I No problem at all, happy to help! Have a great weekend as well.

[Frost_Xue]

I was reading Dirac's "The principle of QM" and bit of confused.
In equation (21) does the number 1 in the sum or outside of the sum? If out side how come the sum over r is 1? X function is the quotient when the factors diveded by single one factor. for example if a*b*c*d then X_b is a*c*d
i

 

[Truecrimson]

1 is outside the sum, and it should be the identity operator, not a number (but this is a very common abuse of notation).

An observable (Hermitian operator) $\xi$ can be decomposed as a sum $\xi = \sum_r c_r |r\rangle \langle r|$ of its orthonormal eigenvectors $\{ |r \rangle \}$ written as projection operators $|r\rangle \langle r|$, and eigenvalues $c_r$.

What Dirac is doing is the converse. How can we write each projection operator as a function of $\xi$ ? Let's first look at $$\chi_r (\xi) = \prod_{q \neq r} ( \xi - c_q \hat{1}), $$ where $\hat{1}$ is the identity operator. Each term in the product annihilates $| q \rangle$ i.e. $(\xi - c_q \hat{1} )|q \rangle = 0$. So it is easy to see in the basis in which $\xi$ is diagonalized that $\chi_r (\xi)$ has a bunch of zeros in the diagonal. In fact, the only vector not annihilated by $\chi_r (\xi)$ is $|r\rangle.$ Moreover, $$\chi_r (\xi) = \prod_{q \neq r} (c_r - c_q)|r\rangle \langle r|.$$ Note that $|r\rangle \langle r|$ could be inside or outside the product. It doesn't matter, because $(|r\rangle \langle r|)^2 = |r\rangle \langle r|$. Therefore, \begin{aligned} \frac{\chi_r (\xi) }{\chi_r (c_r)} &= |r \rangle \langle r| \\ \sum_r \frac{\chi_r (\xi) }{\chi_r (c_r)} &= \hat{1}, \end{aligned} where I used the fact that $\{ |r\rangle \}$ forms a resolution of the identity in the last line.

Perhaps an easier way to say all of this is that, whenever we have the resolution of the identity $\sum_r |r\rangle \langle r| = \hat{1}$, we can expand any vector in the $|r \rangle$ basis by inserting the identity $$ |P\rangle = \hat{1} |P\rangle = \sum_r \langle r|P \rangle |r\rangle. $$

 

[Frost_Xue]

1 is outside the sum, and it should be the identity operator, not a number (but this is a very common abuse of notation).

An observable (Hermitian operator) $\xi$ can be decomposed as a sum $\xi = \sum_r c_r |r\rangle \langle r|$ of its orthonormal eigenvectors $\{ |r \rangle \}$ written as projection operators $|r\rangle \langle r|$, and eigenvalues $c_r$.

What Dirac is doing is the converse. How can we write each projection operator as a function of $\xi$ ? Let's first look at $$\chi_r (\xi) = \prod_{q \neq r} ( \xi - c_q \hat{1}), $$ where $\hat{1}$ is the identity operator. Each term in the product annihilates $| q \rangle$ i.e. $(\xi - c_q \hat{1} )|q \rangle = 0$. So it is easy to see in the basis in which $\xi$ is diagonalized that $\chi_r (\xi)$ has a bunch of zeros in the diagonal. In fact, the only vector not annihilated by $\chi_r (\xi)$ is $|r\rangle.$ Moreover, $$\chi_r (\xi) = \prod_{q \neq r} (c_r - c_q)|r\rangle \langle r|.$$ Note that $|r\rangle \langle r|$ could be inside or outside the product. It doesn't matter, because $(|r\rangle \langle r|)^2 = |r\rangle \langle r|$. Therefore, \begin{aligned} \frac{\chi_r (\xi) }{\chi_r (c_r)} &= |r \rangle \langle r| \\ \sum_r \frac{\chi_r (\xi) }{\chi_r (c_r)} &= \hat{1}, \end{aligned} where I used the fact that $\{ |r\rangle \}$ forms a resolution of the identity in the last line.

Perhaps an easier way to say all of this is that, whenever we have the resolution of the identity $\sum_r |r\rangle \langle r| = \hat{1}$, we can expand any vector in the $|r \rangle$ basis by inserting the identity $$ |P\rangle = \hat{1} |P\rangle = \sum_r \langle r|P \rangle |r\rangle. $$

That was great answer. Thank you so much for the help and time. Sorry to bother you with such trivial question. have a good weekend.