# Showing that a wavefunction can be written as a product

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1. Nov 21, 2016

### TheSodesa

1. The problem statement, all variables and given/known data
Let us look at a 3-dimensional potential box. Show, that the wave function in this situation can be written as the product of 3 single-argument functions.

2. Relevant equations
The 3D Schrödinger equation:

-\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) + V\Psi = E\Psi,

where $\Psi = \Psi(x,y,z)$

3. The attempt at a solution
I'm not at all sure how I'm supposed to show that this is true. I have not worked with partial differential equations at all, so I'm probably not in the right with what I tried to do:

I assumed, that $\Psi(x,y,z) = \Psi_x(x) \Psi_y(y) \Psi_z(z)$, and tried plugging it into the Schrödinger equation as follows:

\begin{gather*}
-\frac{\hbar^2}{2m} &\left( \frac{\partial^2 \Psi(x,y,z)}{\partial x^2} + \frac{\partial^2 \Psi(x,y,z)}{\partial y^2} + \frac{\partial^2 \Psi(x,y,z)}{\partial z^2} \right) + V\Psi(x,y,z) = E\Psi(x,y,z)\\
\\
-\frac{\hbar^2}{2m} & \left( \frac{\partial^2 \Psi_x(x) \Psi_y(y) \Psi_z(z)}{\partial x^2} + \frac{\partial^2 \Psi_x(x) \Psi_y(y) \Psi_z(z)}{\partial y^2} + \frac{\partial^2 \Psi_x(x) \Psi_y(y) \Psi_z(z)}{\partial z^2} \right)
+ V\Psi_x(x) \Psi_y(y) \Psi_z(z) = E\Psi_x(x) \Psi_y(y) \Psi_z(z)\\
\\
-\frac{\hbar^2}{2m} & \left( \Psi_y(y) \Psi_z(z) \frac{\partial^2 \Psi_x(x)}{\partial x^2} + \Psi_x(x) \Psi_z(z) \frac{\partial^2 \Psi_y(y)}{\partial y^2} + \Psi_x(x) \Psi_y(y) \frac{\partial^2 \Psi_z(z)}{\partial z^2} \right)
+ V\Psi_x(x) \Psi_y(y) \Psi_z(z) = E\Psi_x(x) \Psi_y(y) \Psi_z(z)
\end{gather*}

Dividing by $\Psi_x(x) \Psi_y(y) \Psi_z(z)$:
\begin{gather*}

-\frac{\hbar^2}{2m} & \left( \frac{1}{\Psi_x(x)} \frac{\partial^2 \Psi_x(x)}{\partial x^2} + \frac{1}{\Psi_y(y)} \frac{\partial^2 \Psi_y(y)}{\partial y^2} + \frac{1}{\Psi_z(z)} \frac{\partial^2 \Psi_z(z)}{\partial z^2} \right)
+ V = E

\end{gather*}
But... Now what? Is there a way to proceed from here, or did I not pursue the correct path?

Last edited: Nov 21, 2016
2. Nov 21, 2016

### PeroK

Looks good. You just need a bit of insight to see what to do now. Can you remember what you did with separation of $x$ and $t$ functions in 1D?

3. Nov 21, 2016

### TheSodesa

Basically, we did what I just did: Plugged the product into the original equation, which resulted in a similar expression to what I just wrote. Then the trick was to notice, that each side of the equation is dependent on only a single variable. My book claims that this results in both of the sides of the equation being a separation constant $C$, which later turned out to be the energy $E$.

Are you saying I should have taken the time dependence into account? At a quick glance, it doesn't seem that I could use the same trick here.

4. Nov 21, 2016

### PeroK

No, in 1D you had two variables, here you have three. You can use the same trick. To give you a big hint: try setting $y = z = 0$ in your equation.

5. Nov 21, 2016

### TheSodesa

Well, if I do that, since we are talking about an infinite potential outside the box, the wave functions $\Psi_y(y)$ and $\Psi_z(z)$ go to zero at $y = 0$ and $z = 0$ respectively, because of continuity. We are of course assuming, that the coordinates are placed as to allow this to happen.

