One-dimensional free packet states: I. At t=0, the wave function of a free particle of mass m is

One-dimensional free packet states: I. At t=0, the wave...

Key Concepts

Momentum Operator

The momentum operator in quantum mechanics is represented in the position basis by a derivative operator, typically -i? times the derivative with respect to position. This operator's role is to extract momentum information from the wave function and is used to compute the expectation value of the momentum.

Position Operator

The position operator in quantum mechanics acts by multiplying the wave function by the position variable. It represents the observable associated with the particle's position and is used to calculate quantities like the expectation value of the position.

Kinetic Energy Operator

The kinetic energy operator in quantum mechanics is derived from the momentum operator and is given by the square of the momentum operator divided by twice the mass of the particle. It is used to calculate the expectation value of the kinetic energy, reflecting the energy associated with the particle's motion.

Normalization

Normalization is the requirement that the total probability of finding a quantum particle somewhere in space is equal to one. This condition ensures that the wave function is physically meaningful and is achieved by adjusting the constant factors in the wave function so that the integral of its squared modulus over all space equals one.

Wave Function

In quantum mechanics, the wave function is a complex-valued function that contains all the information about the state of a particle or system. It is central to the theory because its squared modulus gives the probability density of finding the particle in a particular region of space.

Expectation Value

The expectation value of an observable in quantum mechanics is the average value of the measurement outcome that one would obtain from an ensemble of identically prepared systems. It is computed by taking the integral of the product of the wave function, the operator corresponding to the observable, and the complex conjugate of the wave function over the entire space.

Free Particle States

Free particle states refer to quantum states where the particle is not under the influence of any external potential, meaning its evolution is solely determined by its kinetic energy. These states are typically described by wave packets or plane waves and serve as fundamental examples in quantum mechanics for studying basic principles such as dispersion and uncertainty.

Transcript

00:03 This is the problem number 22 and the problem is based on the gaussian wave packet.

00:10 A free particle has the initial wave function that is si x0 is equal to a e power minus a x square where a capital a and the a are constant.

00:22 First question is the normalize x0.

00:28 That means the way function for the gaussian way packet at t is equal to zero is given that is si x0 is equal to a e power minus a x square first question normalize x0 that may we have to find out this capital a here this a and a are positive real constant to find out this value of a we have to use the normalization condition and which is minus infinitive to plus infinity, psi x0, psi x0 and dx is equal to 1.

01:32 From here this will be because it is the real constant, this will be a square and here this will be e power minus 2a x square and dx.

01:50 Is equal to 1 and the integration of this will be a square and the value of this integration is square root of pi by suppose this is alpha e power minus elph x square that is pi upon alpha that is pi upon 2a is equal to 1 hence the value of a is equal to 2a upon pi power 1 by 4 and this is the value a plus the normalized b function that t is equal to 0 is x 0 is equal to 2 a upon pi is the normalization constant 1 upon 4 e power minus a x square now in the second part of the question, b part, find xx t.

03:42 And in order to find out this integration, we have to use the integration that is minus infinity to plus infinity e power minus alpha square plus beta x tx is equal to square out of pi upon alpha, e power minus beta square by 4 alpha.

04:09 We have to use this integration in order to find out of the psi xt.

04:17 For this, we have to find out the momentum way function.

04:26 That is let us assume this is pi k is equal to 1 upon 2 pi, pi, minus infinity to plus infinity given way function that is a e power minus a x square and e power minus i k x and d x now substituted the value of the a that is 2 a upon pi 2 a upon pi power 1 by 4 minus infinitive plus in three e power minus a x square plus i k x and d x here alpha is equal to a and beta is equal to i k so that means the value of this integration is 2 a upon phi is e power minus beta square by 4 alpha into pi by alpha where alpha is equal to a and beta is equal to i k now substitute this value here then si sorry fi k is equal to 1 upon 2 pi 2 a upon pi and this beta square that is pi sqa is management it is minus k square e power n minus k square upon 4a and this pi by alpha is a from here this pi to pi can see this pi this pie can see and if you solve this then you will get rearrange this the power of 2a and the pi then you will get is equal to 1 upon 2 pi a power 1 by 4 and e power minus k square by 4a.

