quantum mechanics - Rationale for writing wave function as product of independent wave functions

When solving Schrödinger's equation for a 3D quantum well with infinite barriers, my reference states that: $$\psi(x,y,z) = \psi(x)\psi(y)\psi(z) \quad\text{when}\quad V(x,y,z) = V(x) + V(y) + V(z) = V(z).$$ However, I cannot find any rationale for this statement. It may be obvious, but I would appreciate any elucidation.

Answer

It's because when $$V(x,y,z) = V_x(x) + V_y(y) + V_z(z),$$ (I guess that your extra identity $V(x,y,z)=V(z)$ is a mistake), we also have $$H = H_x + H_y + H_z$$ because $H = (\vec p)^2 / 2m + V(x,y,z)$ and $(\vec p)^2 = p_x^2+p_y^2+p_z^2$ decomposes to three pieces as well.

One may also see that the terms such as $H_x\equiv p_x^2/2m+V_x(x)$ commute with each other, $$[H_x,H_y]=0$$ and similarly for the $xz$ and $yz$ pairs. That's because the commutators are only nonzero if we consider positions and momenta in the same direction ($x$, $y$, or $z$).

At the end, we want to look for the eigenstates of the Hamiltonian $$H|\psi\rangle = E |\psi \rangle$$ and because we have $H = H_x+H_y+H_z$, a Hamiltonian composed of three commuting pieces, we may simultaneously diagonalize them i.e. look for the common eigenstates of $H_x,H_y,H_z$, and therefore also $H$. So given the separation condition for the potential, we may also assume $$H_x |\psi\rangle = E_x |\psi\rangle$$ and similarly for the $y,z$ components. However, the equation above is just a 1-dimensional problem that implies that $|\psi\rangle$ must depend on $x$ as a one-dimensional quantum mechanical energy eigenstate wave function, $$\psi(x) = C\cdot \psi_n(x)$$ which is an eigenstate of $H_x$. This has to hold but the normalization factor is undetermined. We usually say that it's a constant but this statement only means that it is independent of $x$. In reality, it may depend on all observables that are not $x$ such as $y,z$. So a more accurate implication of the $H_x$ eigenstate equation is $$\psi(x,y,z) = C_x(y,z)\cdot \psi_{n_x}(x) .$$ In a similar way, we may show that $$\psi(x,y,z) = C_y(x,z)\cdot \psi_{n_y}(y)$$ and $$\psi(x,y,z) = C_z(x,y)\cdot \psi_{n_z}(z)$$ and by combining these three formulae, we see that the whole function must factorize to a product of functions of $x$ and $y$ and $z$ separately. If you need a rigorous proof of the last simple step, take e.g. the complex logarithms of the three forms for $\psi$ above and compare e.g. the first pair: $$\ln\psi = \ln C_x(y,z) +\ln\psi_{n_x}(x) = \ln C_y(x,z)+\ln \psi_{n_y}(y)$$ Take e.g. the partial derivative of the last equation with respect to $y$: $$\frac{\partial \ln C_x(y,z)}{\partial y} = \frac{\partial \ln\psi_{n_y}(y) }{ \partial y }$$ The other two (1+1) terms are zero because they didn't depend on $y$. The right hand side above only depends on $y$, so the same must be true for the left hand side. I am going to make a simple conclusion but to make it really transparent, let's differentiate the latter equation over $z$, too. The $\psi_{n_y}$ term disappears as well so we have $$\frac{\partial^2 \ln C_x(y,z)}{\partial y\,\partial z} = 0$$ It means that $\ln C_x(y,z)$ must have the form $K_x(y)+L_x(z)$, and $e^{K_x(y)}e^{L_x(z)}$ must be the remaining factors in the wave function.

We say that the wave function in the product form is a "tensor product" of the three independent one-dimensional wave functions and more "operationally", as another user mentioned, the method described above is the method of "separation of variables".