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Schrodinger equation problem

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Homework Statement


Consider the time-dependent Schrodinger equation for a free particle in two spatial dimensions
1zwnbds.jpg

Using the method of separation of variables, determine the wave function ψ(x,y,t)

Homework Equations




The Attempt at a Solution


Not sure how to do the separation here since it is a function of 3 variables, I'm inclined to think that Ψ(x,y,t) = f(x) g(y) j(t) can be assumed, hence substitute it in the equation and separate. Is it correct to assume that?
 

Answers and Replies

  • #2
TSny
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1 I'm inclined to think that Ψ(x,y,t) = f(x) g(y) j(t) can be assumed, hence substitute it in the equation and separate. Is it correct to assume that?
Yes. See what you get.
 
  • #3
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Yes. See what you get.

I would get this
b7nwgl.png

Then what? I can't separate each variable to each side, since there are 3 variables
Correction: in the LHS the g(t) should be g(y)
 
  • #4
TSny
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The next step is standard: divide the equation by Ψ(x, y, t) = f(x)g(y)j(t).
 
  • #5
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The next step is standard: divide the equation by Ψ(x, y, t) = f(x)g(y)j(t).
Ok ok. Then I get this
2mchzfb.png

Then what do I do with it?
 
  • #7
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The problem statement seems to take for granted that you know how to do separation of variables. Have you done separation of variables before?
https://en.wikipedia.org/wiki/Separation_of_variables#Partial_differential_equations
Yes, but I'm new to it. So I believe I should set the whole equation equal to a variable lambda, then the LHS would be
24mds9f.png

And do the same for the RHS where I would get two other separate equations, one in terms of f and one in terms of g? or would it just give:
2jfnehg.png
?
 
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  • #8
TSny
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Yes, but I'm new to it. So I believe I should set the whole equation equal to a variable lambda, then the LHS would be
24mds9f.png
Yes, except that you should think of ##\lambda## as a constant rather than a variable. To see this, note that your equation in post #5 must be satisfied for all allowed values of t, x, and y. So, suppose you arbitrarily pick a value for x and y. The right hand side is then some number. The left hand side is a function of time that must equal that number for any value of t. So, in fact, the left side must be independent of t. That is, the left side must be a constant which you can call ##-\lambda##.

See if you can construct similar arguments to show that the individual terms on the right side of the equation in post #5 must each be equal to a constant (although not necessarily the same constant.)
 
  • #9
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Yes, except that you should think of ##\lambda## as a constant rather than a variable. To see this, note that your equation in post #5 must be satisfied for all allowed values of t, x, and y. So, suppose you arbitrarily pick a value for x and y. The right hand side is then some number. The left hand side is a function of time that must equal that number for any value of t. So, in fact, the left side must be independent of t. That is, the left side must be a constant which you can call ##-\lambda##.

See if you can construct similar arguments to show that the individual terms on the right side of the equation in post #5 must each be equal to a constant (although not necessarily the same constant.)
Ok, then the terms in the RHS would be
e16bdw.png

Where lambda is just an arbitrary constant
Is the problem finished like this?
 
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  • #10
TSny
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OK. But, you should distinguish the three constants by not using the same symbol ##-\lambda## for each one. Also, there is no need to write the constants with a negative sign. In the Wikipedia link, they wrote the constant with a negative sign for later convenience for the particular partial differential equation there. In your case, you will discover whether the constants should be positive or negative as you solve the individual equations.

The three constants are not completely arbitrary at this point. The equation in post #5 gives you a relation that the three constants must satisfy.
 
  • #11
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OK. But, you should distinguish the three constants by not using the same symbol ##-\lambda## for each one. Also, there is no need to write the constants with a negative sign. In the Wikipedia link, they wrote the constant with a negative sign for later convenience for the particular partial differential equation there. In your case, you will discover whether the constants should be positive or negative as you solve the individual equations.

The three constants are not completely arbitrary at this point. The equation in post #5 gives you a relation that the three constants must satisfy.
Ok, then I can say
hw0jdt.png

And so
33vzpso.png

And same for the LHS of the equation in #5
16axxdf.png
 
  • #12
TSny
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OK. These are standard differential equations which you can solve. How is the constant ##\lambda## related to the constants ##B## and ##C##? (Use the equation in post #5)
 
  • #13
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OK. These are standard differential equations which you can solve. How is the constant ##\lambda## related to the constants ##B## and ##C##? (Use the equation in post #5)
From the equation, λ is related to B and C as
λ = B + C
 
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  • #14
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OK. These are standard differential equations which you can solve. How is the constant ##\lambda## related to the constants ##B## and ##C##? (Use the equation in post #5)
Then where do I go from there?
 
  • #15
TSny
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From the equation, λ is related to B and C as
λ = B + C
Yes. So, λ will be determined once the values of B and C are known.
 
  • #16
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Then where do I go from there?
You have 3 ordinary differential equations: one for j(t), one for f(x), and one for g(y). Try to solve them.
 
  • #17
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You have 3 ordinary differential equations: one for j(t), one for f(x), and one for g(y). Try to solve them.
I get really complicated stuff. For example for f(x), I multiply the equation by dx^2 and divide by f(x), then integrate with respect to f, and the rest of the terms are constants, the integration gives f(ln(f)-1). Am I doing something wrong here?
 
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  • #18
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I get really complicated stuff. For example for f(x), I multiply the equation by dx^2 and divide by f(x), then integrate with respect to f, and the rest of the terms are constants, the integration gives f(ln(f)-1). Am I doing something wrong here?
You can't multiply by dx2. You can't "break apart" the second derivative that way.

Simplify the equation by lumping all the constants together. For example, suppose you let ##k_x## be a constant such that ##k_x^2 = \frac{2mB}{\hbar^2}##. Write the differential equation in terms of ##k_x##. You will see that you get a familiar type of ordinary differential equation.
 
  • #19
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You can't multiply by dx2. You can't "break apart" the second derivative that way.

Simplify the equation by lumping all the constants together. For example, suppose you let ##k_x## be a constant such that ##k_x^2 = \frac{2mB}{\hbar^2}##. Write the differential equation in terms of ##k_x##. You will see that you get a familiar type of ordinary differential equation.
I need to take the square root to get the of the second power on df^2/dx^2 ?
 
  • #20
TSny
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I need to take the square root to get the of the second power on df^2/dx^2 ?
:bugeye: No.

Write the differential equation as ##f''(x) = -k_x^2f(x)##, where the constant ##k_x## was defined in post #18. You're looking for a function ##f(x)## such that when you take its second derivative you get back the function multiplied by a negative constant.

Are you familiar with the form of the wavefunction for a free particle moving in one dimension? You might expect a similar form for ##f(x)##.
 
  • #21
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:bugeye: No.

Write the differential equation as ##f''(x) = -k_x^2f(x)##, where the constant ##k_x## was defined in post #18. You're looking for a function ##f(x)## such that when you take its second derivative you get back the function multiplied by a negative constant.

Are you familiar with the form of the wavefunction for a free particle moving in one dimension? You might expect a similar form for ##f(x)##.
Still doesn't look familiar, I know how to solve df^2/dx^2 + df/dx + f = 0 , but this equation seems like df^2/dx^2 = 0, i think its solution would be just c1 f + c2
 
  • #22
TSny
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I know how to solve df^2/dx^2 + df/dx + f = 0
If you can solve this equation, then you should be able to solve d2f/dx2 + b f = 0, where b is a constant.

Note, the second derivative is written d2f/dx2, not df2/dx2. The numerator is not the square of df.
 

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