Postulates of Quantum Mechanics

Postulates of Quantum Mechanics

• All properties of a quantum system can be found with its state vector \(\left|\psi\right>\), which belongs to the complex Hilbert space
• Every observable physical property has its corresponding Hermittian operator. The eigenvalues of its operator are the possible outcome values when measuring the physical property $$M\left| \lambda_n \right> = \lambda_n \left|\lambda_n\right>.$$ The set of eigenvectors span the Hilbert space.
• The probability of measuring \(M\) to have the value \(\lambda_n\) of the quantum system in state vector \(\left|\psi\right>\) is given by $$P = \left| \left<\lambda_\middle|\psi\right> \right|^2$$
• If the result after measuring \(M\) is \(\lambda_n\), then the state vector of the system immediately after the measurment is \(\left|\lambda_n\right>\)
• The evolution of the state of the quantum system is governed by the Schrodinger equation $$H \left|\psi(t)\right> = i\hbar \frac{d\left|\psi(t)\right>}{dt}$$

Compactibility

An operator \(A\) is said to be compactible to an operator \(B\) if \(A\) commutes with \(B\), i.e. $$[A,B] = AB - BA = 0,$$ and incompactible if $$[A,B] \neq 0$$

Schwarz Inequality

Let \(a_n\) and \(b_n\) be two sets of sequences. The Schwarz inequality states that $$\left(\sum^n_i a^2_i\right)\left(\sum^n_i b^2_i\right) \geq \left(\sum^n_ia_ib_i\right)^2$$

Proof
Let S(n) be the statement $$\left(\sum^n_i a^2_i\right)\left(\sum^n_i b^2_i\right) \geq \left(\sum^n_ia_ib_i\right)^2.$$ When \(n=1\), $$L.S. = a^2_1 + b^2_1$$ while $$R.S. = (a_1b_1)^2,$$ so left hand side equals right hand side. S(1) is true. Assume S(k) is true, i.e. $$\left(\sum^k_i a^2_i\right)\left(\sum^k_i b^2_i\right) \geq \left(\sum^k_ia_ib_i\right)^2.$$ When \(n=k+1\), \begin{align} R.S. &= \left(\sum^{k+1}_ia_ib_i\right)^2\\ &= \left(\sum^k_ia_ib_i + a_{k+1}b_{k+1}\right)^2\\ &= \left(\sum^k_ia_ib_i\right)^2 + 2a_{k+1}b_{k+1}\left(\sum^k_{i}b_i^2\right)\left(\sum_ka^2_i\right) + a^2_{k+1}b^2_{k+1} \end{align} \begin{align} L.S. &= \left(\sum^{k+1}_i a^2_i\right)\left(\sum^{k+1}_i b^2_i\right) \\ &= \left(a^2_{k+1} + \sum^k_i a^2_i\right)\left(b^2_{k+1} + \sum^k_i b^2_i\right)\\ &= a^2_{k+1}b^2_{k+1} + a^2_{k+1}\sum^k_{i}b_i^2 + \sum_ka^2_ib^2_{k+1} + \left(\sum^k_i a^2_i\right)\left(\sum^k_i b^2_i\right)\\ &\geq a^2_{k+1}b^2_{k+1} + a^2_{k+1}\sum^k_{i}b_i^2 + b^2_{k+1}\sum_ka^2_i + \left(\sum^k_ia_ib_i\right)^2, \end{align} by S(k). Substitue right hand side into the inequality, \begin{align} L.S. &\geq a^2_{k+1}b^2_{k+1} + a^2_{k+1}\sum^k_{i}b_i^2 + b^2_{k+1}\sum_ka^2_i + \left(\sum^k_ia_ib_i\right)^2 \\ &= \left(\sum^{k+1}_ia_ib_i\right)^2 - 2a_{k+1}b_{k+1}\left(\sum^k_{i}b_i^2\right)\left(\sum_ka^2_i\right) + a^2_{k+1}\sum^k_{i}b_i^2 + b^2_{k+1}\sum_ka^2_i.\\ \end{align} Examining the last four terms, we have \begin{align} & 2a_{k+1}b_{k+1}\left(\sum^k_{i}b_i^2\right)\left(\sum_ka^2_i\right) + a^2_{k+1}\sum^k_{i}b_i^2 + b^2_{k+1}\sum_ka^2_i\\ &= \left(a_{k+1} - 2a_{k+1}b_{k+1}\frac{\sum^k_{i}a_ib_i}{\sum^k_ib_i^2}\right)\sum^k_ib^2_i + b^2_{k+1}\sum^k_ia^2_i\\ &=\left(a_{k+1}-b_{k+1}\frac{\sum^k_ia_ib_i}{\sum_ib^2_i}\right)^2\left(\sum^k_ib^2_i\right) - \left(\sum^k_ib^2_i\right)b^2_{k+1}\left(\frac{\sum^k_ia_ib_i}{\sum_ib^2_i}\right) + b^2_{k+1}\sum^k_ia^2_i\\ &= \left(a_{k+1}-b_{k+1}\frac{\sum^k_ia_ib_i}{\sum_ib^2_i}\right)^2\left(\sum^k_ib^2_i\right) + b^2_{k+1}\left(\frac{-\left(\sum^k_ia_ib_i\right)^2 + \sum^k_ia^2_i\sum^k_ib^2_i}{\sum^k_ib^2_i}\right), \end{align} in which the first term is non-negative as it is a product of two non-negative number, while the second term is also non-negative, by our assumption S(k) is true. Hence, it is a non-negative number. So, we conclude \begin{align} L.S. &= \left(\sum^{k+1}_i a^2_i\right)\left(\sum^{k+1}_i b^2_i\right) \\ &\geq \left(\sum^{k+1}_ia_ib_i\right)^2 - 2a_{k+1}b_{k+1}\left(\sum^k_{i}b_i^2\right)\left(\sum_ka^2_i\right) + a^2_{k+1}\sum^k_{i}b_i^2 + b^2_{k+1}\sum_ka^2_i\\ &\geq \left(\sum^{k+1}_ia_ib_i\right)^2. \end{align} Therefore, S(k+1) is also true. By mathematical induction, S(n) is true for all natural number n.