07:42 Now the complete way function, x t, is defined as 1 upon 2 pi minus infinitive, 2 plus infinitive and phi of k with a superposition of these waves and e power i k x and e power minus i en t by a h bar or here we will write on the e e t by h bar and d k and we know the value of the e e t by h bar and dk and we know the value of the e e e t by h bar and decay and we know the value of to p square by 2m and p is equal to h cross k square by 2m from here we can find e by h bar is equal to h bar k square by 2m now substitute this value here 1 upon 2 pi and the value this 5 k 5 k is 1 upon 2 pi a power 1 by 1 by 1.

09:22 4 and e power minus k square e.

09:35 Power minus k square by 4a, e power i k x and this value is e power minus i and e by hb is h k square by 2m and the t and the integration with respect to dk.

10:18 Okay, now let's rearrange with the power of e.

10:27 It is 1 upon 2 pi, 1 upon 2 pi, a power 1 5 4, and this value will be e power minus 1 by 4a.

10:59 And here k square that means this will be minus is common that is plus i h bar t by 2m and k square k square and this value will be minus i k x and dk so again this integration is of of the form of the e power minus alpha k square plus beta k so this time the value of the beta is ix and alpha is this now the value of this integration will be that is si x t is equal to this whole constant 1 upon 2 pi is square root 1 upon 2 pi a 1 by 4 and into this value e power minus beta square and beta is minus i x and this value will be beta square that is i squared minus minus one and this is already minus that is one and this value will be x square so this will be minus x square upon 4 alpha and 4 alpha is this from 1 by 4a plus i h bar t by 2m into square root of pi by alpha pi by this whole value that is 1 by 4a plus i h bar t by now if you take form 1 by 4 common here then it will 4 to 4 cancel and then it will go up and this this whole solved and finally you will get 2a by pi power 1 by 4 e power minus a x square divided by so this value will be 1 plus and here 2 i h bar a t by m this is the way function if i assume this something some cost rate let u is equal to 2 i h bar at t by m then this value will be that is si of x t is equal to 2 a upon pi power 1 by 4 e bar minus a x square divided by 1 plus i sorry we have to remove this one that is only 2 h bar 80 by m then it will be 1 plus i u this is the wave function at time greater than 0.

15:40 Now in this c part we have to find out the mod of psi xt square, si xt square is equal to psi x t and si xt.

16:05 Let's find the complex conjugate of this one.

16:09 Here it is 2 a upon pi.

16:15 For this real this is real then it will be a root and e power minus a x square upon one minus i u e power minus a x square one plus i u let's solve this one power minus a x square and it will be one plus i u plus one minus i u it will be 1 plus u square and this pencil and this cancel now this value is x t is equal 2 a upon 5 e power minus a x square that is 2 a x square upon 1 plus u square now we have to express this result in terms of the omega where omega is here it is we have to express the result in terms of the omega where omega is square root of a plus 1 plus i u no one plus u square sorry this is one plus one plus u square.

18:29 So here a upon 1 plus a square that is square that is it will be a omega square.

18:38 So hence, si x t is equal to 2 a upon pi, e power minus 2 omega square and x squared.

19:15 One term i think missing and 2 upon pi okay in si x t one term is missing this term is missing here that means one more term is here that is this part is missing here it will be that is 1 by 4 e common that is 1 plus 2 i sq by m and this one term is missing here and it is 1 plus 2 i hatt 80 by let's write this term here so one term is also missing in this term that it will be square root of 1 plus i u and here one term is also missing okay so here we can write one term that is 1 upon i u and one term is 1 minus i u and in square root of this that is 1 plus u square here we can write here that is 1 plus u square and now this term here that is 1 plus 1 u square that is 1 plus 1 upon u square so this term will be a upon omega square a upon omega square...

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