Sandrasa
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Homework Statement
Find the eigenvector of the annhilation operator a.
Homework Equations
a|n\rangle = \sqrt{n}|{n-1}\rangle
The Attempt at a Solution
Try to show this for an arbitrary wavefunction:
|V\rangle = \sum_{n=1}^\infty c_{n}|n\rangle
a|V\rangle = a\sum_{n=1}^\infty c_{n}|n\rangle = \sum_{n=1}^\infty \sqrt{n} c_{n}|n-1\rangle
Define a new constant, k = n - 1, and put this in:
- > \sum_{k=0}^\infty \sqrt{k + 1} c_{k + 1}|k\rangle = \sum_{n=0}^\infty \sqrt{n + 1} c_{n + 1}|n\rangle
This was a trick I found in my notes from a lecture, but I do not quite understand how you can say that k = n in the last part. Is it just because k and n are an arbitrary notation for the basis set? Or is it due to the fact that since the sum goes to infity, the difference of 1 does not matter (since I really defined k to be n - 1)?
When I continue with this, I get:
a( c_{1}|1\rangle + c_{2}|2\rangle + c_{3}|3\rangle + ...) = c( \sqrt{1}c_{1}|0\rangle\ + \sqrt{2}c_{2}|1\rangle + \sqrt{3}c_{3}|2\rangle + \sqrt{4}c_{4}|3\rangle)
I am not really sure where to go next. I tried setting the coeffecient before each basisvector equal, like this:
c\sqrt{1}c_{1} = 0
c_{1} = \sqrt{2}c_{2}c
c_{2} = \sqrt{3}c_{3}c
c_{3} = \sqrt{4}c_{4}c
The problem her is that since the first part does not contain the vector 0|\rangle, either c or c_{1} has to be zero in the first equation, and then everything becomes zero, but I know that the annihilation operator has eigenstates.