00:01
So in this problem we have a quantum harmonic oscillator of frequency omega.
00:05
In the state, psi is equal to 1 over root 3 times the zero state minus root 2 times the one state.
00:15
And for part a, if we conduct a measurement on this state, we need to know which energy eigenvalues we could get and with which probability.
00:30
So here, the possible energy eigenvalues are either e0, which is equal to h -bar omega over 2 with probability b0, which is equal to 1 third, or energy e1, which is equal to e1, which is equal to 3 h -bar.
01:11
Omega over 2 with probability b1 and this is equal to two thirds this is for part a now for part b we need to get the expectation value of the possession operator and the expectation value of the possession operator square so here we need to define the possession operator x in terms of the creation and annihilation operators so here x x hat is equal to square root of h bar over 2 m omega where here m is the mass of the harmonic oscillator and omega is the angular frequency of the harmonic oscillator and here h bar is a planks constant and this is multiplied by a plus which is the creation creation operator plus a minus which is the annihilation operator so here the expectation value of x is equal to the dot product between the state psi and the operator x as follows.
02:27
And this is equal to one -third times square root of h bar over 2m omega multiplied by the pro of the state psi.
02:46
Which is 0 minus root 2 1 this is multiplied by a plus plus a minus and this is multiplied by by the cat operators which is here 0 minus root 2 and here this is equal to 1 3rd times the root of h bar over 2m omega multiplied by the pro state which is 0 minus root 2 1 and now we act with the creation operator on this state so here a plus over 0 which would get the state 1 and here a plus over state 1 which would give root 2 multiplied by root 2 times the state 2 and now we again act with the annihilation operator on this state so a minus over the state 0 would give 0 and here a minus over the state 1 would give 0 0.
04:34
So now we conduct the dot product between the prostate and the katt state.
04:41
So here the dot product between the zero state and one would give 0 and the dot product between 0 and 2 would give 0 but the dot product between the zero state or the pro zero state and the ket 0 state would give 1.
04:57
So this is equal to 1 third times hbar over to m omega multiplied by minus root 2 and here again the only living term here is minus root 2 multiplied by the dot product between the bar state 1 and the cat state 1 which would give minus root 2 and this is equal to minus 2 over 3 multiplied by the square root 3 multiplied by the square root of h bar over m omega and this is the expectation value of the possession operator now we need we need to get the expectation value of the position operator squared so here the position operator squared is equal to h bar over to m omega multiplied by a plus squared plus a minus squared plus a plus a minus plus a minus a plus because here the commutator between the creation and annihilation operator is not the zero so here a plus and a minus will not commute so now the expectation value of x squared is equal to one -third multiplied by hbar over 2 m omega multiplied by the prostate which is 0 minus root 2 1 multiplied by a plus squared plus a minus a minus plus a minus plus a plus a minus over the cat state which is 0 minus root 2 1 and this is equal to h bar over 6m omega multiplied by the prostate which is 0 minus root 2 1 multiplied by here a plus squared over the 0 state would give root 2 multiplied by and a plus squared over the 0 state would give root 2 multiplied by state 2 and a plus squared over the 1 state would give minus root 2 multiplied by root 2 multiplied by root 2 multiplied by root 3 and here the state would go from 1 to 3.
07:58
Now the a minus squared over the 0 state would give 0 and the a minus squared over the state 1 would give 0.
08:12
So the effect of a minus squared over this whole state is zero.
08:19
Now we would act with a plus a minus over this state.
08:24
So a plus a minus over the zero state would give zero.
08:30
And a plus a minus over the state one.
08:35
So the a minus operator over the state one would give the zero state.
08:40
And the a plus operator over the zero state would give the one state.
08:46
So here this is equal to minus root 2 times 1 the state 1.
08:52
And finally for a minus a plus the effect of a plus over the zero state would give the first state and the effect of a minus over the first state would give again the zero state.
09:08
So this is here plus 0.
09:13
Now the effect of a plus over the first state would give the zero state and sorry, it would give the second state multiplied by root 2.
09:25
So here this is minus root 2 multiplied by root 2.
09:31
And the effect of a minus over the state 2 would give minus would give minus would give again root 2 multiplied by the state 1 so here we now conducted the dot product between the prostate and the kit state so here the only dot product that would live with the prostate 0 is this state so here this is equal to h bar over 6m omega multiplied by 1 from here from the top product between 0 and 0 and for the pro 1 state the only living term is these two terms so here plus 2 from here from this top product plus root 2 power 4 which is and this is equal to 7 over 6 hbar over m omega and this is the expectation value of the position operator squared.
10:58
This is for part b, now for part c, we need to get the expectation value of the momentum operator.
11:09
So we first start by defining the momentum operator in terms of the creation and annihilation operators...