Coefficients of a particle in Linear superposition (QM)

[Johnahh]

Homework Statement

a particle is in a linear superposition of two states with energy $$E_0 \ and\ E_1$$
$$|\phi> = A|E_0> + \frac{A}{(3-\epsilon)^{1/2}}|E_1>$$

where:
$$A \ > \ 0, \ 0\ <\ \epsilon \ <\ 3$$

What is the value of A expressed as a function of epsilon

Homework Equations

$$P(E_0) \ +\ P(E_1) = 1\\ P(E_0) = |<E_0|\phi>|^2\\ P(E_1) = |<E_1|\phi>|^2$$

The Attempt at a Solution

My attempt was to normalise the function to find a value for A in terms of epsilon.
$$<E_0|\phi> = A\\ <E_1|\phi> = \frac{A}{\sqrt{(3-\epsilon)}} \\ |<E_1|\phi>|^2 + |<E_0|\phi>|^2 = A^2 + \frac{A^2}{(3-\epsilon)} = 1\\ A^2(1 + \frac{1}{3-\epsilon}) = 1\\ 4A^2 -\epsilon A^2 = 3-\epsilon\\ A^2 = \frac{3-\epsilon}{4-\epsilon}\\ A = \sqrt\frac{3-\epsilon}{4-\epsilon}\\$$

But this does not give me a value of 1 when i put it back in? I'm unsure where I am going wrong - only had 2 QM lectures so my knowledge is limited.

 

[blue_leaf77]

But this does not give me a value of 1 when i put it back in?

I think your answer is correct.

 

[Johnahh]

I think your answer is correct.

Wow, I was being an absolute idiot, and inputting A back in incorrectly, for some reason i was just putting the value of A back into
$$<E_0|\phi> and <E_1|\phi>$$ expecting to get one.

Thankyou lol