Measurement

$$\newcommand{\ket}{\mbox{|#1\rangle}} \newcommand{\bra}{\mbox{\langle #1|}} \newcommand{\cn}{\mbox{\mathbb{C}}} \newcommand{\rn}{\mbox{\mathbb{R}}} \newcommand{\knull}{\mbox{|{\mathbf 0}\rangle }} \newcommand{\bnull}{\mbox{\langle {\mathbf 0}|}} \newcommand{\sprod}{\mbox{\langle #1 | #2 \rangle }} \newcommand{\ev}{\mbox{\langle #1 | #2 | #3 \rangle }} \newcommand{\av}{\langle #1 \rangle} \newcommand{\norm}{\mbox{\Vert #1 \Vert }} \newcommand{\diad}{\mbox{|#1\rangle \langle #2 | }} \newcommand{\tr}{\mbox{Tr\{#1\}}} \newcommand{\sx}{\mbox{\hat{\sigma}_x}} \newcommand{\sy}{\mbox{\hat{\sigma}_y}} \newcommand{\sz}{\mbox{\hat{\sigma}_z}} \newcommand{\smx}{\mbox{\left( \begin{array}{cc} 0 & 1 \\ 1 & 0\end{array} \right)}} \newcommand{\smy}{\mbox{\left( \begin{array}{cc} 0 & -i \\ i & 0\end{array} \right)}} \newcommand{\smz}{\mbox{\left( \begin{array}{cc} 1 & 0 \\ 0 & -1\end{array} \right)}} \newcommand{\mat}{\mbox{\left( \begin{array}{cc} #1 & #2 \\ #3 & #4 \end{array} \right)}} \newcommand{\col}{\mbox{\left( \begin{array}{c} #1 \\ #2 \end{array} \right)}}$$

Unitary Evolution

The evolution of the state of a quantum system is determined by its Hamiltonian operator, $$\hat{H}$$, that is related to the Hamiltonian function of classical mechanics. The procedure that connects these two objects is beyond the scope of these lectures and we will simply assume that the Hamiltonian is given to us and is independent of time (energy is conserved). Let us consider a system that was prepared in a state $$\ket{\psi(0)}$$ at time $$t=0$$. At time $$t>0$$ its state, $$\ket{\psi(t)}$$, will be given by \begin{equation} \ket{\psi(t)}=\hat{U}(t)\ket{\psi(0)}, \label{u_definition} \end{equation} where $$\hat{U}(t)$$ is a unitary operator determined by the Hamiltonian. $$\hat{U}(t)$$ is the evolution operator. We postulate that the evolution is continous, such that $$\hat{U}(t)$$ is a continous operator function of the time $$t$$. Because of continuity the evolution operator has a first derivative and we postulate that it is \begin{equation} \frac{d\hat{U}(t)}{dt}=-\frac{i}{\hbar}\hat{H}\hat{U}(t). \label{u_diff} \end{equation} Beacuse the Hamiltonian is independent of time the evolution operator does not depend on the choice of the initial time point and the following expression holds \begin{equation} \hat{U}(t)=e^{-\frac{i}{\hbar}\hat{H}t}. \label{conservative_evolution} \end{equation} Taking the time derivative of equation \ref{u_definition} and using the evolution postulate \ref{u_diff} we obtain \begin{eqnarray} \frac{d\ket{\psi(t)}}{dt}=\frac{d\hat{U}(t)}{dt}\ket{\psi(0)}=-\frac{i}{\hbar}\hat{H}\hat{U}(t)\ket{\psi(0)}=-\frac{i}{\hbar}\hat{H}\ket{\psi(t)}. \end{eqnarray} This is the evolution equation for states \begin{eqnarray} \frac{d\ket{\psi(t)}}{dt}=-\frac{i}{\hbar}\hat{H}\ket{\psi(t)}. \end{eqnarray} Equation \ref{conservative_evolution} has an important consequence. Since the evolution operator is an exponential function of the Hamiltonian they have the same eigen-spaces so \begin{eqnarray} \hat{H}&=&\sum_{k=1}^m E_k \hat{P}_k, \\ \hat{U}(t)&=&\sum_{k=1}^m e^{-\frac{i}{\hbar}E_k t} \hat{P}_k. \end{eqnarray} This means that the evolution of an arbitrary state of a conservative system is completely determined by the eigen-decomposition of the Hamiltonian. Eigen-states of the Hamiltonian evolve only by a phase factor and are called stationary since a complex phase does not change any of the probability amplitudes. If we want to determine the evolution of an arbitrary state we need to expand it in therms of the Hamiltonian eigen-basis \begin{equation} \ket{\psi(t)}=\sum_{k=1}^m e^{-\frac{i}{\hbar}E_k t} (\hat{P}_k \ket{\psi(0)}). \end{equation} What we just described was the Schrödinger picture in which observables (operators) are stationary and states evolve in time. However, we can pick a different and equivalent view of the unitary space of our system, the Heisenber picture. In it, observables evolve in time while states stay unchanged. For example the time evolution of the average value \begin{eqnarray} \av{\hat{A}}(t)=\ev{\psi(t)}{\hat{A}}{\psi(t)}&=&[\bra{\psi(0)}\hat{U}^\dagger(t)]\hat{A}[\hat{U}(t)\ket{\psi(0)}] \nonumber \\ &=&\bra{\psi(0)}[\hat{U}^\dagger(t)\hat{A}\hat{U}(t)]\ket{\psi(0)}]=\ev{\psi(0)}{\hat{A}(t)}{\psi(0)}. \end{eqnarray} We see that we get the same result if we assume that our intial state stays the same but the observables evolve as \begin{equation} \hat{A}(t)=\hat{U}^\dagger(t)\hat{A}(0)\hat{U}(t). \end{equation}

