A linear map from $\mathbb{C}^n \otimes \overline{\mathbb{C}^n}$ to $M_n(\mathbb{C})$

Question

$\newcommand{\bra}[1]{\left<#1\right|}\newcommand{\ket}[1]{\left|#1\right>}\newcommand{\bk}[2]{\left<#1\middle|#2\right>}$

I am trying to establish some properties of the linear map: $T:\mathbb{C}^n \otimes \overline{\mathbb{C}^n} \to M_n(\mathbb{C})$ given by: $T(\ket{i} \otimes \ket{j}) = \ket{i}\bra{j} \text{ , } 0 \leq i,j \leq n-1$.

Where $\otimes$ denotes the tensor product, $\overline{\mathbb{C}^n}$ is the conjugate space of the complex (finite) Hilbert space $\mathbb{C}^n$, and $M_n(\mathbb{C})$ is the set of $n\times n$ matrices over $\mathbb{C}$.

I want to show the following:

(i) T is a bijection that preserves inner products (under the Hilbert-Schmidt inner product on $M_n(\mathbb{C})$

(ii) $T(\ket{\psi}\otimes\ket{\varphi} = \ket{\psi}\bra{\varphi}$ for every $\ket{\psi}, \ket{\varphi} \in \mathbb{C}^n$.

I think part of my trouble is figuring out the Dirac notation / how the outer product works. I'm not sure exactly what $\ket{i}\bra{j}$ looks like as a matrix (obviously it's a matrix in $M_n(\mathbb{C})$ but I'm not quite sure what the entries look like).

Also, I think part (ii) of the question is an immediate consequence from the fact that $T$ is a linear, inner product preserving bijection, but I'm a bit confused about why $\ket{\varphi}$ is now coming from $\mathbb{C}^n$ rather than $\overline{\mathbb{C}^n}$.

Any help is appreciated.

Answer 1

In my opinion, this is one of several troubles with Dirac's notation. When you see the elements of $\mathbb C^n$ ("kets") as columns vectors, the "bras" are the adjoints, i.e. their conjugate tranposes. So you have, in $\mathbb C^2$, $$ |1\rangle=\begin{bmatrix} 1\\0\end{bmatrix},\ \ \langle 1|=\begin{bmatrix} 1\\0\end{bmatrix}^*=\begin{bmatrix} 1&0\end{bmatrix}. $$ Mathematicians usually use $\{e_1,\ldots,e_n\}$ for the canonical basis of $\mathbb C^n$, and so your map is $$ T(e_i\otimes e_j)=e_ie_j^*. $$ And now you can calculate the entries of the matrix explicitly.

To see that $T$ is a bijection, you just check that $\{e_i\otimes e_j\}_{i,j=1,\ldots,n}$ is a basis of $\mathbb C^n\otimes\mathbb C^n$, and that $\{e_ie_j^*\}_{i,j=1,\ldots,n}$ is a basis of $M_n(\mathbb C)$ (we usually call the matrices $e_ie_j^*$ the "matrix units").

For the inner product, since the aforementioned bases are orthonormal bases, it is enough to check that $T$ preserves the inner product of basis elements. We have \begin{align} \langle T(e_i\otimes e_j),T(e_s\otimes e_t)\rangle &=\langle e_ie_j^*,e_se_t^*\rangle=\operatorname{Tr}((e_se_t^*)^*e_ie_j^*) =\operatorname{Tr}(e_te_s^*e_ie_j^*) =e_s^*e_i\,\operatorname{Tr}(e_te_j^*)\\ &=\delta_{i,s}\,\delta_{j,t}=\langle e_i,e_s\rangle\,\langle e_j,e_t\rangle\\ &=\langle e_i\otimes e_j,e_s\otimes e_t\rangle. \end{align}

As for your last question, $\langle \varphi|\in\overline{\mathbb C^n}$ is the adjoint of $|\varphi\rangle\in\mathbb C^n$. The confusion arises is that the domain of $T$ is $\mathbb C\otimes \mathbb C$. Then, on the right, you take the adjoint of the second coordinate.

