Energy of a superposition state. A particle in an infinite...

Energy of a superposition state. A particle in an infinite square well extending between $x=0$ and $x=L$ has the wave function $$ \Psi(x, t)=A\left(2 \sin \frac{\pi x}{L} e^{-l E, r i n}+\sin \frac{2 \pi x}{L} e^{-i \boldsymbol{h}_2 z(t)}\right) $$ where $E_n=n^2 h^2 / 8 m L^2$. (a) Putting $t=0$ for simplicity, find the value of the normalization factor $A$. (b) If a measurement of the energy is made, what are the possible results of the measurement, and what is the probability associated with each? (c) Using the results of (b) deduce the average energy and express it as a multiple of the energy $E_1$ of the lowest eigenstate. (d) The result of (c) is identical with the expectation value of $E$ (denoted $\langle E\rangle)$ for this state. A procedure for calculating such expectation values in general is based on the fact that, for a pure eigenstate of the energy, the wave function is of the form, $\Psi(x, t)=$ n $\Psi(x) e^{-i E R i n}$, which yields the identity $$ i \hbar \frac{\partial \Psi}{\partial t}=E \Psi $$ Clearly, in this case, $\int \Psi^*$ (ih $\left.a / \partial t\right) \Psi d x=E \int \Psi^* \Psi d x=E$. An extension of this to a state involving an arbitrary superposition of energies suggests the following formula for calculating expectation values of $E$ : $$ \langle E\rangle=\int_{\text {all } x} \Psi^*(i \hbar \partial / \partial t) \Psi d x $$ This procedure is in fact correct. By applying it to the particular wave function of this exercise, verify that the value of $\langle E\rangle$ is identical with the average energy found in (c).

Energy of a superposition state. A particle in an infinite square well extending between $x=0$ and $x=L$ has the wave function
$$
\Psi(x, t)=A\left(2 \sin \frac{\pi x}{L} e^{-l E, r i n}+\sin \frac{2 \pi x}{L} e^{-i \boldsymbol{h}_2 z(t)}\right)
$$
where $E_n=n^2 h^2 / 8 m L^2$.
(a) Putting $t=0$ for simplicity, find the value of the normalization factor $A$.
(b) If a measurement of the energy is made, what are the possible results of the measurement, and what is the probability associated with each?
(c) Using the results of (b) deduce the average energy and express it as a multiple of the energy $E_1$ of the lowest eigenstate.
(d) The result of (c) is identical with the expectation value of $E$ (denoted $\langle E\rangle)$ for this state. A procedure for calculating such expectation values in general is based on the fact that, for a pure eigenstate of the energy, the wave function is of the form, $\Psi(x, t)=$ n $\Psi(x) e^{-i E R i n}$, which yields the identity
$$
i \hbar \frac{\partial \Psi}{\partial t}=E \Psi
$$

Clearly, in this case, $\int \Psi^*$ (ih $\left.a / \partial t\right) \Psi d x=E \int \Psi^* \Psi d x=E$. An extension of this to a state involving an arbitrary superposition of energies suggests the following formula for calculating expectation values of $E$ :
$$
\langle E\rangle=\int_{\text {all } x} \Psi^*(i \hbar \partial / \partial t) \Psi d x
$$

This procedure is in fact correct. By applying it to the particular wave function of this exercise, verify that the value of $\langle E\rangle$ is identical with the average energy found in (c).

Key Concepts

Quantum Superposition

Quantum superposition refers to the principle that a quantum system can exist in a combination of two or more eigenstates simultaneously. In the context of energy measurements, a superposition state implies that the system does not have a single, definite energy before measurement but rather a set of possible energy eigenvalues, each with its associated probability determined by the coefficients in the superposition.

Normalization of Wavefunctions

Normalization is the requirement that the total probability of finding a particle somewhere in space is 1. For any wavefunction, this means that the integral of the absolute square of the wavefunction over the complete space must equal unity. This process ensures that the coefficients of the eigenstates are correctly scaled to be interpreted as probability amplitudes.

Energy Eigenstates and Measurements

In problems such as the infinite square well, the allowed energy levels are quantized, and the corresponding energy eigenstates are the solutions to the Schrödinger equation with specific boundary conditions. When a measurement of energy is performed on a state that is a superposition of these eigenstates, the possible outcomes are the eigenvalues corresponding to each eigenstate, with the measurement probabilities given by the square modulus of the coefficients in the superposition.

Expectation Value in Quantum Mechanics

The expectation value of an observable, such as energy, is the statistical average of the outcomes of many measurements on a quantum system. It can be calculated either by summing the products of each eigenvalue and its corresponding probability or by applying the operator to the state and integrating over space. This method confirms the predictions of quantum mechanics and shows the consistency between the operator formalism and the probabilistic interpretation of measurement outcomes.

