Show that the exchange integral ∫d^3 𝐱 d^3 𝐱^' (Ψ1^*(𝐱) Ψ2(𝐱) Ψ2^*(𝐱^') Ψ1(𝐱^'))/(|𝐱-𝐱^'|) is re

Show that the exchange integral ∫d^3 𝐱 d^3 𝐱^' (Ψ1^*(𝐱)...

Key Concepts

Hermitian Operators

Hermitian operators are those whose matrix elements satisfy the condition that the complex conjugate of an element is equal to the matrix element with the bra and ket interchanged. This property ensures that the eigenvalues and expectation values computed using these operators are real. In the context of the exchange integral, the Coulomb kernel (1/|x - x'|) acts as a Hermitian operator, which is a key factor in demonstrating that the integral is real.

Symmetry Under Coordinate Exchange

The exchange integral involves integration over two spatial coordinates, and the integrand has a form that is symmetric under the exchange of these coordinates. Specifically, the Coulomb kernel depends only on the relative distance between the two points, making it invariant under swapping x and x'. This symmetry allows one to change the integration variables in a way that reveals the equality of the integral with its complex conjugate, thereby ensuring the integral is real.

Complex Conjugation and Integral Properties

Complex conjugation plays a crucial role when analyzing integrals involving complex functions. By applying the complex conjugation operation and relabeling the integration variables using the symmetry of the integrand, one can show that the exchange integral is equal to its complex conjugate. This property directly implies that the value of the integral is real, since a complex number that is equal to its own complex conjugate must be real.

Transcript

00:01 Okay, in this question we have to construct the second stationary state which is si to x a t so in order to solve this question we need to have three things the number one is the initial wave function is given as m omega out of pi h cut raised to the power 1 out of 4 e to the power minus m omega out of 2 h cut x squared and the second thing we need to have is the ladder operator which is equal to 1 out of under 2 h cut m omega into minus iota into momentum plus m omega x and this ladder operator is equal to a 1 out of 2 h cut m omega into minus h cut d upon d x a plus m omega x and the third thing we need to have is the n's stationary state is equal to 1 upon und root n factorial into a plus raised to the power n into sin not.

01:50 For the second stationary state we can write the above equation.

01:56 For second stationary state that is si 2 x we can write above equation as si 2 equal to 1 out of under 2 factorial into a plus square a sineat so to find out this side second stationary state let's first find out this term so we know that a plus into sin not which is equal to one out of under two h cut m omega into minus h cut partial derivative of x plus m omega x into m omega out of pi h cut whole square and not whole square is one out of four e raised to the power minus m omega out of two pi x a square so therefore we can write a plus equal to 1 upon under 2 h cut m omega into m omega out of pi h cut raised to the power 1 out of 4 into minus h cut into partial derivative of x e minus m omega out of 2 h cut x square plus m omega x e raised to the power minus m out of 2 h cut x square so this can also return as this can be written as 1 out of unroot 2 h cut m omega into m omega out of pi h cut raised to the power 1 out of 4 into e raised to the power minus m omega out of 2 h cut x square into minus m omega out of edge cut x into plus m omega into minus m omega out of h cut x into plus m omega into minus m omega out of 2 h cut x square so therefore we can write it one out of unroad 2 h cut m omega into m omega out of pi h cut raised to the power one out of four and this this term and this term are the same so they so therefore we can can write it a 2m omega x e raised to the power minus m omega out of 2 h cut x square so taking the second derivative which is a square sineat equal to is equal to 2m omega upon or 2 h cut m omega into as we have take it outside the parenthesis so that's why these two will get cancelled and we can write omega m omega out of pi h cut into rest to the power 1 out of 4 into minus h cut partial derivative of x plus m omega x into x into x in into a e m omega out of 2 h cut x square so therefore we can write it as 1 out of h cut into m omega out of pi h cut raised to the power 1 out of 4 into minus h cut e raise to the power minus m omega out of 2 h cut x square plus x e raised to the power minus m omega out of 2 h cut x square into minus m omega out of h cut x square plus m omega x square e raised to the power minus m omega two h cut x square so therefore now we can write it as 1 out of h cut into m omega out of pi h cut, raised to the power 1 out of 4 into h cut, e raised to the power minus m omega out of 2h cut, x square plus m omega, x square e raised to the power minus m omega out of 2 h squared x square so therefore a plus a square plus si not equal to 1 upon 2 victoria by putting in the the above equation we get m omega out of 5 h cut into 1 out of 4 into 2 and omega out of edge cut minus 1 e raised to the power minus m omega out of 2 edge cut x square so in the second part we have to find out in the second part we need to graph the three stationary states that is si not si 1 and and and si 2 so si not is equal to m omega out of pi hkut raised to the power 1 out of 4 e minus m omega out of 2 h cut so if we draw the graph of this function then we have a graph like this this is the x -axis and this is the sin not x so this this expression here this expression if we let k is equal to mhmuagout of 2 h card then this expression is e raised to per minus k x square so this expression shows that this entire expression is equal to 0 if we put infinity here so the graph is is equal to zero near the origin of and it shoots up and then it again equal to 0 when the value of x is large then this multiply this term and it become equal to 0 so if x is equal to minus infinity again the expression is equal to 0 and this is the graph of this a sine not and for si1 and for si1 this expression is equal to m omega out of pi h cut raised to the power one out of four unroot two m omega out of h cut multiply by e to the power minus m omega out of 2 h cut x square and this graph should be equal to like this initially this graph is this is x and this is minus infinity plus infinity this is minus infinity also in the last part of the question this part of the question we have to change we have to the orthogonality of the wave function so we can write it as minus infinity to infinity, si not, si 2 d x minus i 2 d x minus infinity to infinity sin not si 1 d x into minus infinity to infinity si 1, si 2 d x is equal to 0.

13:13 As these 2 wave function, which are si not and si 2 and and si 1, are equal to 0 because the wave function are orthogonal, therefore these two wave function are equal to 0...

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