Measurement

The measurement postulate is one of the strangest postulates of quantum mechanics. Many physicist think that it is artificial and extraneous but it cannot be said that there is a consensus on this issue. What it claims is that the act of measurement is a special event that breaks the continuous, unitary, deterministic evolution that we just described. When, at a time $$t$$, a measurement of an observable, $$\hat{A}$$, is performed the outcome is random and the probability that the outcome will be the eigen-value $$\lambda$$ of $$\hat{A}$$ is given by 6 from the previous section. The measurement postulate states that the measurement then instantenously converts the state of the system to the following normalized projection \begin{equation} \ket{\psi(t+)}=\frac{\hat{P_{\lambda}}\ket{\psi(t-)}}{\norm{\hat{P_{\lambda}}\ket{\psi(t-)}}} \label{collapse} \end{equation} where $$\ket{\psi(t-)}$$ and $$\ket{\psi(t+)}$$ are the state of the system immediately before and immediately after the measurement. This is the famous "collapse of the wavefunction". One of the obvious inconsistencies introduced by this postulate is that if we treat the measurement aparatus and the measured system as a new quantum object its state has to evolve continously as described in the previous section. However since the measurement aparatus measured the system at time $$t$$ and induced the random change of \ref{collapse} the measurement postulate implies that the composite system does not evolve deterministically. Without dwelling on these issues we will simply accept this postulate since it has been experimentally verified and is good enough for practical purposes.

