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Coefficients of a particle in Linear superposition (QM)

  • Thread starter Johnahh
  • Start date
88
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1. Homework Statement
a particle is in a linear superposition of two states with energy [tex] E_0 \ and\ E_1 [/tex]
[tex] |\phi> = A|E_0> + \frac{A}{(3-\epsilon)^{1/2}}|E_1>[/tex]

where:
[tex] A \ > \ 0, \ 0\ <\ \epsilon \ <\ 3[/tex]

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

2. Homework Equations
[tex]P(E_0) \ +\ P(E_1) = 1\\
P(E_0) = |<E_0|\phi>|^2\\
P(E_1) = |<E_1|\phi>|^2 [/tex]

3. The Attempt at a Solution
My attempt was to normalise the function to find a value for A in terms of epsilon.
[tex] <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}\\ [/tex]

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.
 
Last edited:
88
0
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
[tex]<E_0|\phi> and <E_1|\phi>[/tex] expecting to get one.

Thankyou lol
 

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