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Find normalised linear combinations that are orthogonal

  • Thread starter g.lemaitre
  • Start date
  • #1
267
1

Homework Statement


I'm a little weary of posting this in this forum. If I post it in the math section it will be answered in about 30 min whereas here it might take about 5 hours, but we'll see.
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Homework Equations


The Attempt at a Solution


Number one, I'm not exactly sure how they get from
[tex]\int \phi_1^* (c_1\phi_1 + c_2\phi_2)d\tau = c_1 + c_2 d [/tex]

I think it's because

[tex] \phi^*\phi = 1[/tex] but I'm not sure.
Number two, I don't understand the following step:
[tex] \int (c_1\phi_1 + c_2 phi_2)*(c_1 \phi_1 + c_2\phi_2)d\tau = c_1^2 + c_2^2 + 2dc_1c_2 [/tex]

why does [itex]\phi[/itex] disappear?
I figure that it must have something to do with the fact that [itex]\phi_1[/itex] is orthogonal which means it = 0
Number three, I can't get step 3. I put the equations as follows:
[tex]
c_1 + c_2d = 0
c_1^2 + c_2^2 + 2 dc_1c_2 = 1
c_1 = -c_2d[/tex]
therefore
[tex]
(-c_2d)^2 + c_2^2 + 2d(-c_2d)c_2 = 1
(-c_2d)^2 + c_2^2 - 2(c_2d)^2 = 1
[/tex]
And then I can go no further.
Number four, I don't understand how
[tex]
\int (\phi_1 + \phi_2)*(c_1\phi_1 + c_2\phi_2)d\tau [/tex]

simplifies to
[tex]
(c_1 + c_2)(1+d)
[/tex]
Number five, what do they mean by 2 and 5 gives
[tex]
\frac{(\phi_1 - \phi_2)}{\sqrt{2-2d)}}
[/tex]
As you can see I'm real clueless with regards to this stuff. I've got a private tutor lined up but I won't be able to meet with him until sometime next week.
 
Last edited by a moderator:

Answers and Replies

  • #2
128
0
From the first sentence of the problem: "[itex]\phi_1[/itex] and [itex]\phi_2[/itex] are normalized eigenfunctions..."

What is the definition of normalized? Specifically, how does the fact that [itex]\phi_1[/itex] and [itex]\phi_2[/itex] are normalized relate to the integrals [itex]\int \phi_1^*\phi_1 d\tau[/itex] and [itex]\int \phi_2^*\phi_2 d\tau[/itex]?

Answer that question and that should tell you everything you need to know to work through the derivation.
 
  • #3
267
1
Normalized means = 1. I already knew that. Still clueless.
 
  • #4
128
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What, specifically, is equal to 1?

Edit: Hint - Your statement in the original post that [itex]\phi_1^* \phi_1 = 1[/itex] is not quite correct. What is the correct equation?
 
Last edited:
  • #5
267
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[tex]\phi_1 \phi_2[/tex]
 
  • #6
128
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Not true. Nothing in the problem set up indicates [itex]\phi_1 \phi_2 =1[/itex] in general. The correct relation between the two is [itex]\int \phi_1^* \phi_2 d\tau = d[/itex].

So, I think I'm seeing several issues here that are causing you problems. First, you need to be a lot more careful about the difference between [itex]\phi[/itex] and [itex]\phi^*[/itex]. The two are not the same, and which terms have the * and which terms don't matters.

Second, you need to go back and review the definition of normalization. I think you have assumed that it means something that it doesn't mean, and that's why you can't evaluate the integrals above.

Third, slow down, take a bit more time, and don't skip any steps when you are working through the problem. Also, when replying on this thread, be as specific as you can; it'll make it easier to diagnose where your misunderstanding is coming from.

To work through each of the integrals in the derivation above, you need to know three different quantities:

[itex]\int \phi_1^* \phi_1 d\tau = ?[/itex]
[itex]\int \phi_2^* \phi_2 d\tau = ?[/itex]
[itex]\int \phi_1^* \phi_2 d\tau = ?[/itex]

Each of the integrals in the derivation can be broken up into pieces containing those three terms.
 
  • #7
267
1
I think I'll just wait until I meet with that private tutor. My problems are just so overwhelming. I think maybe I zipped through the calc, linear algebra and DE prerequisites too quickly. Maybe I'm trying QM without sufficient knowledge.
 
  • #8
128
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Actually, the math in this particular derivation is quite simple, once you have all of your definitions straight. That's why I was asking you about what normalization meant. I don't think this particular derivation is beyond you, I just think you need to slow down, take your time with it, and review the basics.

As another hint: for a normalized (1d) wave function in the position representation, [itex]\phi(x)[/itex], the probability of finding the particle between the points [itex]x_1[/itex] and [itex]x_2[/itex] is given by [itex]P[x_1<x<x_2]=\int_{x_1}^{x_2} \phi^*(x) \phi(x) dx [/itex].

If I take [itex]x_1[/itex] to [itex]-\infty[/itex] and [itex]x_2[/itex] to [itex]\infty[/itex], what's the probability of finding the particle between [itex]x_1[/itex] and [itex]x_2[/itex]? That is,

[itex]P[-\infty<x<\infty]=\int_{-\infty}^{\infty} \phi^*(x) \phi(x) dx = ?[/itex]

Don't try to solve the integral, just use common sense. What's the probability of finding the particle if we search the entirety of the real line?
 
  • #9
267
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as for the odds of finding a particle between negative infinity and plus infinity it has to be 1, nothing can exist beyond the edge of infinity since there is no edge to infinity. I certainly hope the math is not beyond me.
 
  • #10
128
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Exactly. [itex]P[-\infty<x<\infty]=\int_{-\infty}^{\infty} \phi^*(x) \phi(x) dx = 1[/itex]. That is the definition of what it means for a wave function to be normalized.

So, what does that statement imply for the three integrals needed for this problem?

[itex]\int \phi_1^* \phi_1 d\tau = ?[/itex]
[itex]\int \phi_2^* \phi_2 d\tau = ?[/itex]
[itex]\int \phi_1^* \phi_2 d\tau = ?[/itex]
 
  • #11
267
1
I'm guessing they equal 1.
 
  • #12
128
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Not all of them. :) One of the three is given to you in the problem. The other two equal 1.
 
  • #13
267
1
[itex]\int \phi_1^* \phi_2d\tau = 0[/itex] because it's orthogonal, right?
 
  • #14
128
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Nope. Actually, that's the point of the problem. [itex]\phi_1[/itex] and [itex]\phi_2[/itex] are not orthogonal, but rather integrate to a constant, [itex]\int \phi_1^* \phi_2 d\tau = d[/itex]. The question is, given that the are not orthogonal, how would you go about constructing orthogonal functions from them.

That is, you are trying to find some [itex]\bar{\phi}_2 \equiv c_1 \phi_1 + c_2 \phi_2[/itex] such that [itex]\int \phi_1^* \bar{\phi}_2 d\tau = 0[/itex].
 
  • #15
267
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I have to leave now and I will return to this problem after about 20 hours have passed. I appreciate you helping me.
 

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