Continued from Quantum mechanics postulates- part 2
The expectation value or the average value of the reproduced measurements of any observable corresponding to the operator $\hat{Q}$ is defined as,
$$ \langle\hat{Q}\rangle = \int \psi^* \hat{Q} \psi dx $$
From Born's interpretation, the square of the amplitude of the coefficients of the eigen functions in the general expansion gives the probability of finding the particle at its particular eigenvalue. According to this, the expectation value is defined by taking the weighted average i.e. sum of all the possible eigen value with its corresponding probability.
In, bra-ket notation, the expectation value of Q at the state $ \psi$ is given by, $$ \langle \hat{Q}\rangle _{\psi} = \frac{\langle\psi\vert\hat{Q}\psi\rangle}{\langle\psi\vert\psi\rangle} $$
It is as same as, $$ \langle\hat{Q}\rangle = \langle\psi\hat{Q}\vert\psi\rangle = \sum_i \langle\psi\hat{Q}\vert\psi_i\rangle\langle\psi_i\vert\psi\rangle \\~\\ = \sum_i \lambda_i \langle\psi\vert\psi_i\rangle \langle\psi\vert\psi_i\rangle^* = \sum_i \lambda_i \vert\langle\psi\vert\psi_i\rangle\vert^2 = \sum_i \lambda_i \vert{c_i}\vert^2 $$
where I use the fact that $$ \sum_i \vert\psi_i\rangle\langle\psi_i\vert = I $$ and wave function is normalized to unity $$\langle\psi\vert\psi\rangle = 1 $$
We can also prove the above as follows, $$\frac{\langle\psi\vert\hat{Q}\left(\sum_i c_i \psi_i\right)\rangle}{\langle\psi\vert\psi\rangle} = \frac{\langle\sum_ic_i\psi_i\vert\sum_i c_i \lambda_i \psi_i\rangle}{\langle\psi\vert\psi\rangle} \,\,or\,\,\frac{\langle\psi\vert\sum_i\lambda_ic_i\psi_i\rangle}{\langle\psi\vert\psi\rangle}\\~\\ = \sum_i c_i c_i^*\lambda_i \frac{\langle \psi_i\vert\psi_i\rangle}{\langle\psi\vert\psi\rangle}\,\,or\,\,\sum_i\lambda_i \frac{\langle\psi\vert\sum_ic_i\psi_i\rangle}{\langle\psi\vert\psi\rangle} \\~\\ = \sum_i \lambda_i \vert{c_i}\vert^2 $$ which is the weighted average of all eigen values.
we used the fact $$\langle\psi\vert\psi_i\rangle = c^*_i\,\,\,\,or\,\,\,\,\langle\psi_i\vert\psi\rangle = c_i\\~\\\rightarrow\,\,\,\,\langle\psi\vert\sum_ic_i\psi_i\rangle = \sum_i c_i c^*_i$$ and the wave function is normalized to unity and eigen states corresponding to distinct eigenvalues are orthonormal, $$ \langle\psi_i\vert\psi_j\rangle = \delta_{ij}$$
Now, we arrived our final part of these postulates which involves Schrodinger equation given by, The time evolution of wave function is given by,$$ i\hbar\frac{\partial\vert\psi(t)\rangle}{\partial{t}} = H\vert\psi(t)\rangle $$ which can be represented in position basis as, $$i\hbar\frac{\partial\psi(x,t)}{\partial{t}} = \left[\frac{-\hbar^2}{2m} \frac{\partial^2}{\partial{x^2}} + V(x)\right] \psi(x,t) $$ where $$\psi(x,t) = \langle{x}\vert\psi(t)\rangle$$
which is the usual wave equation we encounter in most of our Quantum Mechanics problems.
The expectation value or the average value of the reproduced measurements of any observable corresponding to the operator $\hat{Q}$ is defined as,
$$ \langle\hat{Q}\rangle = \int \psi^* \hat{Q} \psi dx $$
From Born's interpretation, the square of the amplitude of the coefficients of the eigen functions in the general expansion gives the probability of finding the particle at its particular eigenvalue. According to this, the expectation value is defined by taking the weighted average i.e. sum of all the possible eigen value with its corresponding probability.
In, bra-ket notation, the expectation value of Q at the state $ \psi$ is given by, $$ \langle \hat{Q}\rangle _{\psi} = \frac{\langle\psi\vert\hat{Q}\psi\rangle}{\langle\psi\vert\psi\rangle} $$
It is as same as, $$ \langle\hat{Q}\rangle = \langle\psi\hat{Q}\vert\psi\rangle = \sum_i \langle\psi\hat{Q}\vert\psi_i\rangle\langle\psi_i\vert\psi\rangle \\~\\ = \sum_i \lambda_i \langle\psi\vert\psi_i\rangle \langle\psi\vert\psi_i\rangle^* = \sum_i \lambda_i \vert\langle\psi\vert\psi_i\rangle\vert^2 = \sum_i \lambda_i \vert{c_i}\vert^2 $$
where I use the fact that $$ \sum_i \vert\psi_i\rangle\langle\psi_i\vert = I $$ and wave function is normalized to unity $$\langle\psi\vert\psi\rangle = 1 $$
We can also prove the above as follows, $$\frac{\langle\psi\vert\hat{Q}\left(\sum_i c_i \psi_i\right)\rangle}{\langle\psi\vert\psi\rangle} = \frac{\langle\sum_ic_i\psi_i\vert\sum_i c_i \lambda_i \psi_i\rangle}{\langle\psi\vert\psi\rangle} \,\,or\,\,\frac{\langle\psi\vert\sum_i\lambda_ic_i\psi_i\rangle}{\langle\psi\vert\psi\rangle}\\~\\ = \sum_i c_i c_i^*\lambda_i \frac{\langle \psi_i\vert\psi_i\rangle}{\langle\psi\vert\psi\rangle}\,\,or\,\,\sum_i\lambda_i \frac{\langle\psi\vert\sum_ic_i\psi_i\rangle}{\langle\psi\vert\psi\rangle} \\~\\ = \sum_i \lambda_i \vert{c_i}\vert^2 $$ which is the weighted average of all eigen values.
we used the fact $$\langle\psi\vert\psi_i\rangle = c^*_i\,\,\,\,or\,\,\,\,\langle\psi_i\vert\psi\rangle = c_i\\~\\\rightarrow\,\,\,\,\langle\psi\vert\sum_ic_i\psi_i\rangle = \sum_i c_i c^*_i$$ and the wave function is normalized to unity and eigen states corresponding to distinct eigenvalues are orthonormal, $$ \langle\psi_i\vert\psi_j\rangle = \delta_{ij}$$
Now, we arrived our final part of these postulates which involves Schrodinger equation given by, The time evolution of wave function is given by,$$ i\hbar\frac{\partial\vert\psi(t)\rangle}{\partial{t}} = H\vert\psi(t)\rangle $$ which can be represented in position basis as, $$i\hbar\frac{\partial\psi(x,t)}{\partial{t}} = \left[\frac{-\hbar^2}{2m} \frac{\partial^2}{\partial{x^2}} + V(x)\right] \psi(x,t) $$ where $$\psi(x,t) = \langle{x}\vert\psi(t)\rangle$$
which is the usual wave equation we encounter in most of our Quantum Mechanics problems.
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