7.31 Hermitian and Quantum States

Linear Operators on Hilbert Space

Linear

\(M:A\to B\) is linear if \(M(a|\psi\rang+b|\phi\rang)=aM|\psi\rang+bM|\phi\rang\) for all \(a,b\in \mathbb{F},\forall |\psi\rang,|\phi\rang\in A\)

Matrix

Since \(|y\rang =\begin{pmatrix} 0\\\vdots\\1\\0\\\vdots\\0 \end{pmatrix}\) here \(1\) is \(y\)-location and \(\lang x|=(0,...,0,1,0,...,0)\) here \(1\) is \(x\)-location

Then \(|y\rang \lang x|=\begin{pmatrix} 0&\cdots&0\\ \vdots&1&\vdots\\ 0&0&0 \end{pmatrix}\) here \(1\) is in the column \(x\) and row \(y\)

Then \(\lang y|\cdot M|x\rang =M_{yx}\) where \(M=\sum_{x\in[m]}\sum_{y\in[n]}M_{yx}|y\rang\lang x|\) (This is the basis with coefficients)

Then \(M|\psi\rang=\sum_{x}\sum_{y}M_{yx}|y\rang\lang x|\psi\rang=\sum_{y}\underbrace{\left(\sum_{x} M_{yx}\lang x|\psi\rang\right)}_{number}|y\rang\)

Also clearly that \(M=\sum_{x\in[m]}\sum_{y\in[n]}M_{yx}|y\rang\lang x|\) and \(M^*=\sum_{x\in[m]}\sum_{y\in[n]}\overline{M}_{yx}|x\rang\lang y|\)

Adjoint

The adjoint of a linear operator \(M:A\to B\) is itself a linear operator \(M^*:B\to A\)

\[ \langle\phi|M\psi\rangle=\langle M^{*}\phi|\psi\rangle,\forall\psi\in A,\phi\in B \]

Unitary Operator

\(U:A\to A,U^*U=I^A\) and \(\dim (A):=|A|\) where \(I^{A}= \begin{pmatrix} 1 & \cdots & 0 \\ \vdots & 1 & \vdots \\ 0 & \cdots & 1 \end{pmatrix}=\sum^{|A|}_{x=1}|x\rang\lang x|\)

Note that the identity operator \(I^A : A \to A\) is a positive semi-definite operator (i.e. an operator with all eigenvalues strictly greater than zero)

Prove: If \(\left\{|v_x\rang\right\}^{|A|}_{x=1}\) is an orthonormal basis, then \(I^A=\sum^{|A|}_{x=1}|v_x\rang \lang v_x|\)

Proof

Method 1：

Assume \(|\psi\rangle = a_{1}|v_{1}\rangle + \dots + a_{|A|}|v_{|A|}\rangle\), then \(\sum_{x=1}^{|A|}|v_{x}\rangle\langle v_{x}||\psi\rangle = |v_{1}\rangle a_{1} + \dots + |v_{|A|}\rangle a_{|A|}= |\psi\rangle\)

Method 2:

Since \(\lang v_x||\psi\rang =\lang v_x|\psi\rang\) is a scalar, then we can move \(\lang v_x|\psi\rang\) in front of \(|v_x\rang\): \(\left(\sum_{x\in[m]}|v_{x}\rangle\langle v_{x}|\right)|\psi\rangle=\sum_{x\in[m]} \langle v_{x}|\psi\rangle|v_{x}\rangle=|\psi\rangle\)

Therefore, \(\sum_{x \in [m]} |v_x\rangle\langle v_x| = I^A\) for any orthonormal basis \(\{|v_x\rangle\}_{x \in [m]}\) of A.

Isometry

Definition 1

Let \(V:A\to B\) such that \(V^{*}V=I^{A}\)(clearly \(|B|\geq |A|\)), then \(V\) is called isometry.

