The Laws of Quantum Mechanics

1. In Newtonian Mechanics the motion of a particle under the influence of a potential $V(\vec{r},y)$ is described by the particle's trajectory $\vec{r}(t)$. There are different trajectories corresponding to different possible states of the particle.

   In Quantum Mechanics there is a different wavefunction for every state of a particle. For example consider three different states of a particle referred to as states $1$, $2$ and $3$ with wave functions $\psi_1(x,t)$, $\psi_2(x,t)$ and $\psi_3(x,t)$ respectively.

2. In Newtonian Mechanics the trajectory is determined by solving Newton's equation of motion
   

   $$m \frac{d^2 \vec{r}}{d^2 t}=-\vec{\nabla} V(\vec{r},t)$$ (20.1)

   
   

   In Quantum Mechanics the wavefunction is governed by Schrodinger's equation
   

   (20.2)

   
   
   For a particle free to move only in one dimension along the $x$ axis we have
   

   $$i \hbar \frac{\partial}{\partial t} \psi(x,t) = \frac{-\hbar... ...c{\partial^2}{\partial x^2} \psi (x,t) + V (x,t)\psi (x,t) \,.$$ (20.3)

   
   
   Here we shall only consider time independent potentials $V(x)$. Applying the method of separation of variables we take a trial solution
   

   (20.4)

   
   
   whereby the Schrodinger's equation is
   

   $$X i \hbar \frac{dT}{dt} = \frac{-\hbar^2}{2m} T \frac{d^2 X}{dx^2} + V(x) X T$$ (20.5)

   
   
   which on dividing by ${XT}$ gives
   

   $$i \hbar \frac{1}{T} \frac{dT}{dt} = \frac{-\hbar^2}{2m} \frac{1}{X} \frac{d^2X}{dx^2} + V(x) = E$$ (20.6)

   
   
   The first term is a function of $t$ alone whereas the second term is a function of $x$ alone. It is clear that both terms must have a constant value if they are to be equal for all values of $x$ and $t$. Denoting this constant as $E$ we can write the solution for the time dependent part as
   

   $$T(t)=A e^{- i E t/\hbar} \,.$$ (20.7)

   
   
   The $x$ dependence has to be determined by solving
   

   $$\frac{\hbar^2}{2m} \frac{d^2 X}{dx^2} = - [E-V(x)] X\,.$$ (20.8)

   
   
   The general solution can be written as
   

   $$\psi(x,t)= A e^{- i E t/\hbar} \, X(x)$$ (20.9)

   
   
   where the $t$ dependence is known and $X(x)$ has to be determined from equation (20.8).

   Free particle The potential $V(x)=0$ for a free particle. It is straightforward to verify that $X(x)=e^{i p x/\hbar}$ with $p=\pm \sqrt{2 m E}$ satisfies equation (20.8). This gives the solution
   

   $$\psi(x,t) = A e^{-i (E t - p x)/\hbar} \,.$$ (20.10)

   
   

   The is a plane wave with angular frequency $\omega = E/\hbar$ and wave number $k= p/\hbar$ where and $p$ are as yet arbitrary constant related as $E = p^2/2m$. This gives the wave's dispersion relation $\omega= \hbar k^2/2 m$.

   Here different value of $E$ will give different wavefunctions. For example $p_1$ and $p_2$ are different constants with $E_1=p_1^2/2m$ and $E_2=p_2^2/2m$, then
   

   $$\psi_1(x,t)=A_1 e^{-i (E_1 t - p_1 x)/\hbar}$$ (20.11)

   
   
   and
   

   $$\psi_2(x,t)=A_2 e^{-i (E_2 t - p_2 x)/\hbar}$$ (20.12)

   
   
   are two different wavefunctions corresponding to two different states of the particle.

