- #1
Kara386
- 208
- 2
In my lecture notes, it says that
##\left\langle l \right| A_{nm} \left| \psi \right\rangle = \sum_{n,m} A_{nm} \left\langle m \right| \left|\psi \right\rangle \left\langle l \right| \left| n \right\rangle##
##=\sum_{n,m} A_{nm}\left\langle m \right| \left| \psi \right\rangle \delta_{ln}##
##= \sum_{m} A_{lm} \left\langle m \right| \left|\psi \right\rangle##
And the notes also state that a basis is a set of vectors {##\left| n \right\rangle##}, where n = 1,2,3,...,N. I'm not sure what that means, but presumably this is what ##\left| m \right\rangle## and ##\left| n \right\rangle## are in the summations above? If ##\left| n \right\rangle = 1,2,3...N## and these bra-ket things are like vectors, wouldn't that mean that ##\left| n \right\rangle =
\left(
\begin{array}{c}
1\\
2\\
...\\
N\\
\end{array}
\right)
##? I know that's not true. I think it's meant to be a column vector of unit vectors like the i,j,k used in Cartesian. I don't see how a column vector of unit vectors equates to {##\left| n \right\rangle##}, where n = 1,2,3,...,N.
Essentially, I don't understand the manipulation above at all. I don't know where the delta comes from or why you can write an operator ##A## as ##A = \sum_{n,m} A_{nm} \left| n \right\rangle \left\langle m \right|##, which I presume is what's happened here? I think it's meant to be obvious because there's no explanation, but it isn't obvious to me.
I'd really appreciate any help - it's quite a long question! :)
##\left\langle l \right| A_{nm} \left| \psi \right\rangle = \sum_{n,m} A_{nm} \left\langle m \right| \left|\psi \right\rangle \left\langle l \right| \left| n \right\rangle##
##=\sum_{n,m} A_{nm}\left\langle m \right| \left| \psi \right\rangle \delta_{ln}##
##= \sum_{m} A_{lm} \left\langle m \right| \left|\psi \right\rangle##
And the notes also state that a basis is a set of vectors {##\left| n \right\rangle##}, where n = 1,2,3,...,N. I'm not sure what that means, but presumably this is what ##\left| m \right\rangle## and ##\left| n \right\rangle## are in the summations above? If ##\left| n \right\rangle = 1,2,3...N## and these bra-ket things are like vectors, wouldn't that mean that ##\left| n \right\rangle =
\left(
\begin{array}{c}
1\\
2\\
...\\
N\\
\end{array}
\right)
##? I know that's not true. I think it's meant to be a column vector of unit vectors like the i,j,k used in Cartesian. I don't see how a column vector of unit vectors equates to {##\left| n \right\rangle##}, where n = 1,2,3,...,N.
Essentially, I don't understand the manipulation above at all. I don't know where the delta comes from or why you can write an operator ##A## as ##A = \sum_{n,m} A_{nm} \left| n \right\rangle \left\langle m \right|##, which I presume is what's happened here? I think it's meant to be obvious because there's no explanation, but it isn't obvious to me.
I'd really appreciate any help - it's quite a long question! :)
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