partial u} = \frac{1}{ \alpha}\frac{\partial}{\partial x}$$ Thus, $$\alpha^2\frac{\partial^2}{\partial u^2} =\frac{\partial^2}{\partial x^2}$$
Thus, we can write kinetic energy as :
$$\langle T_f\rangle= \alpha^2\langle \psi(u)|(-\frac{\hbar}{2m} \nabla_u^2)|\psi(u)\rangle = \alpha^2\langle T_i \rangle$$
However, if I write this same thing through integration, I'm facing a problem.
$$\langle T_i \rangle = \int\psi^*(x)(-\frac{\hbar}{2m} \nabla_x^2)\psi(x)dx$$ Similarly, we have :
$$\langle T_f \rangle = \int\psi^*(u)(-\frac{\hbar}{2m} \nabla_x^2)\psi(u)dx$$
As we have seen, $$\nabla_x^2 = \alpha^2\nabla_u^2 \space\space\space\& \space\space\space dx=\frac{du}{\alpha}$$
Plugging these two values in, and noting that $u$ is just a dummy variable, we have :
$$\langle T_f \rangle = \int\psi^*(u)(-\alpha^2\frac{\hbar}{2m} \nabla_u^2)\psi(u)\frac{du}{\alpha} = \alpha \langle T_i\rangle$$
Even though the two notations are equivalent, there are giving me different answers. Can someone guide me as to where I'm making a mistake, and how should I deal with problems such as these?