Quantum mechanics in multiple dimensions

[aaaa202]

Suppose we have a 2d problem (for instance a 2d box). You look for separable solutions in x and y. But it seems to me that the solutions are in a way determined by how we choose the separation constant. Do we know anything a priori that tells us if the separation constant should be negative or positive?
Consider for instance the 2d box. The separation constant for the x and y solution should have opposite sign, yet the physical situation is exactly the same. What is wrong here?

 

[Morgoth]

the wrong choice of a separation constant will lead you in unwanted behavior of your solutions...
So generally you want to avoid infitities and of course you want to apply some boundary conditions...

 

[aaaa202]

But I just don't see how you can determine it. Consider for instance the 2D box. You would be crazy not to think that the wave functions in x and y are the same. The SE is:

HY(y)X(x) = EY(y)X(x)

but now comes a choice:

In which of the separated equation does E go? And in which of the separated equation do we put a minus sign and a plus sign? The minus sign will of course be fatal since you then get a solution with an exponential rather than and exp(ix)

 

[Jorriss]

But I just don't see how you can determine it. Consider for instance the 2D box. You would be crazy not to think that the wave functions in x and y are the same. The SE is:

HY(y)X(x) = EY(y)X(x)

but now comes a choice:

In which of the separated equation does E go? And in which of the separated equation do we put a minus sign and a plus sign? The minus sign will of course be fatal since you then get a solution with an exponential rather than and exp(ix)

The separation constant goes with both.

The way it works is f(x)=g(y) => f(x) = c = g(y). Both functions equal the separation constant.

Second, you don't assign a sign to the separation constant - it's just a constant and the value of it will come out in the math.

 

[aaaa202]

So suppose you have the 2d box.

Inside it you will have an SE of the form:

-hbar^2/2m(∂_x² +∂_y² )XY = EXY

So we could separate variables and get the equation:

-hbar^2/2m(∂_x²Y = (K+E)X

-hbar^2/2m(∂_y²Y = -KY

How am I to determine what the values of these constants K should be? Of course from the symmetry you want the solutions in each dimension to be the same, but with this choice of letting E go into either x and y it doesn't seem like you get that.

 

[bp_psy]

So suppose you have the 2d box.

Inside it you will have an SE of the form:

-hbar^2/2m(∂_x² +∂_y² )XY = EXY

So we could separate variables and get the equation:

-hbar^2/2m(∂_x²Y = (K+E)X

-hbar^2/2m(∂_y²Y = -KY

How am I to determine what the values of these constants K should be? Of course from the symmetry you want the solutions in each dimension to be the same, but with this choice of letting E go into either x and y it doesn't seem like you get that.

There seems to be something odd about your separation of variables.

The way it works is:
-hbar^2/2m(∂_x² +∂_y² )XY = EXY=(E_x+E_y)XY

then :
((1/X)(∂_x²X+ 2mE_x/hbar^2) +((1/Y)(∂_y²Y+ 2mE_y/hbar^2) =0

In order for this to be 0 for all x ,y both terms must equal 0 independently.
∂_x²X =-2mE_x/hbar^2 X

∂_y²Y =-2mE_y/hbar^2 Y

E_x and E_y a are your energies associated to each coordinate and are determined by your boundary conditions. There is no arbitrary choice.

 

[aaaa202]

Okay so you start with the two separate equations and add them. Hmm I guess that makes physical sense, but I have just seen the other version used for e.g. the hydrogen atom.
To recap on this form what I do is say:

-hbar^2/2m(∂_x² +∂_x² )XY = EXY

Divide by XY and you get:

-hbar^2/2m(1/X∂_x²X +1/Y∂_x² )Y = E

And since the sum of the X and Y term is a constant, they must each be equal to a constant. You can either let the E go into the X or Y equation. Doing so for X you end up with the equations from my previous post.
I have seen this used for the hydrogen atom where you let the E-term go into the radial equation. On the other hand, it is later showed how this separation was actually physically equivalent to constructing simultaneous eigenstates of L², Lz, H, so maybe this explains it?

I hope I have made my problem clear - I don't see how you just from the math can determine anything about the separation constants.

 

[jtbell]

In which of the separated equation does E go?

You can do it either way. It doesn't make any difference in the final result. Using your 2-dimensional box example start with:

$$-\frac{\hbar^2}{2m} \frac{1}{X} \frac{\partial^2 X}{\partial x^2}
-\frac{\hbar^2}{2m} \frac{1}{Y} \frac{\partial^2 Y}{\partial y^2} = E$$

Method 1: Put E with the y-term and define a second separation constant

$$-\frac{\hbar^2}{2m} \frac{1}{X} \frac{\partial^2 X}{\partial x^2} =
\frac{\hbar^2}{2m} \frac{1}{Y} \frac{\partial^2 Y}{\partial y^2} + E = F$$

This leads to the two separated equations

$$-\frac{\hbar^2}{2m} \frac{\partial^2 X}{\partial x^2} = FX\\
-\frac{\hbar^2}{2m} \frac{\partial^2 Y}{\partial y^2} = (E - F)Y$$

Define $E_x = F$ and $E_y = E - F$. This leads to $E = E_x + E_y$ and

$$-\frac{\hbar^2}{2m} \frac{\partial^2 X}{\partial x^2} = E_x X\\
-\frac{\hbar^2}{2m} \frac{\partial^2 Y}{\partial y^2} = E_y Y$$

Method 2: Put E with the x-tern and define a third separation constant

$$-\frac{\hbar^2}{2m} \frac{1}{Y} \frac{\partial^2 Y}{\partial y^2} =
\frac{\hbar^2}{2m} \frac{1}{X} \frac{\partial^2 X}{\partial x^2} + E = G$$

This leads to the two separated equations

$$-\frac{\hbar^2}{2m} \frac{\partial^2 Y}{\partial y^2} = GY\\
-\frac{\hbar^2}{2m} \frac{\partial^2 X}{\partial x^2} = (E - G)X$$

Define $E_y = G$ and $E_x = E - G$. This leads to $E = E_x + E_y$ and

$$-\frac{\hbar^2}{2m} \frac{\partial^2 Y}{\partial y^2} = E_y Y\\
-\frac{\hbar^2}{2m} \frac{\partial^2 X}{\partial x^2} = E_x X$$

which are the same equations as in method 1.

 

[aaaa202]

okay of course. There's just one tiny thing that bothers me still - namely how do you determine if the constants are positive or negative? After all that has a lot to say for the final wave functions?

 

[Jano L.]

From the boundary condition that on the boundary of the square, the function is 0. If $F$ was negative, the equation for x would not longer have oscillatory solutions (equation of an harmonic oscillator) but would have exponentially rising solutions, which are non-zero everywhere and cannot satisfy $X = 0$ on the boundary.