Intro to probability density QM

[Taylor_1989]

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.

 

[andrewkirk]

I think they just want you to plot and label the three points with coordinates ##(A,A/\sqrt{3-\varepsilon})##, on the number plane, for three different values of ##\varepsilon##, in each case replacing the value of ##A## by the value obtained from the formula you derived above.

 

[Taylor_1989]

Ah I see, can I ask what the point in this, as I understand then that these plots would not be a solution to the equation. So when would it be appropriate to find the normalising coefficient

 

[DrClaude]

Ah I see, can I ask what the point in this, as I understand then that these plots would not be a solution to the equation.

What equation are you talking about?

So when would it be appropriate to find the normalising coefficient

Always? One normally wants to work with normalized states.

The question is strangely worded. I agree with @andrewkirk that they want you to plot ##| \phi \rangle## as a real vector on a Cartesian graph with ##|E_0\rangle## and ##|E_1\rangle## as the axes, but this doesn't really represent "the location of the state ##| \phi \rangle## in the Hilbert space."

 

[Taylor_1989]

I am talking about the equation that need to be plotted so: ##|\phi>## with my obtained value of A. I just don’t know what the question is really trying to show me. I mean I understand the the probability should always up to one, but in this case they do not for the certain value of epsilon, so why do it? Also what is Hilbert space exactly, how would I draw this on Hilbert space even if it possible.

 

[DrClaude]

I mean I understand the the probability should always up to one, but in this case they do not for the certain value of epsilon, so why do it?

I don't understand why you say that the probabilities don't add up to 1. Using your equation for ##A##, what is ##|\phi\rangle## for ##\epsilon=0,1,2##?

Also what is Hilbert space exactly, how would I draw this on Hilbert space even if it possible.

The Hilbert space is an abstract vector space, see https://en.wikipedia.org/wiki/Hilbert_space

 

[Taylor_1989]

Can you please ignore that they do added up to 1, iv just been thinking about this problem so much and trying to understand it that, I kept doing basic arithmetic errors. Sorry for the confusion.