Time Evolution and Phase Factors

Time evolution in quantum mechanics is governed by the Schrödinger equation, leading to time-dependent phase factors for each energy eigenstate. These phase factors, typically expressed as exponential terms (e.g., exp(-iEt/?)), do not affect the probability density of a state, but they are essential in preserving the overall coherence of the superposition and in determining interference effects between different states.

Transcript

00:01 In this question, we have a particle in an infinite square wall with this initial wave function.

00:09 Si x 0 is equal to a times x x plus psi 2x.

00:20 So for infinite square well, infinite square well problem.

00:36 So the diagram looks like this, a from 0 to a.

00:42 And the potential is infinity.

00:47 And then, so the re -function, side 1x is 2 over a, sine pi x over a, and then side 2x is root 2 over a, sine 2 pi x over a.

01:08 Okay, we have e1 equals to pi square, hbars square over 2m a square and then you have e2 equals to uh 4 pi square hbar square divide by 2m a square okay so um first in there are five parts in the question okay so in part a you want to find a the normalization constant okay so we'll use the normalization condition which states that the integral from negative infinity to infinity mod big side x zero square the x is equal to 1 okay so so what you are going to have is mod a square okay side 1 plus side 2 star times side 1 plus side 2 okay and the x and then this is equal to 1 and then you are going to have mod side 1 square plus mod side 2 square plus um side 1 star side 2 plus side 2 star side 1 and d x it goes to 1 okay since all the functions are real so um and then we have our all the energy eigen functions are normalized so this is equals to one and we have mod side 2 square the x is also equals to 1 and then side 1 and side 2 are orthogonal to each other so we have side 1 side 2 the x is equal to 0 and similarly inside two side one okay so the above the line above we'll get two times mod a square is equal to one which means that mod a is equal to 1 over root 2 okay so this is our normalization constant for part a okay in pride b you want to find the way function in time and also the probability density at time t okay so um the so we are given this as the initial way function okay and we want to find the way function at time t okay what we need to do is just to attach the face sector into the negative i e1 t over h bar and then to each term, k for side 2 will just be negative i .e 2t over h bar.

04:47 Okay, and then we define omega as e1 over h bar, so which is pi square h bar over 2m a square.

05:00 So we can simplify the side xt to become side 1 x e to the minus i omega t plus and then e2 is four times of e1 so the omega will be four times minus i for omega t okay so this is the we function at time t okay then if you want to find the probability density okay which is mod square okay so so yeah, so just record, just remember that sine 1 x is root 2 over a, sine pi x over a.

06:00 A, side 2x is root 2 over a, sine 2 pi x over a.

06:09 Okay, yeah, so i'm not putting the exact functions here yet, but if you want to substitute that we can just substitute.

06:19 So if you do the mod square then you have half.

06:27 So just okay complete.

06:33 Okay, so the mod square will be side star, right? okay, so the 1 over 2 will become half.

06:47 Okay, and then you have si 1x.

06:51 Yeah, because of the space, the spatial way function is real so i'm not going to put a star but the time base factor from minus i becomes i and then side 2x into the i for omega t so this is the side star and then the side without the star will be you can just copy okay and then then you just multiply.

07:39 So you have side 1 square.

07:42 So this term multiply with this terms, the exponential term cancel out.

07:48 And then you also have the side 2 square as well.

07:56 These two terms, there's no time dependence.

08:00 And then you have side 1, side 2, e to the minus i 3 omega and then you also have the side one side two, e to the i 3 omega t.

08:17 Okay, then you continue the algebra.

08:31 So here you can pull out the side 1 and side 2.

08:35 You have e to the i 3 omega t plus e to the minus i 3 omega t.

08:41 Okay, this can be rewritten as 2 % using the oil formula cosine 3 omega t okay and if you want to substitute the functions inside here so you have half times so there's a 2 over a okay outside and then sine square pi x over a plus sine square 2x over a plus plus 2xxx sine 2 pi x over a, cosine 3 omega t and then you can simplify further so this is the we function the probability density of the function at time t okay go to parsi, pasi, you want to find the expectation value of position.

10:42 Okay, so expectation value of position is, according to the definition, siza, x, sye, and since x doesn't operate, x can be moved everywhere, so you actually get this x times mod, side x t star that's i square mod square okay so then you just need to copy what you have in the previous question okay to here okay go from zero to a x times one over a okay sine square pi x over a plus sine square two pi x over a plus 2, sine, pi x over a, sine 2 pi x over a, cosine 3 omega t, and then b x, okay.

11:57 So here there are three integrals that we need to do.

12:01 Okay, i'm just going to do them one by one.

12:05 It makes it easier.

12:07 Okay, so 0 to a x, sine square, x over a the x this gives us a square over 2 this is also the same for x sine square 2 pi x over a x okay and then the third integral which is 0 to a x here sign high x over a 2 pi x over a, cosine 3 omega t, the x.

12:55 Then if you use, so the cosine two omega t doesn't participate in the integration is a constant in this case...

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