Hilbert space setting for quantum mechanics

• A Hilbert space $\mathit{H}$ is a complete normed vector space over $\mathbb{C}$ :

  1.

  $\mathit{H}$ is a vector space over $\mathbb{C}$

  2.

  There is an inner product
  $\langle{\cdot}\vert{\cdot}\rangle$ : $\mathit{H}$ x $\mathit{H}\rightarrow \mathbb{C}$
  which is conjugate linear:
  $\langle{v}\vert{w}\rangle = \overline{\langle{w}\vert{v}\rangle}$
  $\langle{v}\vert{\alpha w}\rangle = \alpha \langle{v}\vert{w}\rangle$ for $\alpha \in \mathbb{C}$
  $\langle{v+w}\vert{z}\rangle = \langle{v}\vert{z}\rangle + \langle{w}\vert{z}\rangle$
  $\langle{v}\vert{v}\rangle \ge 0$ and $\langle{v}\vert{v}\rangle = 0$ iff v = 0

  3.

  From the inner product, as usual, we define the norm of a vector:
  $\Vert v \Vert ^ 2 = \langle{v}\vert{v}\rangle$

  4.

  $\mathit{H}$ is complete with respect to the norm.

• We will typically use the bra/ket notation:
  $\vert{v}\rangle$ is a vector in $\mathit{H}$, and
  $\langle{v}\vert$ is the covector which is the conjugate transpose of v.
• This notation also allows us to represent the outer product of a vector and covector as $\vert{v}\rangle \langle{w}\vert$, which, for example, acts on a vector $\vert{z}\rangle$ as $\vert{v}\rangle \langle{w}\vert{z}\rangle$. For example, if {v₁,v₂} is an orthonormal basis for a two-dimensional Hilbert space, $\vert{v_1}\rangle\langle{v_2}\vert$ is the transformation that maps $\vert{v_2}\rangle$ to $\vert{v_1}\rangle$ and $\vert{v_1}\rangle$ to (0, 0)^T since
  
  $$\begin{array}{l} \vert{v_1}\rangle\langle{v_2}\vert\vert{v_2... ... \left(\begin{array}{c}0\\ 0\end{array}\right).\\ \end{array}$$
  
  
  Equivalently, $\vert{v_1}\rangle\langle{v_2}\vert$ can be written in matrix form where $\vert{v_1}\rangle = (1, 0)^T$, $\langle{v_1}\vert = (1, 0)$, $\vert{v_2}\rangle = (0, 1)^T$, and $\langle{v_2}\vert = (0, 1)$. Then
  
  $$\vert{v_1}\rangle\langle{v_2}\vert = \left(\begin{array}{c}1\... ... (0, 1) = \left(\begin{array}{cc}0&1\\ 0&0\end{array}\right).$$
  
  

• A unitary operator $U : \mathit{H}\to \mathit{H}$ is a linear mapping whose conjugate transpose is its inverse: $U^\dag = U^{-1}$
• Unitary operators are norm preserving:
  $\Vert Uv \Vert ^2 = \langle{v}\vert U^\dag U \vert{v}\rangle = \langle{v}\vert{v}\rangle = \Vert v \Vert ^2$
• We will think of a quantum state as a (normalized) vector $\vert{v}\rangle \in \mathit{H}$. For math folks, we are in effect working in Complex projective space, normalizing to 1 so that the probabilities make sense.
• The dynamical evolution of a quantum system is expressed as a unitary operator acting on the quantum state.

• Eigenvalues of a unitary matrix are of the form $e ^ {i\omega}$ where $\omega$ is a real-valued angle. A unitary operator is in effect a rotation.
• Just for reference, a typical expression of Schrödinger's equation looks like
  
  $$\left[-\frac{\hbar^2}{2m_e}\bigtriangledown^2+V(x,y,z)\right]\Psi = i\hbar\frac{\partial}{\partial t} \Psi$$
  
  
  with general solution
  
  $$\Psi(x,y,z,t)=\sum_{n=0}^\infty c_n\Psi_n(x,y,z)\exp\left(\frac{-iE_nt}{\hbar}\right)$$
  
  
  where $\Psi_n(x,y,z)$ is an eigenfunction solution of the time independent Schrödinger equation with E_n the corresponding eigenvalue. The inner product, giving a time dependent probability, looks like
  
  $$P(t) = \int\overline\Psi\Psi dv.$$
  
  

• Another way to think of this is that we have to find the Hamiltonian ${\cal H}$ which generates evolution according to:
  
  $$i\hbar\frac{\partial}{\partial t}\vert{\Psi(t)}\rangle={\cal H }\vert{\Psi(t)}\rangle.$$
  
  
  In our context, we will have to solve for ${\cal H}$ given a desired U:
  
  $$\vert{\Psi_f}\rangle=\exp\left(-\frac{i}{\hbar}\int{\cal{H}}dt\right)\vert{\Psi_0}\rangle= U\vert{\Psi_0}\rangle$$
  
  
  A solution for ${\cal H}$ always exists, as long as the linear operator U is unitary.
• A measurement consists of applying an operator O to a quantum state v. To correspond to a classical observable, O must be Hermitian, $O^\dag = O$, so that all its eigenvalues are real. If one of its eigenvalues $\lambda$ is associated with a single eigenvector $u_\lambda$, then we observe the value $\lambda$ with probability $\vert \langle{v}\vert{u_\lambda}\rangle \vert ^ 2$ (i.e., the square of the length of the projection along $u_\lambda$).

• In general, if there is more than one eigenvector $u_\lambda$ associated with the eigenvalue $\lambda$, we let $P_\lambda$ be the projection operator onto the subspace spanned by the eigenvectors, and the probability of observing $\lambda$ when the system is in state v is $\Vert P_\lambda v\Vert ^ 2$.
• Most projection operators do not commute with each other, and are not invertible. Therefore, we can expect that the order in which we do measurements will matter, and that doing a measurement will irreversibly change the state of the quantum system.