- #1
Taylor_1989
- 402
- 14
Homework Statement
Q: A particle is in a linear superposition of two states with energies: ##E_0##& ##E_1##
$$|\phi>=A|E_0>+\frac{A}{\sqrt{3-\epsilon}}|E_1>$$
(a) What is the value of A ? Express your answer as a function of ##\epsilon##
(b) Use your expression to plot A vs ##\epsilon##
(c) Show by a diagram the location of the state $$|\phi>$$ on the Hilbert space, using ##E_0##& ##E_1## as the basis vectors for ##\epsilon= 0,1## & ##2##
Homework Equations
The Attempt at a Solution
I am new to this notation so, if I have used the wrong notation or there is a more easier way please tell me. All so I assumed that ##E_0=(i+0j)##&##E_1=(0i+j)##
a) To calculate the value of A I did the following:
$$P(E_0)=<E_0|\phi>=(A^*<E_0|+0)(A|E_0>+\frac{A}{\sqrt{3-\epsilon}}|E_1>)=A* \times A \times 1=|A|^2 $$
$$P(E_1)=<E_1|\phi>= (0+(\frac{A}{\sqrt{3-\epsilon}})^*)(A|E_0>+\frac{A}{\sqrt{3-\epsilon}}|E_1>))=(\frac{A}{\sqrt{3-\epsilon}})* \times \frac{A}{\sqrt{3-\epsilon}}=\frac{A^2}{3-\epsilon}$$
$$P(E_0)+P(E_1)=1$$
After simplifying I got the following:
$$A=\sqrt{\frac{3-\epsilon}{4-\epsilon}}$$
b) I used desomos the bit with in the the shaded area is what I make it.
c) this is where my problem lies. I am slightly confused because for different values of epsilion the one gives the probabilities do not add up to 1 so before I plot them do I have to workout an normalisation constant before I plot them. As I said I am very new and this is an intro course I am on, so I am learning as I am going.
Any advice would be appreciated.
I have also attched photo of actual question just incase I have missed anything out.