Notice that if we let \(\vec{a}=\sum^n_ia_i\vec{e}_i\) and \(\vec{b}=\sum^n_ib_i\vec{e}_i\) be two vectors of \(n\) dimensions and \(\vec{e}_i\) is the basis vectors, then \(\left(\sum^n_ia_ib_i\right)^2\) is the inner products of the two vectos and \(\left(\sum^k_i a^2_i\right)\), \(\left(\sum^k_i b^2_i\right)\) are the inner product with themselves. Then, in terms of vector representation, the Schwarz Inequality can be written as $$\left< f \middle| f \right>\left< g \middle| g \right> \geq |\left< f \middle| g \right> |^2.$$

Uncertainty Principle

Proof taken from Griffiths. The standard derivation of an observable \(A\) is given by \begin{align} \sigma^2_A+ &= \left< (\hat{A} - \left< A \right> )^2\right> \\ &= \left< \Psi \right| (\hat{A} - \left< A \right> )^2 \left| \Psi \right> \\ &= \left< (\hat{A} - \left< A \right> ) \Psi \middle| (\hat{A} - \left< A \right> )\Psi \right> \\ &= \left< f \middle| f \right>, \end{align} where \(f \equiv (\hat{A}- \left< A \right>)\Psi\). Similarly, for another observable \(B\), $$\sigma_B^2 = \left< g \middle| g\right>,$$ where \(g \equiv (\hat{B}- \left< B \right>)\Psi\). By Schwarz inequality, $$\sigma^2_A\sigma^2_B = \left< f \middle| f \right> \left< g \middle| g \right> \geq | \left< f \middle| g \right> |^2.$$ For any complex number \(z\), $$|z|^2 = [\text{Re}(z)]^2 + [\text{Im}(z)]^2 \geq [\text{Im}(z)]^2 = \left[\frac{1}{2i}(z-z^*)\right]^2.$$ This is also true for \(\left< f \middle| g \right>\), so $$\sigma^2_A\sigma^2_B \geq | \left< f \middle| g \right> |^2 \geq \left(\frac{1}{2i}[\left< f \middle| g \right> - \left< g \middle| f \right>]\right)^2.$$ Compute the first term on the right hand side inside the bracket, we have \begin{align} \left< f \middle| g \right> &= \left< (\hat{A} - \left< A \right> )\Psi \middle| (\hat{B} - \left< B \right> )\Psi\right> \\ &= \left< \Psi \right| (\hat{A} - \left< A \right> ) (\hat{B} - \left< B \right> )\left|\Psi\right>\\ &= \left< \Psi \right| (\hat{A}\hat{B} - \hat{A}\left< B \right> - \hat{B}\left< A \right> + \left< A \right> \left< B \right> ) \left|\Psi\right>\\ &= \left< \hat{A}\hat{B} \right> - \left< B \right>\left< A \right> - \left< A \right> \left< B \right> + \left< A \right> \left< B \right>\\ &= \left< \hat{A}\hat{B} \right> - \left< A \right> \left< B \right>, \end{align} where \(\left< A \right>\) and \(\left< B \right>\) are real numbers so they can be taken out from \(\left< \middle| \Psi \right>\) and they commute. Similarly, we have $$\left< g \middle| f \right> = \left< \hat{B}\hat{A} \right> - \left< A \right> \left< B \right>.$$ Thus, $$\left< f \middle| g \right> - \left< g \middle| f \right> = \left< \hat{A}\hat{B} \right> -\left< \hat{B}\hat{A} \right> = \left< \left[ \hat{A},\hat{B} \right] \right>.$$ Substitute back, we have $$ \bbox[5px,border:2px solid #666] { \sigma^2_A\sigma^2_B \geq \left(\frac{1}{2i}\left< \left[\hat{A},\hat{B}\right] \right>\right)^2. } $$ This is called the uncertainty principle.