$\require{cancel} \newcommand{\Ket}[1]{\left|{#1}\right\rangle} \newcommand{\Bra}[1]{\left\langle{#1}\right|} \newcommand{\Braket}[1]{\left\langle{#1}\right\rangle} \newcommand{\Rsr}[1]{\frac{1}{\sqrt{#1}}} \newcommand{\RSR}[1]{1/\sqrt{#1}} \newcommand{\Verti}{\rvert} \newcommand{\HAT}[1]{\hat{\,#1~}} \DeclareMathOperator{\Tr}{Tr}$

Hilbert Space Constructions

^{First created in July 2018}

This page is about constructing Hilbert Spaces of higher dimensions from lower ones.

In this context, let the basis of $V$ be $A=\{\Ket{\alpha_1},\Ket{\alpha_2},\ldots,\Ket{\alpha_n}\}$ and the basis of $W$ be $B=\{\Ket{\beta_1},\Ket{\beta_2},\ldots,\Ket{\beta_m}\}$.

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$V\oplus W$ denotes the direct sum of two vector spaces $V$ and $W$.
When $U=V\oplus W$, we say that $V\oplus W$ is a decomposition of $U$.
The basis of $V\oplus W$ will be $A\cup B$, and $\dim(V\oplus W)=\dim(V)+\dim(W)$.
If $V$ and $W$ are inner product spaces, $(\Bra{v_2}\oplus\Bra{w_2})~~(\Ket{v_1}\oplus\Ket{w_1})=\Braket{v_2\Verti v_1}+\Braket{w_2\Verti w_1}$. $\quad\Big(?\Braket{v\Verti w}=0\text{ for all }v\in V\text{ and }w\in W\Big)$.

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$V\otimes W$ denotes the Kronecker product of two vector spaces $V$ and $W$.
The basis of $V\otimes W$ will be $\left\{\Ket{\alpha_i}\otimes\Ket{\beta_j}~:~i=1,2,\ldots,n;j=1,2,\ldots,m\right\}$, and $\dim(V\otimes W)=\dim(V)\times\dim(W).$

Ref: Kronecker product

$(A\otimes B)~(C\otimes D)=(AC)\otimes(BD).~~~~ (A\otimes B)^{-1}=(A^{-1}\otimes B^{-1}).~~~~ (A\otimes B)^\dagger=(A^\dagger\otimes B^\dagger).$

$\Ket{vw}:=\Ket v\otimes\Ket w.~~~ \Bra{wv}:=\Bra w\otimes\Bra v.$

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$\Ket{vw}^\dagger=\Bra{vw}.$

Example:

Let $\Ket v=[1,2i,3]^T$, and $\Ket w=[4i,5]^T,~~ \Ket{vw}=[4i,5,-8,10i,12i,15]^T,~~ \Ket{vw}^\dagger=[-4i,5,-8,-10i,-12i,15] .$

$\Bra{wv} =\Bra{w}\otimes\Bra{v} =[-4i,5]\otimes[1,-2i,3] =[-4i,-8,-12i,5,-10i,15] \ne\Ket{vw}^\dagger .$

$\Bra{vw} =\Bra{v}\otimes\Bra{w} =[1,-2i,3]\otimes[-4i,5] =[-4i,5,-8,-10i,-12i,15] =\Ket{vw}^\dagger .$

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$\Ket{v_2w_2}\Bra{v_1w_1} =(\Ket{v_2}\otimes\Ket{w_2})~(\Bra{v_1}\otimes\Bra{w_1}) =\Ket{v_2}\Bra{v_1}\otimes\Ket{w_2}\Bra{w_1}$.

$\Braket{v_1w_1\Verti v_2w_2} =\Bra{v_1w_1}\Ket{v_2w_2} =(\Bra{v_1}\otimes\Bra{w_1})~~(\Ket{v_2}\otimes\Ket{w_2}) =\Braket{v_1\Verti v_2}\Braket{w_1\Verti w_2}.$

$\displaystyle (\Ket{v_1}+\Ket{v_2})\otimes\Ket{w}=\Ket{v_1}\otimes\Ket{w}+\Ket{v_2}\otimes\Ket{w}.\\ \Ket{v}\otimes(\Ket{w_1}+\Ket{w_2})=\Ket{v}\otimes\Ket{w_1}+\Ket{v}\otimes\Ket{w_2}.\\ (a\Ket{v})\otimes\Ket{w}=\Ket{v}\otimes(a\Ket{w})=a(\Ket{v}\otimes\Ket{w}) .$

$\Ket v\Bra w=\Ket v\otimes\Bra w.$

If the bases of $V$ and $W$ are orthonormal, so will that of $V\otimes W$.

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Example: Let basis of $V$ be $\{\Ket0,\Ket1\}$ and of $W$ be $\{\Ket{00},\Ket{01},\Ket{10},\Ket{11}\}$.
$\Ket{v}=a_0\Ket0+a_1\Ket1$ and $\Ket{w}=b_{00}\Ket{00}+b_{01}\Ket{01}+b_{10}\Ket{10}+b_{11}\Ket{11}$.
$\Ket{v}\otimes\Ket{w}= \left(a_0 b_{00}\Ket{000}+a_0 b_{01}\Ket{001}+a_0 b_{10}\Ket{010}+a_0 b_{11}\Ket{011}\right)+ \left(a_1 b_{00}\Ket{100}+a_1 b_{01}\Ket{101}+a_1 b_{10}\Ket{110}+a_1 b_{11}\Ket{111}\right)$.

Some shorthand examples:

• $\Ket{101}=\Ket1\otimes\Ket{01}$.
• We may use decimal inside the Ket based on $n$-qubit context.
  e.g. $\Ket{5}$ as $\Ket{101}$, but in a 4-qubit system $\Ket{13}=\Ket{1101}$ and $\Ket{5}$ would mean $\Ket{0101}$.
• The normalisation of the coefficients are optional but recommended. (It is considered the same state before and after normalisation.)
• Sometimes the Braket notation might be omitted too, leaving the state bared, like 00+11.

Notes:

1. The state space for an $n$-qubit system has $2^n$ dimensions.

2. Not all elements in $V\otimes W$ can be written as $\Ket{v}\otimes\Ket{w}$. (Those that can are sometimes written as $\Ket{v}\Ket{w}$, or even $\Ket{vw}$, omitting $\otimes$.)

3. If $\Ket{\psi}\in V\otimes W$ cannot be written as $\Ket{v}\otimes\Ket{w}$, we call that $\psi$ is in an entangled state.

4. Entanglement is basis dependent. $\Ket{\psi}$ may not be entangled (factorisable) if decomposed into a different basis.

5. Multiple vectors may refer to the same state due to normalisation and standardisation (relative phase). If you only consider vectors with the first (left most) non-zero coefficient being positively real (no phase), they are one-one corresponding to unique state.