If \(|B|>|A|\), then \(V\) can only be isometry not unitary. Thus we call it isometry

If \(|B|=|A|\), then \(V\) is isometry(unitary in particular). Conventionally, we call it unitary

Definition 2

\(V:A\to B\) is isometry if \(\forall |\psi\rang \in A,\lang V\psi|V\psi\rang =\lang \psi|\psi\rang\), that is \(\forall |\psi\rang\in A,||V\psi||=||\psi||\) (preserve the norm)

Proof of equivalence

\(1\Rightarrow2)\,\,\)\(\forall|\psi\rangle\in A,\langle V\psi|V\psi\rangle=\langle\psi|V^{*}V|\psi\rangle =\langle\psi|I^{A}|\psi\rangle=\lang \psi|\psi\rang\)

Homework 1: \(2\Rightarrow1)\,\,\langle\psi|(V^{*}V-I^{A})|\psi\rangle=\langle\psi|V^{*}V|\psi \rangle-\langle\psi|I^{A}|\psi\rangle=\langle V\psi|V\psi\rangle-\langle\psi|I^{A} |\psi\rangle=\langle\psi|\psi\rangle-\langle\psi|\psi\rangle=0\)

Then the eigenvalues of \(V^*V-I^A\) are all \(0\), then \(V^*V-I^A=0\)

Conclusion: \(\lang \psi|\phi\rang = \lang \psi|I|\phi\rang=\lang \psi| V^{*}V|\phi\rang=\lang V\psi|V\phi\rang\)

Eigenvalue of Unitary

\(U|\psi\rang =e^{i\theta}|\psi\rang\), that is \(U=\sum^d_{x=1}e^{i\theta_x}|\psi_x\rang\lang\psi_x|\)

So, it says the norm of eigenvalues of a unitary operator must be \(1\), that is \(||\lambda||=1\)

Given a orthonormal basis \(\{|\psi_x\rang\}_{x\in [d]}\) of \(A\) with \(U|\psi_x\rang =e^{i\theta_x}|\psi_x\rang\) (since O.N. basis)

Hermitian and Positive Matrices

Hermitian(Self-adjoint)

\(H^*=H\) and \(H=\sum_{x\in[d]}\lambda_{x}|\psi_{x}\rang \lang \psi_{x}|\) is spectral decomposition of \(H\) where \(\{|\psi_x\rang \}_{x\in [d]}\) is orthonormal eigen basis of \(A\)

Same reason as above, we can get \(H|\psi_{x}\rangle=\lambda_{x}|\psi_{x}\rangle\)

We denote by \(\mathrm{Tr}[H]\) the trace of Hermitian operator \(H: A \rightarrow A\).
That is, \(\mathrm{Tr}[H]:=\sum_{x \in[m]}\langle x|H| x\rangle\)
This is independent on the choice of the orthonormal basis \(\{|x\rangle\}\) of \(A\).

Positive semi-definite Matrices

\(P:A\to A\) is positive semi-definite iff \(\lang \psi|P|\psi\rang \geq 0,\forall |\psi\rang \in A\)

Notation: \(P\geq 0\)

Propostion

\(P\) is positive semi-definite iff \(P\) is Hermitian and eigenvalue are non-negative

Proof

\(\Rightarrow\)) \(\langle\psi|P|\psi\rangle=\overline{\langle\psi|P|\psi\rangle}=(\langle\psi|P|\psi \rangle)^{*}=(|\psi\rangle^{*}P|\psi\rangle)^{*}=(P|\psi\rangle)^{*}|\psi\rangle= |\psi\rangle^{*}P^{*}|\psi\rangle= \lang\psi|P^{*}|\psi\rangle\)

Then \(\lang \psi|(P^{*}-P)|\psi\rang=0\), then \(\langle\psi|i(P^{*}-P)|\psi\rangle=0\)

Let \(i(P^*-P)=H\), then easily \(H^{*}=H\)

Then \(\lambda \lang \psi|\psi\rang =0\), since \(\lang \psi|\psi\rang \neq 0\), then \(\lambda =0\), then \(H=0\), then \(P^*=P\)

Since \(P\) is hermitian, then it's diagonalizable and has a eigen basis \(B\).

