Showing that a wavefunction can be written as a product

[TheSodesa]

Homework Statement

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.

Homework Equations

The 3D Schrödinger equation:
\begin{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,
\end{equation}
where ##\Psi = \Psi(x,y,z)##

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?

 

[PeroK]

But... Now what? Is there a way to proceed from here, or did I not pursue the correct path?

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?

 

[TheSodesa]

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?

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.

 

[PeroK]

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.

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.

 

[TheSodesa]

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.

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?

 

[PeroK]

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?

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.

 

[TheSodesa]

You misinterpreted my hint!

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:
[tex]
-\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)
[/tex]
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?

 

[TheSodesa]

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.

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?

 

[PeroK]

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?

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.

 

[TheSodesa]

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.

Thank you so much.

 

[TheSodesa]

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.

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.

 

[PeroK]

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.

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.

 

[TheSodesa]

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.

Ah, thanks for clarifying.