And yes, part (ii) follows as you say: if $\varphi=\sum_{s=1}^n \varphi_se_s$ and $\psi=\sum_{t=1}^n \psi_te_t$, then \begin{align} T(\varphi\otimes\psi) &=\sum_{s,t} \varphi_s\overline{\psi_t}\,T(e_s\otimes e_t)\\ &=\sum_{s,t} \varphi_s\overline{\psi_t}\,e_se_t^*\\ &=\left(\sum_s\varphi_se_s\right)\,\left(\sum_t\psi_te_t\right)^*\\ &=\varphi\psi^*. \end{align}

Answer 2

Since Martin Argerami already gave the answer, the purpose of my post is to clarify the Dirac notation.

Let $|\varphi\rangle$ denotes a vector in some complex inner product space $(V, \langle\cdot \mid \cdot\rangle)$ and $\langle \psi|$ denotes a dual vector in $V^\ast$. Here $ \langle \psi|$ maps $V$ to $\mathbb{C}$ via $\langle \psi|\Big(|\varphi\rangle \Big)=\langle \psi \mid \varphi\rangle \in \mathbb{C}$.

Matrix Representation of Vectors: If $\mathcal{B}=\{|j\rangle\}_{j=1}^n$ is an orthonormal basis for $V$ and $\mathcal{B}^\ast=\{\langle j|\}_{j=1}^n$ is the dual basis for $V^\ast$ (i.e. $\langle i\mid j\rangle = \delta_{i, j}$, the Kronecker delta), then we see that \begin{align} |\varphi\rangle = \sum^n_{j=1}\alpha_j |j\rangle=\sum^n_{j=1} \langle j\mid\varphi\rangle |j\rangle \ \ \text{ and } \ \ \langle \varphi| = \sum^n_{j=1} \beta_j\langle j| = \sum^n_{j=1}\langle\varphi\mid j\rangle \langle j| \end{align} or in matrix form we have the array of complex numbers \begin{align} [|\varphi\rangle]_\mathcal{B} = \begin{pmatrix} \langle 1\mid \varphi\rangle\\ \langle 2 \mid \varphi\rangle\\ \vdots\\ \langle n\mid \varphi\rangle \end{pmatrix} \in \mathbb{C}^n \ \ \text{ and } \ \ [\langle \varphi|]_{\mathcal{B}^\ast} = \begin{pmatrix} \langle \varphi\mid 1\rangle, \langle \varphi\mid 2\rangle, \ldots, \langle \varphi\mid n\rangle \end{pmatrix} \in (\mathbb{C}^n)^\ast. \end{align} By the conjugate symmetry of the complex inner product ($\langle \varphi\mid \psi\rangle = \overline{\langle \psi\mid \varphi\rangle}=: \langle \psi^\ast\mid \varphi^\ast\rangle$), we see that the two arrays of numbers are related to one another via conjugate transpose. Hence it follows \begin{align} \langle \psi \mid \varphi\rangle = \sum^n_{j=1} \langle \psi|j\rangle \langle j\mid\varphi\rangle = \sum^n_{j=1} \overline{\langle j\mid \psi\rangle} \langle j\mid\varphi\rangle. \end{align}

Matrix Representation of Some Operators: Next, by rewriting $|\varphi\rangle$, one can get \begin{align} |\varphi\rangle = \sum^n_{j=1}|j\rangle\langle j\mid \varphi\rangle =: \left(\sum^n_{j=1}|j\rangle\langle j| \right)|\varphi\rangle \end{align} where $|j\rangle \langle j|$ are defined to be linear transformations from $V\rightarrow V$ via $|j\rangle \langle j|\Big(|\varphi\rangle \Big)=\langle j\mid \varphi\rangle|j\rangle$. Then it follows that \begin{align} \sum^n_{j=1}|j\rangle\langle j| = I \end{align} called the resolution of the identity. Also, note that matrix representation of $|j\rangle\langle j|$ in $\mathcal{B}$ is given by \begin{align} [|1\rangle\langle 1|]_\mathcal{B} = \begin{pmatrix} 1\\ 0 \\ \vdots\\ 0 \end{pmatrix} \begin{pmatrix} 1 & 0& \cdots & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 & \cdots & 0\\ 0 & & & \\ \vdots & &\mathbf{0} & \\ 0 & & & \end{pmatrix} \in M_n(\mathbb{C}). \end{align} Likewise for $j=2, \ldots, n$. Also, we called $|j\rangle \langle j|$ the projection onto $|j\rangle$.

