Confusion Regarding a Spectral Decomposition

[ARoyC]

Hi. I am not being able to understand how we are getting the following spectral decomposition. It would be great if someone can explain it to me. Thank you in advance.

 

[vanhees71]

It's simply a non-sensical equation. On the one side you write down a matrix, depicting matrix elements of an operator and on the other the operator itself. Correct is
$$\hat{A}=v_3 |0 \rangle \langle 0| + (v_1-\mathrm{i} v_2) |0 \rangle \langle 1| + (v_1+\mathrm{i} v_2) |1 \rangle \langle 0| - v_3 |1 \rangle \langle 1|.$$
The matrix elements in your matrix are then taken with respect to the basis ##(|0 \rangle,|1 \rangle)##.
$$(A_{jk})=\langle j|\hat{A}|k \rangle, \quad j,k \in \{0,1 \}.$$
To see this, simply use ##\langle j|k \rangle=\delta_{jk}##. Then you get, e.g.,
$$A_{01}=\langle 0|\hat{A}|1 \rangle=v_1-\mathrm{i} v_2.$$

 

[ARoyC]

It's simply a non-sensical equation. On the one side you write down a matrix, depicting matrix elements of an operator and on the other the operator itself. Correct is
$$\hat{A}=v_3 |0 \rangle \langle 0| + (v_1-\mathrm{i} v_2) |0 \rangle \langle 1| + (v_1+\mathrm{i} v_2) |1 \rangle \langle 0| - v_3 |1 \rangle \langle 1|.$$
The matrix elements in your matrix are then taken with respect to the basis ##(|0 \rangle,|1 \rangle)##.
$$(A_{jk})=\langle j|\hat{A}|k \rangle, \quad j,k \in \{0,1 \}.$$
To see this, simply use ##\langle j|k \rangle=\delta_{jk}##. Then you get, e.g.,
$$A_{01}=\langle 0|\hat{A}|1 \rangle=v_1-\mathrm{i} v_2.$$

Oh! Then we can go to the LHS of the equation from the RHS. Can't we do the reverse?

 

[vanhees71]

Sure:
$$\hat{A}=\sum_{j,k} |j \rangle \langle j|\hat{A}|k \rangle \langle k| = \sum_{jk} A_{jk} |j \rangle \langle k|.$$
The mapping from operators to matrix elements with respect to a complete orthonormal system is one-to-one. As very many formal manipulations in QT, it's just using the completeness relation,
$$\sum_j |j \rangle \langle j|=\hat{1}.$$

 

[ARoyC]

Sure:
$$\hat{A}=\sum_{j,k} |j \rangle \langle j|\hat{A}|k \rangle \langle k| = \sum_{jk} A_{jk} |j \rangle \langle k|.$$
The mapping from operators to matrix elements with respect to a complete orthonormal system is one-to-one. As very many formal manipulations in QT, it's just using the completeness relation,
$$\sum_j |j \rangle \langle j|=\hat{1}.$$

How are we getting the very first equality that is A = Σ|j><j|A|k><k| ?

 

[Haborix]

$$\hat{A}=\hat{1}\hat{A}\hat{1}=\left(\sum_{j} |j \rangle \langle j|\right)\hat{A}\left(\sum_{k} |k \rangle \langle k|\right)=\sum_{j,k} |j \rangle \langle j|\hat{A}|k \rangle \langle k| = \sum_{jk} A_{jk} |j \rangle \langle k|.$$

 

[ARoyC]

$$\hat{A}=\hat{1}\hat{A}\hat{1}=\left(\sum_{j} |j \rangle \langle j|\right)\hat{A}\left(\sum_{k} |k \rangle \langle k|\right)=\sum_{j,k} |j \rangle \langle j|\hat{A}|k \rangle \langle k| = \sum_{jk} A_{jk} |j \rangle \langle k|.$$

Oh, okay, thanks a lot!