📂Hilbert Space

Unitary Operator

Definition

Let the Hilbert space $H$ and the bounded linear operator $T : H \to H$ be bijective, and satisfy the following; then it is called a unitary operator.

$$ T^{\ast} = T^{-1} $$

$T^{\ast}$ is the adjoint operator of $T$, and $T^{-1}$ is the inverse operator of $T$.

Explanation

The conditions of the definition can be restated as follows.

$$ TT^{\ast} = I = T^{\ast}T $$

$I$ is the identity operator. The definition of the adjoint operator is as follows.

$$ \braket{T\mathbf{v}, \mathbf{w}} = \braket{\mathbf{v}, T^{\ast}\mathbf{w}} $$

Since the adjoint of the adjoint is the operator itself, the inverse of a unitary operator is also unitary.

$$ (T^{-1})^{\ast} = (T^{\ast})^{\ast} = T \implies T^{-1}(T^{-1})^{\ast} = I $$

In finite dimensions, i.e., in terms of matrices, $T^{\ast}$ is the conjugate transpose matrix of $T$, and $T^{\ast}T$ is the identity matrix. In other words, a unitary operator is a generalization of a unitary matrix.

Properties¹

Let $H$ be a Hilbert space, and let $U : H \to H$ and $V : H \to H$ be unitary operators. Then the following hold.

(a) $U$ being unitary is equivalent to the following holding. $$ \braket{U \mathbf{v}, U \mathbf{w}} = \braket{\mathbf{v}, \mathbf{w}}, \quad \forall \mathbf{v}, \mathbf{w} \in H $$ (a-1) A unitary operator is an isometric mapping. $$ \| U \mathbf{v} \| = \| \mathbf{v} \|, \quad \forall \mathbf{v} \in H $$

(a-2) If $H \ne \{0\}$, then the operator norm of $U$ is given by $$ \| U \| = 1 $$

(b) $U^{-1}$ is also unitary.

(c) $UV = U \circ V$ is also unitary.

(d) $U$ is a normal operator.

Proof

(a)

$(\implies)$ Assume $U$ is unitary. Then by the definition of the adjoint operator, the following holds.

$$ \braket{U \mathbf{v}, U \mathbf{w}} = \braket{\mathbf{v}, U^{\ast}(U \mathbf{w})} = \braket{\mathbf{v}, I\mathbf{w}} = \braket{\mathbf{v}, \mathbf{w}} $$

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$(\impliedby)$ Conversely, assume $\braket{U \mathbf{v}, U \mathbf{w}} = \braket{\mathbf{v}, \mathbf{w}}$ holds. Also, by the definition of the adjoint operator, the following holds.

$$ \braket{U \mathbf{v}, U \mathbf{w}} = \braket{\mathbf{v}, U^{\ast}(U \mathbf{w})} $$

Then we obtain

$$ \braket{\mathbf{v}, \mathbf{w}} = \braket{\mathbf{v}, U^{\ast}(U \mathbf{w})} \implies \braket{\mathbf{v}, \mathbf{w}} - \braket{\mathbf{v}, U^{\ast}(U \mathbf{w})} = 0 $$

$$ \implies \braket{\mathbf{v}, \mathbf{w} - U^{\ast}(U \mathbf{w})} = 0, \quad \forall \mathbf{v}, \mathbf{w} \in H $$

Property of the zero vector

$$ \forall \mathbf{x}\in X,\ \left\langle \mathbf{x},\mathbf{y} \right\rangle = 0 \implies \mathbf{y}=\mathbf{0} $$

By the property of the zero vector in an inner-product space, the following holds.

$$ \mathbf{w} - U^{\ast}U \mathbf{w} = \mathbf{0} $$

Therefore the following holds.

$$ U^{\ast}U \mathbf{w} = \mathbf{w} \implies U^{\ast}U = I $$

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(b)

It suffices to show $(U^{-1})^{\ast} = (U^{-1})^{-1}$. From the assumption that $U$ is unitary and from properties of the adjoint operator, we get the following.

$$ (U^{-1})^{\ast} = (U^{\ast})^{\ast} = U = (U^{-1})^{-1} $$

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(c)

Proof 1

Since $U$ is unitary, by (a) we have

$$ \braket{U(V\mathbf{x}), U(V\mathbf{y})} = \braket{(V\mathbf{x}), (V\mathbf{y})}, \quad \forall \mathbf{x}, \mathbf{y} \in H $$

Since $V$ is unitary, the following holds.

$$ \braket{V\mathbf{x}, V\mathbf{y}} = \braket{\mathbf{x}, \mathbf{y}}, \quad \forall \mathbf{x}, \mathbf{y} \in H $$

Therefore the following holds, and by (a) $UV$ is unitary.

$$ \braket{UV\mathbf{x}, UV\mathbf{y}} = \braket{\mathbf{x}, \mathbf{y}}, \quad \forall \mathbf{x}, \mathbf{y} \in H $$

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Proof 2

Properties of the adjoint operator

$$ (UV)^{\ast} = V^{\ast}U^{\ast} $$

Properties of the inverse operator

$$ (UV)^{-1} = V^{-1}U^{-1} $$

By the properties of the adjoint operator and the inverse operator,

$$ (UV)^{\ast} = V^{\ast}U^{\ast} = V^{-1}U^{-1} = (UV)^{-1} $$

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