# quantum mechanics in multiple dimensions

by aaaa202
Tags: dimensions, mechanics, multiple, quantum
 P: 867 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?
 P: 122 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...
 P: 867 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)
PF Patron
P: 990

## quantum mechanics in multiple dimensions

 Quote by 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)
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.
 P: 867 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.
P: 425
 Quote by 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.

The way it works is:
-hbar^2/2m(∂x2 +∂y2 )XY = EXY=(Ex+Ey)XY

then :
((1/X)(∂x2X+ 2mEx/hbar^2) +((1/Y)(∂y2Y+ 2mEy/hbar^2) =0

In order for this to be 0 for all x ,y both terms must equal 0 independently.
x2X =-2mEx/hbar^2 X

y2Y =-2mEy/hbar^2 Y

Ex and Ey a are your energies associated to each coordinate and are determined by your boundary conditions. There is no arbitrary choice.
 P: 867 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.
Mentor
P: 10,687
 Quote by aaaa202 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.
 P: 867 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?
 P: 969 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.

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