Then \(\forall|\psi\rangle\in B,\langle\psi|P|\psi\rangle=\langle\psi|\lambda|\psi\rangle =\lambda||\psi||\geq0\) since definition

\(\Leftarrow\)) Since \(P\) is hermitian, then it's diagonalizable and has a eigen basis \(B\).

Then \(\forall|\psi\rangle\in B,\langle\psi|P|\psi\rangle=\langle\psi|\lambda|\psi\rangle =\lambda||\psi||\geq0\) since eigenvalue is non-negative

Then \(\forall|\psi\rangle\in A,\langle\psi|P|\psi\rangle\geq0\)

Homework 2: Show that \(P\geq 0\) iff \(P=M^*M\) for some complex matrix \(M\)

Proof

\(\Rightarrow\)) Since \(P\geq 0\), then \(P\) is Hermitian and eigenvalue are non-negative, then \(P\) is diagonalizable

Then \(P=CDC^{*}\) with \(D= \begin{pmatrix} \lambda_{1} & 0 & \cdots & 0 \\ 0 & \lambda_{2} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \lambda_{n} \end{pmatrix}\) and \(D=\sqrt{D}\cdot \sqrt{D}\)

Let \(\sqrt{D}=E= \begin{pmatrix} \sqrt{\lambda_{1}} & 0 & \cdots & 0 \\ 0 & \sqrt{\lambda_{2} }& \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots &\sqrt{\lambda_{n}} \end{pmatrix}\) this is valid since eigenvalue are non-negative

Then clearly \(E^*=E\), then \(P=C(EE^*)C^{*}=(CE)\cdot(CE)^*=M^*M\)

\(\Leftarrow\)) \(\forall|\psi\rangle\in A,\langle\psi|P|\psi\rangle=\langle\psi|M^{*}M|\psi\rangle =(M|\psi\rangle)^{*}(M|\psi\rangle)=||M|\psi\rang||^2\geq0\)

Homework 3: Show that \(P\geq 0\) iff \(MPM^*\geq 0\) for all complex matrix \(M\)

Proof

\(\Rightarrow\)) Since \(P\geq 0\), then \(P=CC^*\) for some complex matrix

Then \(MPM^{*}=MCC^{*}M^{*}=(MC)(MC)^{*}\), then by homework 2, we get \(MPM^*\geq 0\)

\(\Leftarrow\)) Since \(MPM^*\geq 0\), then take \(M=I\), then \(P\geq 0\)

Decomposition of Hermitian Matrices

Given \(H:A\to A\) and \(H\) is Hermitian where \(d:=|A|\), then

\[ H=\sum_{x\in [d]}\lambda_{x}|\psi_{x}\rang \lang \psi_{x}|=\underbrace{\sum_{x\in [d],\lambda_x\geq 0}\lambda_x|\psi_x\rang\lang \psi_x|}_{H_+}-\underbrace{\sum_{x\in [d],\lambda_x<0}|\lambda_x||\psi_x\rang\lang \psi_x|}_{H_-} \]

Then \(H_+\geq 0,H_-\geq 0\) and \(H=H_{+}-H_{-}\) and \(H_+H_-=H_-H_+=0\) (Since orthonormal basis and \(H_+,H_-\) are disjoint)

Homework 4: Show that \(H_\pm=\frac{|H|\pm H}{2}\) where \(|H|:=\sqrt{H^*H}=\sqrt{H^2}\)

Proof

Since \(H\) is Hermitian, then \(H=UDU^{*}\), then \(|H|=\sqrt{UD^{2}U^{*}}\)

Then \(|H|=U|D|U^{*}\), then \(|H|=H_++H_-\) and we know \(H=H_+-H_-\)

Then \(H_\pm=\frac{|H|\pm H}{2}\)

