Quantum Mechanics Intermediates

Tensor Product Space

This part has been moved to Tensor Product Space

Density Matrix

The density matrix operator is

\[\hat\rho(t) = \ket{\psi(t)}\bra{\psi(t)}.\]

The state \(\ket{\psi(t)}\) can be projected onto a set of basis \(\ket{\phi_n}\),

\[\ket{\psi(t)} = \sum_{n} C_m(t) \ket{\phi_n}.\]

Using the basis \(\ket{\phi_n}\), the density matrix is represented as \(\rho_{mn}\)

\[\begin{split}\hat\rho(t) &= \sum_m \sum_n C_m C_n^* \ket{\phi_m} \bra{\phi_n} \\ &= \sum_m \sum_n \rho_{mn} \ket{\phi_m} \bra{\phi_n}\end{split}\]

For pure states, density matrix and kets carries the same information.

Density matrix can be used to denote mixed states easily. For mixed states, we write down the density matrix using a mixture of pure states \(\ket{\psi_i(t)}\)

\[\hat \rho(t) = \sum_k p_k \ket{\psi_k(t)}\bra{\psi_k(t)}.\]

Mixed states are like mixture models in statistics.

Angular Momentum

Angular Momentum

For an new operator, we would like to know

  1. Commutation relation: with their own components, with other operators;
  2. Eigenvalues and their properties;
  3. Eigenstates and their properties;
  4. Expectation and classical limit.

Definition of Angular Momentum

In classical mechanics, angular momentum is defined as

\[\vec L = \vec X \times \vec P .\]

One way of defining operator is to change position and momentum into operators and check if the operator is working properly in QM. So we just define

\[\hat {\vec L} = \hat {\vec X}\times \hat{\vec P}.\]

It is Hermitian. So it can be an operator. We also find

\[\hat{\vec L}\times \hat{\vec L} = i \hbar \hat{\vec L}\]
\[\left[\hat L_i,\hat L_j\right] = \sum_k i\epsilon_{ijk}\hat L_k .\]

More generally, we can define angular momentum as

\[\left[\hat J_i, \hat J_j\right] = i\hbar \sum_k \epsilon_{ijk} \hat J_k\]

We can prove that

\[\left[ \hat J^2,\hat J_z \right] = 0.\]

So they can have the same eigenstates

\[\hat J_z \ket{\lambda m} = m\hbar \ket{\lambda m}\]
\[\hat J^2 \ket{\lambda m} = \lambda^2 \hbar^2 \ket{\lambda m}\]

To find the constraints on these eigenvalues, we can use positive definite condition of certain inner porducts, such as,

\[\bra{\psi} \hat J_+ \hat J_- \ket{\psi} \geq 0\]
\[\bra{\psi} \hat J_- \hat J_+ \ket{\psi} \geq 0\]


\[\hat J_{\pm} = \hat J_x \pm i \hat J_y\]

and we have

\[\left[\hat J_+, \hat J_-\right] = 2 \hbar \hat J_z\]
\[\left[\hat J_z, \hat J_{\pm} \right] = \pm \hbar \hat J_{\pm}.\]

It’s easy to find out that

\[\hat J_z (\hat J_{\pm}\ket{\lambda m}) = (m\pm 1) \hbar (\hat J_{\pm} \ket{\lambda m})\]

i.e., \(\hat J_{\pm}\ket{\lambda m}\) is eigenstate of \(\hat J_z\).

Follow the plan of finding out the bounds through these positive inner products, we can prove that

\[\hat J^2\ket{jm} = j(j+1)\hbar^2 \ket{jm}\]
\[\hat J_{\pm}\ket{jm} = \sqrt{j(j+1)-m(m\pm 1)} \hbar \ket{j,m\pm 1}\]

Eigenstates of Angular Momentum

As we have proposed, the eigenstates of both \(\hat J_z\) and \(\hat{\vec J}^2\) are \(\ket{j,m}\), where \(j=0,1,2,\cdots\) and \(m=-j,-j+1,\cdots, j-1,j\).

We can also find out the wave function in \({\ket{\theta,\phi } }\) basis. Before we do that, the definition of this basis should be made clear. This basis spans the surface of a 3D sphere in Euclidean space and satisfies the following orthonormal and complete condition.

\[\int \mathrm d \Omega \braket{\theta',\phi'}{\theta,\phi} = \delta(\cos\theta'-\cos\theta,\phi'-\phi) \int \mathrm d \Omega \ket{\theta',\phi'}\bra{\theta,\phi} = 1\]

Now we have an arbitary state \(\ket{\psi}\),

\[\begin{split}\ket{\psi} &= \sum _ {l,m} \psi _ {lm}\ket{l,m} \\ &= \sum _ {l,m} \int \mathrm d \Omega \ket{\theta',\phi'}\bra{\theta,\phi} \psi _ {lm}\ket{l,m} \\ &= \sum _ {l,m} \int \mathrm d \Omega \ket{\theta',\phi'} (\braket{\theta,\phi}{l,m} ) \psi _ {lm} \\\end{split}\]

Then we define


which is the spherical harmonic function.


\[\begin{split}\ket{\psi} &= \sum _ {l,m} \psi _ {lm} \int \mathrm d \Omega Y_l^m(\theta,\phi) \ket{\theta',\phi'} \\\end{split}\]

So as long as we find out what \(\psi _ {lm}\) is, any problem is done.