Coefficients of a Superposition of State Vectors?

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Homework Statement


(f) At t = 0, a particle of mass m trapped in an infinite square well of width L is in a superposition of the first excited state and the fifth excited state, ψs(x, 0) = A (3φ1(x) − 2iφ5(x)) , where the φn(x) are correctly-normalized energy eigenstates with energies En. Which of the following values of A give a properly normalized wavefunction? ##\frac { 1 }{ \sqrt { 5 } }## ##\frac { i }{ 5 } ## ##\frac { i }{ \sqrt { 13 } }## ##\frac { 1 }{ 13 } ## None of these

Homework Equations


A normalized wave function: ##A^{ 2 }\sum _{ i }{ a^{ * }_i a_i } =1##.

The Attempt at a Solution


Since, ##\sum _{ i }{ a^{ * }_{ i }a_{ i } } =9-4=5##, ##A=\frac { 1 }{ \sqrt { 5 } } ##.

Does this look correct?

Thnx,
Chris
 

Answers and Replies

  • #2
blue_leaf77
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That does not look correct:wink:.
The basic definition of linear combination (or superposition) is the sum of all weighted basis states, that means if there is a negative sign in front of a particular term, that sign must belong to the expansion coefficient of that corresponding term.
 
  • #3
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I have a negative sign there. a1*a1=9 because it is real. (-2i)*=2i, so (-2i)*x(2i)=-4 for index 5. I am not sure I follow your concern.

Chris
 
  • #4
blue_leaf77
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(-2i)*x(2i)
Shouldn't it be (-2i)*x(-2i) = 4 ?
 
  • #5
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That would be (-2i)*x(-2i)=(2i)x(-2i)=4. Oh, ok. Simple mistake. I should have wrote it out explicitly.

It is now ##A=\frac{1}{\sqrt{13}}## which is not one of the choices. That always makes me feel insecure :D

Chris
 
  • #6
blue_leaf77
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The coefficient ##A## is in general complex, thus instead of ##A^2## you should write it ##|A|^2##. Among the choices, which will give you ##|A|^2 = \frac{1}{13}##?
 
  • #7
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But it is asking for ##A##, not ##{|A|}^2##. Is it not?

Thanks,
Chris
 
  • #8
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I emailed the prof who wrote this test. Yes, I see that A is in general complex, so i/sqrt(13) would satisfy ##{|A|}^2=\frac{1}{13}##. Man, I am so rusty. A lot of brain sweat for what seemed to be such an easy question at first.

Thanks,
Chris
 
  • #9
blue_leaf77
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so i/sqrt(13) would satisfy ##{|A|}^2=\frac{1}{13}##.
Yes that's right.
 
  • #10
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On part g) of this question:
(g) Given the wavefunction ψs, what is the probability of measuring the energy to be E6 at t = 0? Circle one:

The correct answer would be zero, correct? That is because the coefficient in front of a state vector at level 6 would be zero. It is not included in the original wave function correct?

Thanks,
Chris
 
  • #11
blue_leaf77
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Yes the probability should be zero.
 
  • #12
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This one should be zero too, no? The probability of finding a particle and a particular point would would be infinitesimal. You have to integrate over some range on the axis to have a finite probability.

(h) Given the wavefunction ψs, what is the probability density of finding the particle in the middle of the box at time t = 0?

0 3 9 9 Undetermined

Thanks for helping me. I have no answers to this material, and I want to make sure I have these concepts down solid.

Thanks,
Chris
 
  • #13
George Jones
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Are "probability" and "probability density" the same thing?
 
  • #14
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Good point. Density is per unit length in a 1d scenario. Let me go back and work on it some more.

Chris
 
  • #15
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I would be taking the derivative of ##|\psi|^2## and evaluating it at L/2, correct? I get zero because there are sine functions in both terms after finding mod squared, and taking the derivative. I am using his version of the infinite square well, where x1=0 and x2=L. See: http://ocw.mit.edu/courses/physics/8-04-quantum-physics-i-spring-2013/exams/MIT8_04S13_exam1.pdf

My ##|\psi|^2## is: ##|\psi |^{ 2 }=\frac { 1 }{ 13 } \left[ 9\frac { 2 }{ L } { sin }^{ 2 }(\frac { 2\pi }{ L } x)+4\frac { 2 }{ L } { sin }^{ 2 }(\frac { 6\pi }{ L } x) \right] ##

Thanks,
Chris
 
  • #16
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I would be taking the derivative of ##|\psi|^2## and evaluating it at L/2, correct? I get zero because there are sine functions in both terms after finding mod squared, and taking the derivative. I am using his version of the infinite square well, where x1=0 and x2=L. See: http://ocw.mit.edu/courses/physics/8-04-quantum-physics-i-spring-2013/exams/MIT8_04S13_exam1.pdf

My ##|\psi|^2## is: ##|\psi |^{ 2 }=\frac { 1 }{ 13 } \left[ 9\frac { 2 }{ L } { sin }^{ 2 }(\frac { 2\pi }{ L } x)+4\frac { 2 }{ L } { sin }^{ 2 }(\frac { 6\pi }{ L } x) \right] ##

Thanks,
Chris

Never, mind. I don't know why I was thinking of taking the derivative. Mod squared is the probability density. The units are correct for that too. However, this mod square evaluated at x=L/2 should still be zero because the sine functions evaluated at multiples of ##\pi## are equal to zero.

Correct?

Thanks,
Chris
 

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