Quantum state

If \(P\geq 0\) and \(\text{Tr}[P]=1\), then \(P\) is a quantum state or a density matrix

There are two states of a Quantum:

Pure Quantum State: \(\text{rank}P=1\)(definition)

\(P=|\psi\rang \lang \psi|\) where \(\lang \psi|\psi\rang =1\) since \(\text{Tr}[P]=\text{Tr}\left[|\psi\rang \lang \psi|\right]=\lang \psi|\psi\rang =1\)

Example

\(P= \begin{bmatrix} 1/2 & i/2 \\ -i/2 & 1/2 \end{bmatrix}\)

First by trace, we know the sum of two eigenvalues are \(1\)

Then by determinant \(\det P=\frac{1}{4}-\left(\frac{i}{2}\right)\left(\frac{-i}{2}\right)=0\), then there is \(1\) as eigenvalue, then another eigenvalue is \(0\)

Actually, we define \(|+i\rangle:=\frac{1}{\sqrt{2}}(|0\rangle+i|1\rangle)\)

\(P=|+i\rangle\lang+i|=\frac{1}{\sqrt{2}}(|0\rangle+i|1\rangle)\frac{1}{\sqrt{2}}( \langle0 |-i\langle1|)=\frac{1}{2}[|0\rangle\langle0|-i|0\rangle\langle1|+i|1\rangle \langle 0|+|1\rangle\langle1|]\) which is pure clearly
Mixed Quantum State: Not pure

Example

\(P=\frac{1}{4}|0\rang\lang 0|+\frac{3}{4}|1\rang\lang1|\) this is spectral decomposition of \(P\)

\(=\frac{1}{2}|u\rang\lang u|+\frac{1}{2}|v\rang\lang v |\) this is Pure-state decomposition of \(P\)

where \(|u\rang= \sqrt{\frac{1}{4}}|0\rang +\sqrt{\frac{3}{4}}|1\rang\) and \(|v\rang= \sqrt{\frac{1}{4}}|0\rang -\sqrt{\frac{3}{4}}|1\rang\)

Homework 5: Show \(\text{Tr}\left[|\psi\rang \lang \phi|\right]=\lang \phi|\psi\rang\quad\forall|\psi\rang,\lang\phi|\in A\)

Proof

Assume \(|\psi\rang =\begin{pmatrix} a_1\\a_2\\\vdots\\a_n \end{pmatrix}\) and \(\lang \phi|=\begin{pmatrix} b_1&b_2&\cdots&b_n \end{pmatrix}\), then \(|\psi\rangle\langle\phi|=\begin{pmatrix} a_1b_1&a_1b_2&\cdots&a_1b_n\\ a_2b_1&a_2b_2&\cdots&a_2b_n\\ \vdots&\vdots&\ddots&\vdots\\ a_nb_1&a_nb_2&\cdots&a_nb_n\\ \end{pmatrix}\)

Then \(\text{Tr}[|\psi\rangle\langle\phi|]=\sum_{i=1}^{n}a_{i}b_{i}\), also \(\langle\phi|\psi\rangle=\sum_{i=1}^{n}a_{i}b_{i}=\text{Tr}[|\psi\rangle\langle\phi |]\)

Proposition

Theorem

\(P\geq 0\) and \(\text{Tr}(P^2)=\text{Tr}(P)=1\) iff \(P\) is pure

Proof

\(\Rightarrow\))

Let \(\lambda_1,...,\lambda_d\) are eigenvalues of \(P\), then \(\text{Tr}P^2=\text{Tr}P\Rightarrow\sum^d_{x=1}\lambda_x^2=\sum^d_{x=1}\lambda_x=1\)

Then \(\sum^d_{x=1}\lambda_x(1-\lambda_x)=0\).