3. In Quantum Mechanics the superposition of two different solutions $\psi_1(x,t)$ and $\psi_2(x,t)$ corresponding to two different states of the particle is also a solution of the Schördinger's equation, an example being
   

   (20.13)

   
   
   This can be generalized to a superposition of three, four and more states. In the continuum we have
   

   $$\psi (x,t) = \int \limits^\infty_{-\infty} A(p) e^{-i (E(p)t - px)} \, dp$$ (20.14)

   
   

4. What happens when we make a measurement? We have already discussed what happens when we measure a particle's position. In Quantum Mechanics it is not possible to predict the particle's position. We can only predict probabilities for finding the particle at different positions. The probability density $\rho(x,t)=\psi(x,t) \psi^{*}(x,t)$ gives the probability $d P(x,t)$ of finding the particle in the interval $dx$ around the point $x$ to be $d P(x,t)=\rho(x,t) \, dx$. But there are other quantities like momentum which we could also measure. What happens when we measure the momentum?

   InQuantum Mechanics there is a Hermitian Operator corresponding to every observable dynamical quantity like the momentum, angular momentum, energy , etc.

   What is an operator? An operator $\hat{O}$ acts on a function $\psi(x)$ to give another function $\phi(x)$.
   

   $$\hat O \psi (x) = \phi (x)$$ (20.15)

   
   
   We consider an example where $\hat O = \frac{d}{dx}$
   

   $$\hat O \sin x= \frac{d}{dx} \sin x = \cos x \,.$$ (20.16)

   
   
   Consider another example where $\hat O = 2$ so that
   

   $$\hat O \psi(x)= 2 \psi(x)= \phi(x)$$ (20.17)

   
   
   What is a Hermitian operator? We do not go into the definition here, instead we only state a relevant, important property of a Hermitian operators. Given an operator $\hat O$, a function $\psi(x)$ is an eigenfunction of $\hat O$ with eigenvalue $\lambda$ if
   

   $$\hat O \psi(x) = \lambda \psi(x)$$ (20.18)

   
   

   As an example consider
   

   $$\hat O = - i \hbar \frac{d}{dx} ~ \mbox{and} ~ \psi(x) = e^{ikx}$$ (20.19)

   
   
   we see that
   

   $$\hat O \psi(x) = \hbar k \psi(x) \,$$ (20.20)

   
   
   The function $e^{i k x}$ is an eigenfunction of the operator $- i \hbar \frac{d}{dx}$ with eigenvalue $\hbar k$. As another example for the same operator we consider the function
   

   $$\psi(x) = \cos (kx)$$ (20.21)

   
   
   we see that
   

   $$\hat O \psi = - i \hbar \frac{d}{dx} \cos (kx) = i \hbar k \sin(kx) \,.$$ (20.22)

   
   
   This is not an eigenfunction of the operator.

   Hermitian operators are a special kind of operators all of whose eigenvalues are real.

   We present the Hermitian operators corresponding to a few observable quantities.

   Position - x $\rightarrow$ Operator $\hat x$
   

   $$\hat x \psi (x,t) = x \psi (x,t)$$ (20.23)

   
   
   Momentum - p $\rightarrow$ Operator $\hat p$
   

   $$\hat p \psi (x,t) = - i\hbar \frac{\partial \psi}{\partial x}$$ (20.24)

   
   
   Hamiltonian $H(p,x,t) \rightarrow$Operator $\hat H$
   

   $$\hat H \psi (x,t) = i \hbar \frac{\partial}{\partial t} \psi (x,t)$$ (20.25)

   
   
   The eigenvalues of the Hamiltonian $\hat H$ correspond to the Energy.

   Let us return to what happens when we make a measurement. Consider a particle in a state with wavefunction $\psi(x,t)$. We measure its momentum $p$. There are two possibilities
   A.   If $\psi(x,t)$ is an eigenfunction of $\hat p$. For example
   

   $$\psi(x,t) = A_1 e^{-i (E_1t-p_{_1} x)/\hbar}$$ (20.26)

   
   
   
   

   $$\hat p \psi = p_{_1} \psi$$ (20.27)

   
   
   We will get the value $p_{_1}$ for the momentum The measurement does not disturb the state and after the measurement the particle continues to be in the same state with wavefunction $\psi(x,t)$.

