Wavefunction expansion coefficients

In summary, Daniel is summarizing the content of the previous conversation. Daniel is stating that the lowest possible energy that can be measured is the integer of first nonzero coefficient in the expansion \psi = \Sigma \limits_{n=1}^{\infty} a_n \phi_n, where \phi_n = \sqrt{\frac{2}{a}}\sin(\frac{n\pi x}{a}) are the basis functions given in eq (4.15) from the book. Daniel also states that the two states represented by psi1 and psi2 have different a_n^2 values, indicating that they are physically different. Patrick is asking for clarification on a mistake that Daniel is making
  • #1
nd
11
0
I'm working in Liboff, 4e, QM, page 114, problem 4.35.
An electron in a 1-D box with walls at x= 0,a is in the state [tex]\psi(x) = A[/tex] for [tex]x\in (0,a/2)[/tex] and [tex]\psi(x) = -A[/tex] for [tex]x\in (a/2,a)[/tex]. What is the lowest possible energy that can be measured?

From my understanding, the answer to this question will be the integer of first nonzero coefficient in the expansion [tex]\psi = \Sigma \limits_{n=1}^{\infty} a_n \phi_n[/tex], where [tex]\phi_n = \sqrt{\frac{2}{a}}\sin(\frac{n\pi x}{a})[/tex] are the basis functions given in eq (4.15) from the book (the eigenstates for the 1D box Hamiltonian). I do this and I get [tex]a_n = \frac{\sqrt{2}}{n\pi}(1+\cos(n\pi) - 2\cos(n\pi/2))[/tex]. Now correct me if I'm wrong, but is it not true that [tex]\psi(x)=A[/tex] for [tex]x\in (0,a)[/tex] represents the same state since only the square of the wavefunction is given significance? In that case, however, I get [tex]a_n=\frac{\sqrt{2}}{n\pi}(1-\cos(n\pi))[/tex]. It is my understanding that [tex]a_n^2[/tex] represents the probability of measuring the particle to be in the state [tex]\phi_n[/tex]. But in these two cases, we will get different [tex]a_n^2[/tex] indicating that the two states are physically different.

Can anyone point out my mistake?
 
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  • #2
Advice:read more on Fourier series.U could see post #5 from here https://www.physicsforums.com/showthread.php?p=571801#post571801 which deals exactly with a function of type [itex] \psi [/itex] in your problem.

Your problem should reduce to the question:does the fundamental state [itex] \phi_{1} [/itex]-which has the lowest possible energy-have a nonzero probability ...?

U'll need to compute the Fourier coefficients of that wavefunction (whose sq.moduli give probabilities),so that's why i advised you to read.

Daniel.
 
  • #3
nd said:
Now correct me if I'm wrong, but is it not true that [tex]\psi(x)=A[/tex] for [tex]x\in (0,a)[/tex] represents the same state since only the square of the wavefunction is given significance?

No, that is not true. It is not because |psi1|^2 = |psi2|^2 that psi1 and psi2 represent the same physical states. In order for two descriptions A and B to represent the same physical state, you have to have that ALL POSSIBLE measurements cannot make any difference. |psi1| = |psi2| only means that all possible POSITION measurements cannot make a difference. But a momentum measurement can, for instance.

cheers,
Patrick.
 

What are "wavefunction expansion coefficients"?

Wavefunction expansion coefficients are mathematical quantities used to describe the probability amplitudes of a quantum mechanical system. They are used in the mathematical representation of wavefunctions, which describe the physical properties of particles at the quantum level.

How are wavefunction expansion coefficients calculated?

Wavefunction expansion coefficients are typically calculated using mathematical methods such as the Schrödinger equation, which describes the time evolution of quantum systems. These coefficients can also be determined experimentally through measurements of various physical properties of the quantum system.

What do wavefunction expansion coefficients tell us about a quantum system?

Wavefunction expansion coefficients provide information about the probability of finding a particle in a particular state or location within a quantum system. They can also give insight into the energy levels and other physical properties of the system.

What is the significance of the expansion coefficients in quantum mechanics?

In quantum mechanics, the wavefunction expansion coefficients play a crucial role in understanding the behavior and properties of particles at the quantum level. They allow us to make predictions about the behavior of particles and the outcomes of experiments, and have been fundamental in the development of the field of quantum mechanics.

How do wavefunction expansion coefficients relate to the superposition principle?

The superposition principle states that a quantum system can exist in multiple states simultaneously, and the wavefunction expansion coefficients represent the probability amplitudes of these states. In other words, the expansion coefficients allow us to calculate the probability of a particle being in a specific state when it is in a state of superposition.

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