Energy In Quantum Mechanics (Pythagorean theorem) In the Schrodinger Equation for a free electron in three dimensions, can the energy eigenvalue E always be broken up into x y and z components such that $E^2 = E_x^2 + E_y^2 + E_z^2$? What is the reasoning behind the answer?
 A: For this to work it must be possible to break up the Schrodinger equation into three independent equations:
$$
-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x_i^2}\psi_i(x_i)+V(x_i)\psi_i(x_i)=E_i\psi_i(x_i). \tag{1}
$$
with $x_1=x,x_2=y,x_3=z$, and this can happen only if the potential function $V(x_1,x_2,x_3)$ can be broken as a sum $V_1(x_1)+V_2(x_2)+V_3(x_3)$ of potentials each one independent from the other.  The case of a free particle is the one where $V_i(x_i)=0$, meaning that the potential along $x_1$ (it is $0$ in this direction) is independent of the potential in $x_2$ (it is also $0$ in this direction).
In particular, using separation of variables with $V=0$, we have
$$
-\frac{\hbar^2}{2m}\sum_i\frac{\partial^2}{\partial x_i^2}\psi(x_1,x_2,x_3)=E\psi(x_1,x_2,x_3)
$$
where $E=E_1+E_2+E_3$ and $E_i$ is the eigenvalue for Eq.(1) and $\psi(x_1,x_2,x_3)=\psi_1(x_1)\psi_2(x_2)\psi_3(x_3)$, with $\psi_i(x_i)$ the solution to (1) associated with the eigenvalue $E_i$.
A: Okay, it's a long answer, but it can be found everywhere. I'll go quickly through it, tell me if you need more clarification.
The euqation is
$$ -\frac{\hbar^2}{2m} \left(\frac{\partial^2 \psi}{\partial x^2}+\frac{\partial^2 \psi}{\partial y^2}+\frac{\partial^2 \psi}{\partial z^2}\right)=E\psi$$
First idea: variable separation ansatz. You can write
$$\psi=\phi_x(x)\cdot\phi_y(y)\cdot\phi_z(z)$$
So you'll have
$$ -\frac{\hbar^2}{2m} \left(\frac{\partial^2 \phi_x}{\partial x^2}\phi_y\phi_z(z)+\frac{\partial^2 \phi_y}{\partial y^2}\phi_x\phi_z+\frac{\partial^2 \phi_z}{\partial z^2} \phi_x\phi_y\right)=E\ \phi_x\phi_y\phi_z$$
Now you divide everything by $\psi$ so you have
$$ -\frac{\hbar^2}{2m} \left(\phi_x^{-1}\frac{\partial^2 \phi_x}{\partial x^2}+\phi_y^{-1}\frac{\partial^2 \phi_y}{\partial y^2}+\phi_z^{-1}\frac{\partial^2 \phi_z}{\partial z^2} \right)=E$$
In other words,
$$ -\frac{\hbar^2}{2m} \left(\frac{\phi_x''}{\phi_x}+\frac{\phi_y''}{\phi_y}+\frac{\phi_z''}{\phi_z} \right)=E$$
So these are three functions giving a constant. The second idea is that 
$$ -\frac{\hbar^2}{2m} \frac{\phi_x''}{\phi_x} = E - \frac{\phi_y''}{\phi_y}-\frac{\phi_z''}{\phi_z}$$
Something that depends on $x$ giving something that doesn't depend on $x$ means that the whole RHS must be a constant, which you call $k_x$
You do the same for the other two coordinates. Hence you have three equations of the form
$$ -\frac{\hbar^2}{2m} \phi_x'' = k_x \phi_x $$
So they are three independent "HO-equations".