The Statistical Operator

Vectors in Hilbert space for which equation 11 from the previous section correctly describes outcome probabilites are called pure states. This means that we know that the system is in this state with absolute certainty and all porbabilities come from quantum mechanics, not from our lack of knowledge. If we only know a set of possible states and the classical probabilities that the system is one of them the system is no longer described by a pure state but by a statistical operator (density matrix) and we say that it is a mixture. For example, if we prepare a certain number of spins to be in the $$\ket{+}$$ state and the same number of spins to be in the $$\ket{-}$$ state and then physically mix the two ensembles the resulting ensemble is described by a mixture, not a pure state. We only know that there is classical $$50\%$$ chance that a spin is in the state $$\ket{+}$$ or $$\ket{-}$$ without performing any measurements. This ensemble is not described by the linear combination $$(1/\sqrt{2})(\ket{+}+\ket{-})$$, but by a statistical operator that will be defined in the following paragraphs. We first rewrite 11 from the previous section as \begin{equation} \ev{\psi}{\hat{A}}{\psi}=\tr{\diad{\psi}{\psi}\hat{A}}, \end{equation} since we can always find an orthonormal basis in which $$\ket{\psi}$$ is one of the the basis vectors so \begin{equation} \tr{\diad{\psi}{\psi}\hat{A}}=\bra{\psi}[\diad{\psi}{\psi}\hat{A}]\ket{\psi}=\sprod{\psi}{\psi}\ev{\psi}{\hat{A}}{\psi}=\ev{\psi}{\hat{A}}{\psi} \end{equation} Let us now consider a system which can be in any of the states $$\{\ket{\psi}_1,\ldots,\ket{\psi}_l\}$$ with classical probailities $$\{p_1,\ldots,p_l\}$$ such that $$\sum_{k=1}^l p_k=1$$. The average value of some observable $$\hat{A}$$ is then the classical average of the quantum expectation values for each possible state \begin{equation} \av{\hat{A}}=\sum_{k=1}^l p_k \ev{\psi_l}{\hat{A}}{\psi_l}=\sum_{k=1}^l p_k \tr{\diad{\psi_k}{\psi_k}\hat{A}}=\tr{\hat{\rho}\hat{A}}, \end{equation} where we used the linearity of the trace to define the statistical operator (density matrix) \begin{equation} \hat{\rho}=\sum_{k=1}^l p_k \diad{\psi_k}{\psi_k}. \end{equation} The general expression for the expectation value of an observable is then \begin{equation} \av{\hat{A}}=\tr{\hat{\rho}\hat{A}} \end{equation} with $$\hat{\rho}$$ defined above. For pure states only one of the probabilities, $$p_k$$, is nozero and equal to 1 since we know the state of the system with absolute certainty.

Thermal Equilibrium Statistical Operator

As a postulate of quantum mechanics and in analogy to the Gibbs distribution, the statistical operator of a system in thermal equilibrium is given by \begin{equation} \hat{\rho}(\beta)=\frac{e^{\beta \hat{H}}}{\tr{e^{\beta \hat{H}}}} \label{equilibrium} \end{equation} where $$\beta=1/kT$$, with $$k$$ being the Boltzmann constant and $$T$$ the temperature. We now consider in more detail the thermal equilibrium of a system of $$N$$ identical, noninteracting particles.The states of this system belong to the tensor product of $$N$$ identical copies of the one particle Hilbert space, $$\cal{H}$$. \begin{equation} \cal{H}\otimes\cal{H}\otimes \ldots \otimes \cal{H}. \end{equation} The Hamiltonian for this system is \begin{equation} \hat{H}=\hat{H}_1(1)+\hat{H}_1(2)+\ldots +\hat{H}_1(N). \end{equation} where $$\hat{H}_1(i)$$ is the same one-particle Hamiltonian acting on the space of the $$i$$-th particle. For this particular case \ref{equilibrium} simplifies since \begin{equation} e^{\beta \hat{H}}=e^{\beta \sum_{k=1}^N \hat{H}_1(k)}=\prod_{k=1}^N e^{\beta \hat{H}_1(k)}. \end{equation} The trace is then \begin{equation} \tr{e^{\beta \hat{H}}}=\tr{\prod_{k=1}^N e^{\beta \hat{H}_1(k)}}=\prod_{k=1}^N \tr{e^{\beta \hat{H}_1(k)}}=\left[ \tr{e^{\beta \hat{H}_1}} \right]^N \end{equation} where $$\hat{H}_1$$ is any of the one particle operators since they all have the same trace trace. Finally \ref{equilibrium} can be written as \begin{equation} \hat{\rho}(\beta)=\frac{\prod_{k=1}^N e^{\beta \hat{H}_1(k)}}{\left[ \tr{e^{\beta \hat{H}_1}} \right]^N}. \end{equation}