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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.

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3 Answers 3

up vote 6 down vote accepted

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".

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Great answer, thanks! (The V(x,y,z,) = V(z) wasn't a mistake, but the special case for 1D QW.) –  Halftrack Oct 23 '12 at 12:24

This method is called "separation of variables", and it is one of several strategies for finding solutions to multi-dimensional field problems in physics.

It's big advantage is that solving N 1-dimensional differential equations is generally easier than solving 1 N-dimensional problems (N>1), but it is contingent on there being a "uniqueness theorem" for the category of problems that you are looking at. Happily many common field problems in physics have such a theorem.

To show that the condition on the potential is required for the Schrödinger equation simply make the separation, and expand. If the potential has the above form, you can write the LHS of the equation in three terms: $$ \left( \frac{-\hbar^2}{2m} \frac{\partial^2}{\partial x} \psi(x) + V(x) \right)\psi(y)\psi(z) $$ and you can clearly rewrite the whole into three separate conditions like $$ \left( \frac{-\hbar^2}{2m} \frac{\partial^2}{\partial x} \psi(x) + V(x) \right) = i\hbar{}\frac{\partial}{\partial t} \psi(x) . $$ On the other had, if the potential can not be written in this way, you can not get the LHS into the form you need and you can not proceed along these lines.

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Let me just emphasize that you can find a lot of solutions of the equation that do not satisfy the statement in the question (and cannot be presented as such products), but you can use this statement to find the basis of the set of the solutions of the equation. So any solution can be presented as a linear combination of functions satisfying the statement.

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How do you reconcile this with the answer of @lubos-motl? Can you give any examples? –  Halftrack Oct 23 '12 at 12:28
    
Dear akhmeteli and @Halftrack, akhmeteli must mean lots of solutions to the time-dependent Schrodinger equation: solutions may be obtained as linear superpositions of the separated solutions with different energies $E$. For the time-independent "eigenvalue" equation with a single well-defined energy eigenvalue, the separation of variables lists all the solutions (unless there is degeneracy in the spectrum). –  Luboš Motl Oct 23 '12 at 16:11
    
@Halftrack and Luboš Motl: Judging by Luboš Motl's comment, he agrees that the statement is not always satisfied even for the time-independent Schrödinger equation if there is degeneracy. Let me also add that it is not quite obvious from your (Halftrack's) question that it was the time-independent Schrödinger equation that you had in mind, at least I did miss that. –  akhmeteli Oct 24 '12 at 0:46
    
I was only looking to solve the time independent Schrödinger equation, but you are absolutely correct: my question does not state so. If there is degeneracy, how would that break the derivation by @lubos-motl? –  Halftrack Oct 24 '12 at 1:03
    
@Halftrack: I did not study the derivation in detail, but it seems OK. It is important though to understand what he derived. As far as I can judge, he proved that there is a set of eigenstates of the Hamiltonian in the form of your statement (products of three functions), and any eigenstate is a linear combination of eigenstates from that set. However, when the spectrum is degenerate, there are at least two different eigenstates in that set having the same eigenvalue, and linear combinations of such eigenstates are also eigenstates, but they do not have to have the form of your statement. –  akhmeteli Oct 24 '12 at 1:49

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