Question

3.5 Show that the operator*p is not Hermitian (where k and / are positive integers), but that the combination (x"d + pl#9/2 is Hermitian.

I've been told properties of commutators are the method of solving this, I'm just unsure of the implementation.

Properties such as: [AB, C] = A[B, C] + [A,C]B and    [A, B] = AB - BA   and   [x, p] = i * h(bar)    and   [p, x² ] = -2i(hbar)x

Answer 1

The property we can use here is that

and $[x,p_x]=i\hbar$ and $[p_x,x]=-i\hbar$

The property we can use here is that [A, B^n] = n [A, B] B^n - 1 and [x, P_x] = i tshe and [P_x, x] = -i tshe [x^k, p^l_k] = kx^k - 1 [x, p^l_x] = kx^k - 1 [x, p_x] l p^l - 1_x = i tshe k l x^k - 1 p^l - 1_x notequalto 0 thatis (1) Since it is not equal to zero so it is not Hermitian. Now solving for the other combination. 1/2 [x^k p^l_x + p^l_x x^k] = 1/2 [x^k p^l_x] + 1/2 [p^l_x, x^k] Now we can use the same formulation as above. The first term is the same as equation (1) multiplied by 1/2. And in second term x and px are interchanged. In that case, using the property mention above we will come out with a minus sign. Therefore implies i tshe kl x^k - 1 p^l - 1_x - i tshe kl p^l_x x^k - 1 = 0 Therefore it is Hermitian.

$[x^k,p_x^l]=kx^{k-1}[x,p_x^l]=kx^{k-1}[x,p_x]lp_x^{l-1}=i\hbar klx^{k-1}p_x^{l-1}\ne 0$ .................(1)

Since it is not equal to zero so it is not Hermitian.

Now solving for the other combination.

$\frac{1}{2}[x^kp_x^l+p_x^lx^k]=\frac{1}{2}[x^kp_x^l]+\frac{1}{2}[p_x^l,x^k]$

Now we can use the same formulation as above. The first term is the same as equation (1) multiplied by 1/2. And in second term x and px are interchanged. In that case, using the property mention above we will come out with a minus sign. Therefore

$\Rightarrow i\hbar klx^{k-1}p_x^{l-1}-i\hbar klp_x^{l-1}x^{k-1}=0$

Therefore it is Hermitian.