For the orbital angular-momentum operator 𝐋=𝐱 ×𝐩, derive (3.218), that is ⟨𝐱^'|Lz| α⟩=-i ħ(∂)/(∂

step 1 So in this question we have find some products. Now we have an explanation. So there is some pro

For the orbital angular-momentum operator 𝐋=𝐱 ×𝐩, derive...

For the orbital angular-momentum operator $\mathbf{L}=\mathbf{x} \times \mathbf{p}$, derive (3.218), that is $$ \left\langle\mathbf{x}^{\prime}\left|L_{z}\right| \alpha\right\rangle=-i \hbar \frac{\partial}{\partial \phi}\left\langle\mathbf{x}^{\prime} \mid \alpha\right\rangle $$ by using the standard spherical coordinate transformation $$ x^{\prime}=r \cos \phi \sin \theta, \quad y^{\prime}=r \sin \phi \sin \theta, \quad z^{\prime}=r \cos \theta . $$

For the orbital angular-momentum operator $\mathbf{L}=\mathbf{x} \times \mathbf{p}$, derive (3.218), that is
$$
\left\langle\mathbf{x}^{\prime}\left|L_{z}\right| \alpha\right\rangle=-i \hbar \frac{\partial}{\partial \phi}\left\langle\mathbf{x}^{\prime} \mid \alpha\right\rangle
$$
by using the standard spherical coordinate transformation
$$
x^{\prime}=r \cos \phi \sin \theta, \quad y^{\prime}=r \sin \phi \sin \theta, \quad z^{\prime}=r \cos \theta .
$$

Key Concepts

Orbital Angular Momentum Operator

This concept involves defining the angular momentum as the cross product of the position and momentum vectors (L = x × p) in classical mechanics and then promoting it to a quantum operator. It is essential for understanding how rotational motion is quantized and how angular momentum operators act on quantum states.

Momentum Operator as a Differential Operator

In quantum mechanics, the momentum operator is represented by the differential operator -i?? in the position basis. This representation is used to convert classical expressions into differential operators that act on wavefunctions, which is a key step in deriving the action of angular momentum operators.

Spherical Coordinate Transformation

Spherical coordinates (r, ?, ?) are used to exploit the symmetry of problems with rotational invariance. The transformation from Cartesian coordinates to spherical coordinates allows the angular momentum operator, particularly its z-component, to be expressed in terms of the azimuthal angle ?, leading to the derivative -i?(?/??) acting on the wavefunction.

Representation of Operators in the Coordinate Basis

This concept refers to expressing quantum operators as differential operators when acting on wavefunctions in a given basis. In the context of the problem, it involves representing Lz as a differential operator that generates rotations about the z-axis, hence connecting it to the derivative with respect to ?.

Transcript

00:01 So in this question we have find some products.

00:06 Now we have an explanation.

00:16 So there is some products like this.

00:19 Let's we've always starting j, vector j, which is equal to vector l plus vector s.

00:38 Vector l plus vector s.

00:44 So it is reactive j square which is equal to l squared plus s square, s square, s square plus two, l vector dot s vector, rift, l vector dot s vector, which is equal to half j square minus l square, minus s square, minus x square.

02:30 So now we haven't further explanation like this l vector minus l vector dot s vector which is equal to hcaf square over two so the large bracket is there i into i into i plus one into minus l l plus 1 minus s into s minus s minus s minus 1 s plus 1 so we can also write this l dot s vector which is equal to double mode l and again double mode s vector cost theta so we can say that cost theta which is equal to l vector dot s vector over cap square root l into l plus 1 dot root s into s plus 1.

05:35 So we can also say that j into j plus v minus l plus minus s into s plus 1 over 2 into l plus 1 0 into l plus 1 dot again root into s plus 1 so now we have putting p vector b 1 over 2 so 1 over 2 we have found this so which is equal to theta cause inverse over into 1 by 2 bracket 3 over 2 minus 2 minus 1 over 2 into 3 over 2 so the bracket close over 2 dot under root 1 over 2 into 3 over 2 it close so we can write this cause inverse over minus 100 to 2 over 3 let it close which is equal to 145 degree this is the first result of t 1 over 2 so now we have second results t 3 word so now we have putting the values so we know that theta is equal to cause inverse over 3 by 2 into 5 by 2 into minus 1 over 2 into 3 by 2.

11:29 So over 2 under root 2 into 3 by 2...

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