(e) Show that the momentum operators p??, p?? and p?z all...

(e) Show that the momentum operators p??, p?? and p?_z all commute with each other. (f) The angular momentum operator is defined as L? = r? × p? that have components L?? = y?p?_z ? z?p??, L?? = z?p?? ? x?p?_z, L?_z = x?p?? ? y?p?? Calculate the commutator [L??, L??], [L??, L?_z] and [L?_z, L??]. And what pattern do you find? (g) Define L?² = L??² + L??² + L?_z². Compute [L?_z, L?²]. (h) In class, we have defined Hermitian operators. Is the angular momentum operator Hermitian?

Transcript

00:01 Hi, this is a question based on commutators.

00:04 So, here we have the first commutator.

00:07 So, px ,vy which are the linear momentum operator.

00:11 So, we have to prove that px ,py is 0.

00:14 So, let us use poisson's bracket that is px ,py which is equal to dou px by dou x into dou x or into dou py by dou px minus dou px by dou py comma dou py by dou x plus dou px by dou y into dou py by dou y minus dou px by dou py.

00:50 So, this is dou py by dou py divided by dou px.

01:03 So, dou px this is dou py by into dou px divided by dou y.

01:11 So, the differentiation of this term is 0 minus 0 plus 0 minus 0.

01:18 Therefore, we have the commutator px ,py which is equal to py ,pz which is equal to pz ,px is equal to 0.

01:31 Then this is the first problem.

01:35 In the second question, we have to show the commutator lx ,ly which is the angular momentum commutator.

01:41 So, we have lx is equal to ypz minus zpy and ly is equal to zpx minus xpz and lz is equal to xpy minus ypx.

01:56 Then we have lx ,ly is equal to so this is ypz minus zpy comma ly is zpx minus xpz.

02:11 So, now we have to take the common.

02:14 So, in this step ypz is taken outside and this is written and then zpy is taken and this is written.

02:20 Then it each split it each two term is splitted to four terms.

02:25 So, first these two term then again term.

02:27 So, first these two term then again these two term.

02:30 Now we have so taking y outside we have pz comma zpx plus again y is taken inside then pz is written on the right side.

02:43 Similarly, this is done for other three elements.

02:46 So, here we can see that it is pz comma pz.

02:49 So, this commutation is 0 and again we have z comma z commutation.

02:58 So, here it is also 0.

03:00 So, this is the remaining term and this is the remaining term.

03:03 So, we have to write this as zpy comma xpz and in the next part py is taken outside that is plus z comma xpz into py.

03:17 So, these are the four terms.

03:19 Now here we have the commutation of so here we have the commutation of so it is 0 and here we have the commutation of py comma pz.

03:32 So, this is 0 and here we have pz comma px.

03:36 So, this is 0...

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