   B. If $\psi(x,t)$ is not an eigenfunction of $\hat p$. For example
   

   $$\psi(x,t) = \left( \frac{3}{5} \right) e^{-i (E_1t-p_1 x)/\h... ...} + \left( \frac{4}{5} \right) e^{-i (E_2t - p_{_2} x )/\hbar}$$ (20.28)

   
   
   On measuring the momentum we will get either $p_{_1}$ or $p_{_2}$. The probability of getting $p_{_1}$ is $\left( \frac{3}{5} \right)^2$ and probability of getting $p_{_2}$ is $\left( \frac{4}{5} \right)^2$ The measurement changes the wavefunction.

   In case we get $p_{_1}$ the wavefunction will be changed to
   

   $$\psi(x,t) = A_1 e^{-i (E_1t-p_{_1} x)}$$ (20.29)

   
   
   and in case we get $p_{_2}$ - the wavefunction will be changed to
   

   $$\psi(x,t) = A_2 e^{-i (E_2 t-p_{_2} x)} \,.$$ (20.30)

   
   
   If we measure $p$ again we shall continue to get the same value in every successive measurement.

   In general, if
   

   $$\psi(x,t)= c_1 \psi_1 (x,t) + c_2 \psi_2 (x,t)$$ (20.31)

   
   
   such that
   

   (20.32)

   
   
   On measuring $O$
   

   $$P_{_1} = \frac{\vert c_1\vert^2}{\vert c_1\vert^2 + \vert c_... ...= \frac{\vert c_2\vert^2}{\vert c_1\vert^2 + \vert c_2\vert^2}$$ (20.33)

   
   
   are are probabilities of getting the values $O_1$ and $O_2$ respectively. We can now interpret the solution
   

   $$\psi(x,t) = A_1 e^{-i(E_1t-p_{_1} x )}\,.$$ (20.34)

   
   
   This corresponds to a particle with momentum $p_{1}$ and energy $E_1$. We will get these value however many times we repeat the measurement .

   What happens if we measure the position of a particle whose wavefunction $\psi(x,t)$ is given by equation (20.34)? Calculating the probability density
   

   $$\rho(x,t) = \vert A_1\vert^2$$ (20.35)

   
   
   we see that this has no $x$ dependence. The probability of finding the particle is equal at all points. This wave function has no position information.

5. Consider many identical replicas of a particle in a state $\psi(x,t)$ as shown below. We measure the particle's momentum $p$. This is done independently on all the replicas and the result recorded. The outcome cannot be exactly predicted unless $\psi(x,t)$ is an eigenfunction of $\hat p$. Let us consider a general situation where this is not the case. As shown below, a different value of momentum will then be measured for each of the particle.

   
   

   $$\begin{array}{ccc} \fbox{$\psi$} \to p_{_1} & \fbox{$\psi$} ... ...\fbox{$\psi$} \to p_{_1} & \fbox{$\psi$} \to p_{_2} \end{array}$$ (20.36)

   
   
   What is the expectation value of the momentum? This can be determined after the experiment is performed as
   

   $$\langle p \rangle=\frac{\sum \limits_{i=1}^{n}p_i N_i}{N}$$ (20.37)

   
   
   where $N_i$ is the number of times the momentum occurs and $N=\sum \limits_{i=1}^{n} N_i$. In Quantum Mechanics it is possible to predict this using the particle's wavefunction as
   

   $$\langle p \rangle=\int \limits_{-\infty}^{\infty} \psi^{*}(x,t) \, \hat p \, \psi(x,t) \, d x \,.$$ (20.38)

   
   
   Similarly, it is possible to predict the variance in the momentum values
   

   $$\langle (\Delta p)^2 \rangle = \int \psi^* (x,t) \left( \hat p- \langle p \rangle \right)^2 \psi (x,t) dx$$

   $$= \int \psi^* {\hat p}^2 \psi dx - 2 \langle p \rangle \int \psi^* \hat p \psi dx + \langle p \rangle^2 \int \psi^* \psi dx$$

   $$= \langle p^2 \rangle - 2 \langle p \rangle^2 + \langle p \rangle^2 = \langle p \rangle^2 - \langle p \rangle^2$$ (20.39)

   
   
   The uncertainty in the momentum can be calculated as $\Delta p = \sqrt{\langle ( \Delta p)^2 \rangle}$