Since \(P\geq 0\)\, then eigenvalues are non-negative. Since \(\text{Tr}(P)=1\), then eigenvalues are \(\leq 1\)

Then \(0\leq \lambda_x\leq 1\), then \(\lambda_{x}(1-\lambda_{x})=0,\forall x\)

Then there is only one eigenvalue is \(1\), the others are \(0\)

Since \(P\) is positive semi-definite, it is also Hermitian, then it is diagonalizable and have a eigenspace.

By spectral decomposition: \(P = \sum_{x=1}^d \lambda_x |\psi_x\rangle\langle\psi_x|\) with eigenvalues and eigenvectors

Then \(P = |\psi_1\rangle\langle\psi_1|\). Then \(P\) is pure

\(\Leftarrow\))

Since \(P\) is pure, then \(P\geq 0\), \(\text{Tr}[P]=1\) and \(P^2=P\).

Then \(P\geq 0\) and \(\text{Tr}(P^2)=\text{Tr}(P)=1\)

Homework 6: If \(P\) is Hermitian and \(\text{Tr}P^{3}=\text{Tr}P^{2}=\text{Tr}P=1\), then \(P\) is pure state

Proof

Let \(\lambda_1,...,\lambda_d\) are eigenvalues of \(P\), then \(\text{Tr}P^{3}=\text{Tr}P^{2}=\text{Tr}P=1\Rightarrow\sum_{x=1}^{d}\lambda_{x}^{3} =\sum_{x=1}^{d}\lambda_{x}^{2}=\sum_{x=1}^{d}\lambda_{x}=1\)

Then \(\sum_{x=1}^{d}\lambda_{x}^{2}(1-\lambda_{x})=0\)

Since \(\text{Tr}P^2=1\), then \(\lambda_x^{2}\leq 1\), then \(-1\leq \lambda_x\leq 1\)

Then \(1-\lambda_{x}\geq 0\), thus we have \(0\leq \lambda_{x}^{2}\leq 1\) and \(1-\lambda_{x}\geq 0\)

Then there is only one eigenvalue is \(1\), the others are \(0\)

Then the following process is same...

The space of linear operators

Definition

We denote \(\mathcal{L}(A,B)\) be a set of all linear transformations from \(A\) to \(B\) and \(\mathcal{L}(A):=\mathcal{L}(A,A)\)

\(\mathcal{L}(A,B)\) is itself a Hilbert space. Recall: H.S. inner product: \(M,N\in\mathcal{L}(A,B)\) and \(\text{Tr}[M^*N]:=\lang M|N\rang\)

Relation of Hermitian set and Hilbert space

\(\text{Herm}(A)\nsubseteq\mathcal{L}(A)\) because if \(H\in \text{Herm}(A)\), then \((cH)^*=\bar cH^*=\bar cH\)

Thus \(\text{Herm}(A)\) is a Hilbert space over the real number but matrix can have complex coefficients

To summarize, \(\text{Herm}(A)\) is a real Hilbert space.

Theorem

\(\dim(\text{Herm}(A))=|A|^{2}\)

Proof

For an Hermitian matrix, there are two parts.

One is entries on the diagonal. Clearly, the entries should be real. Thus there are \(n\) such bases.

Another is entries off the diagonal. Since the matrix is Hermitian, then we can construct the pair.

For example, \(H(a,b)=x+iy=\overline{H(b,a)}\), thus for each pair \(\{a,b\}\), we need two real bases.

And there are \(1+2+...+(n-1)=\frac{n(n-1)}{2}\) such pairs

Thus totally we have \(n+2\times \frac{n(n-1)}{2}=n^2=|A|^2\)

Example

If \(|A|=2,\dim(\text{Herm}(A))=4=|A|^2\)

For \(\text{Herm}(\mathbb{C^2})\) we have basis of Pauli matrices
\(\sigma_{0},I： \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\) \(\sigma_{1},X： \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\) \(\sigma_{2},Y: \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}\) \(\sigma_{3},Z: \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\)

If we use real number combination, we can get any Hermitian matrix in \(\mathbb{C}^2\)

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