Use the eigenfunction of Problem 24 to calculate the following expectation values, and comment o

Use the eigenfunction of Problem 24 to calculate the...

Key Concepts

Variance and Uncertainty

Variance, calculated from the expectation values of an operator and the operator squared, measures the spread or dispersion of measurement outcomes around the mean. In conjunction with the uncertainty principle, it articulates the fundamental limits to the precision with which pairs of observables, such as position and momentum, can be simultaneously known.

Position and Momentum Operators

The position (x) and momentum (p) operators are central to quantum mechanics. They represent the basic measurable properties of a quantum system, and their expectation values provide important information about the state’s spatial distribution and motion. Their mathematical representations and commutation relations form the foundation for many quantum phenomena.

Hermitian Operators

In quantum mechanics, observables are associated with Hermitian operators, which have the property of yielding real eigenvalues and ensuring orthogonality of eigenfunctions. This guarantees that any measurement of a physical quantity will result in a real number, making the operators physically meaningful.

Expectation Values

The expectation value is a fundamental concept representing the average outcome of a measurement of an observable when a system is in a certain quantum state. It is calculated using the integral of the product of the complex conjugate of the wavefunction, the corresponding operator, and the wavefunction itself, summing over all space.

Eigenfunctions

Eigenfunctions are specific solutions to the operator equations in quantum mechanics where an operator acting on an eigenfunction yields the same function multiplied by a constant (the eigenvalue). These functions form a complete set that can be used to represent any quantum state, making them vital for understanding measurable quantities and the dynamics of a system.

Transcript

00:01 Hello everyone so according to this given problem we need to use the wave function of the previous question this so in the previous question the way function si for n equal to 3 is written as root over 2 by a sine 3 pi x by a where a is the width of the infinite square well potential now in order to find the average value of x so the average value of x can be written as si star x psi dx and integrating over from zero to a now if we write it we will get 2 by a outside x here and this will become sine square 3 pi x by a right so this is the odd integral right because of the odd integral the integration value is zero so we can directly write the average value of x is equal to zero similarly for number so part number two we need to find the average value of p so average value of p can be find using the operator operator of p so operator of p is written as a minus ih cross d'll by d x x x so this will become 2 by a integration uh minus i h cross and uh after the si star is simply just a sign of 3 pi x by a and the differentiation of sign will give you 3 pi by a outside and a sign will become cause of 3 pi x by a and dx.

01:52 Now take the things outside so this will become 2 by a multiplied 3 pi by a and minus of ih cross and this is sign of 3 pi x by a and cost of 3 pi x by a d x and the integration again from 0 to a so this is the known integral formula in which we write integration of sign of an x plus cos of and sorry cos n x d x and integrating from the value over some 0 to let say 0 to pi right and the integration value is 0 right so again we will get the average value of p is 0 now for part number 3 we need to find the average value of x square so the average value of x square can be calculated using the same formula we just have to replace the value of x with x square so we will write x square sine square three pi x by a dx is equal to zero so this will become two by a i'm writing sine square three pi x by a in terms of cost so we know that the cost of two a x is written as 1 minus 2 sine square a x so we can use this identity so we will get 1 minus cost of 2 a x let me just scroll it up 1 minus cost 2 a x divided by 2 is equals to sine square a x the potential.

03:54 I'm just showing you in the formula only.

03:57 Right.

03:58 So this is integration 0 to a writing it over.

04:01 This is 1 by 2 minus cost of 2 times a is basically 3 pi by a.

04:08 This is the constant and x and again we will divide it with 2 and integrated with respect to dx.

04:18 Okay.

04:19 So when you integrate the first part you will get a by two minus now when you integrated here you will get a by six by outside and integration of course will give you sign so this will become sign 6 by a x and integrate from 0 to a right so this is 2 by a a by 2 minus so when you put the value of a here and zero so this will again will give you zero because sine and pi is zero and sine zero is also zero right so the average value of x square is just just a moment we have x square as well right so this is x square here and x square here okay so because of the x square we can't use this thing i just miss the x square so let me just do this again so this is 2 by a and integration of x square by 2 will give you x cube by 6 and integrating from 0 to a and minus integration x square x square let's take 1 by 2 outside so this is cost of 6 pi x by a d x and this is integrating from 0 to a as well right so 2 by a so this will become a cube by 6 minus 1 by a here and the integration 0 to a x square cost of 6x by a dot solid x okay now let me just do this integration only because this is we have to use integration by parts here so 0 to a x square so 0 to a x square cost of 6x by a d x so let's take x square x square first and integrating cost of 6x a so this will give you a by 6 pi 6 pi sine 6x by a and minus the differentiation of x square will give you 2 times x and a by 6 pi the integration of cost of 6x by a we'll use sign of 6 pi x by a dx now here we need to put the limits again 0 to a so when you put the limit a you will get a cube by 6 pi here and because of the sign part the sign 6 pi will also give you 0 and sine 0 also will give you 0.

07:20 So 0 minus 0.

07:21 So the first part is 0 minus.

07:24 So this will become a by 3 pi outside.

07:27 And again we have to do the integration of x sine 6xxa.

07:31 So let's take x in the differentiation part and integrating sine 1.

07:36 So we will get minus of a by 6 pi.

07:41 Cause of 6x by a right and again minus integration a differentiation of x will give you one and the integration of sine 6x will give you minus a 6 pi so i'm taking plus here and when you you have to do the integration again of course so you will get a by 6 pi whole square and sine square sorry sign of xxxxxxx x by a and again zero to a because of the sign part after putting the limits this term will again give you zero here we need to put the limit again so this part is zero minus a by three pi and let's put the limit so this will give you minus of a square by six pi and cost of six pi will give you one minus when you put zero so this term this the second part will become zero so this is just a cube and 18 pi square.

08:51 Okay...

Please give Ace some feedback