Our equation then becomes:
\begin{gather*}
-\frac{\hbar^2}{2m} & \left( \frac{1}{\Psi_x(x)} \frac{\partial^2 \Psi_x(x)}{\partial x^2} \right) + V = E
\end{gather*}

At these specific coordinates of $y$ and $z$ the potential $V(x,y,z)$ is also only a function of $x$. Therefore our equation becomes:
\begin{gather*}
-\frac{\hbar^2}{2m} & \left( \frac{1}{\Psi_x(x)} \frac{\partial^2 \Psi_x(x)}{\partial x^2} \right) + V(x) = E(x)\\
-\frac{\hbar^2}{2m} & \left( \frac{1}{\Psi_x(x)} \frac{\partial^2 \Psi_x(x)}{\partial x^2} \right) = E(x) - V(x)
\end{gather*}
Now the question is, does this hold in general? Could I replace $-\frac{\hbar^2}{2m} \left( \frac{1}{\Psi_x(x)} \frac{\partial^2 \Psi_x(x)}{\partial x^2} \right)$ with $E(x) - V(x)$ in the original equation?

6. Nov 21, 2016

### PeroK

You misinterpreted my hint! I just choose $y = z = 0$ because it was the easiest to write down. Try setting $y = y_0, z = z_0$.

You're missing something much more obvious. There is no complicated maths involved.

7. Nov 21, 2016

### TheSodesa

As per usual.

My brain really wants to go the complicated math route for some reason. If $y$ and $z$ constant, that means we are only moving along the $x$-axis. This means our wave functions don't change in those directions, and their derivatives are zero. Also, our potential $V = V(x,y_0,z_0)$, and energy $E = E(x,y_0,z_0)$. The wave equation in this case is:
$$-\frac{\hbar^2}{2m} \left( \frac{1}{\Psi_x(x)} \frac{\partial^2 \Psi_x(x)}{\partial x^2} \right)+ V(x,y_0,z_0) = E(x,y_0,z_0)$$
So at certain points $y=y_0$ and $z = z_0$, our energy only depends on the x-coordinate. If we chose the $x$-coordinate to be constant, and let $y$ vary, we would get the same result, as with $z$ as well.

I'm again hesitating to say what this means in general. Does this lead to us being able to write $V(x,y,z)$ as $V(x) + V(y) + V(z)$, and the same with the energy $E$? How does this help us?

8. Nov 21, 2016

### TheSodesa

Wait, have I just solved it after all? The fact that our problem was reduced to a single variable problem as long as we let the other variables be constant is what I was looking for, wasn't it? Or have I misunderstood this again?

9. Nov 21, 2016

### PeroK

That's it. One of your problems I think is that you got confused by the second derivatives. These are just functions of $x,y,z$. So, you could rewrite your equation with, say, $f(x) = (\frac{\partial \Psi_x^2}{\partial x^2})/\Psi_x$, and $g(y), h(z)$ similarly.

Then, it's more obvious that $g(0), h(0)$ or $g(y_0), h(z_0)$ are just some constants, and you have an equation for $f(x)$ involving nothing but constants.

10. Nov 21, 2016

### TheSodesa

Thank you so much.

11. Nov 21, 2016

### TheSodesa

I also wanted to say, that it wasn't so much the derivatives, but the potential $V(x,y,z)$ and energy $E(x,y,z)$ still being functions of 3 variables even if moved them to the other side of the equation. So even if I solved for $f(x)$, say, it wouldn't actually be a function of just $x$, at least in my head.

It took a lot of thinking to realize, that we chose $y_0$ and $z_0$ at random, which negated any loss of generality. Actually suggesting to let them be equal to zero really threw me off.

12. Nov 21, 2016

### PeroK

This process works because $V$ and $E$ are constant.

You only needed one value of $y$ and $z$. They didn't need to be arbitrary. Although, as $g$ and $h$ also turn out to be constant, it didn't matter what values you chose.

13. Nov 21, 2016

### TheSodesa

Ah, thanks for clarifying.