In the more general case of $|\varphi\rangle \langle \psi|:V\rightarrow V$, it's not hard to see that \begin{align} [|\varphi\rangle \langle \psi|]_\mathcal{B} =&\ \begin{pmatrix} \langle 1\mid \varphi\rangle\\ \langle 2 \mid \varphi\rangle\\ \vdots\\ \langle n\mid \varphi\rangle \end{pmatrix} \begin{pmatrix} \langle \psi\mid 1\rangle & \langle \psi\mid 2\rangle & \ldots &\langle \psi\mid n\rangle \end{pmatrix}\\ =& \begin{pmatrix} \langle 1\mid \varphi\rangle \langle \psi\mid 1\rangle & \langle 1\mid \varphi\rangle\langle \psi\mid 2\rangle & \cdots & \cdots & \langle 1\mid \varphi\rangle \langle \psi\mid n\rangle \\ \langle 2\mid \varphi\rangle \langle \psi\mid 1\rangle & \langle 2\mid \varphi\rangle\langle \psi\mid 2\rangle & \cdots & \cdots & \langle 2\mid \varphi\rangle \langle \psi\mid n\rangle \\ \vdots & \vdots & \ddots & \vdots & \vdots\\ \langle n-1\mid \varphi\rangle \langle \psi\mid 1\rangle & \cdots & \cdots & \langle n-1\mid \varphi\rangle \langle \psi\mid n-1\rangle & \langle n-1\mid \varphi\rangle \langle \psi\mid n\rangle \\ \langle n\mid \varphi\rangle \langle \psi\mid 1\rangle & \cdots & \cdots & \langle n\mid \varphi\rangle \langle \psi\mid n\rangle & \langle n\mid \varphi\rangle \langle \psi\mid n\rangle \\ \end{pmatrix}\in M_n(\mathbb{C}) \end{align} where $\langle i\mid \varphi\rangle \langle \psi\mid j\rangle$ are called the matrix elements of $|\varphi\rangle\langle \psi|$. Moreover, we see that $\{|i\rangle \langle j|\}$ forms a basis for $M_n(\mathbb{C})$.

Next, note that we can also write \begin{align} [|\varphi\rangle \langle \psi|]_\mathcal{B} =&\ \begin{pmatrix} \langle 1\mid \varphi\rangle\\ \langle 2 \mid \varphi\rangle\\ \vdots\\ \langle n\mid \varphi\rangle \end{pmatrix} \begin{pmatrix} \overline{\langle 1 \mid \psi\rangle} & \overline{\langle 2\mid \psi\rangle} & \ldots & \overline{\langle n\mid \psi\rangle} \end{pmatrix}\\ =&\ [|\varphi\rangle]_\mathcal{B}\otimes_o [|\psi^\ast\rangle]_\mathcal{B} =[|\varphi\rangle\otimes_o |\psi^\ast\rangle]_\mathcal{B} \end{align} where $\otimes_o$ indicates the outer product of two arrays. In short, we see that \begin{align} \mathbb{C}^n\otimes_0 (\mathbb{C}^n)^\ast=[V\otimes_o V^\ast]_\mathcal{B}=M_n(\mathbb{C}) \end{align} or equivalently in a more abstract language \begin{align} V\otimes V^\ast \simeq \text{Hom}(V, V). \end{align}

Answer: As pointed out by Martin Argerami, what you want to show is that $V\otimes V \simeq V\otimes V^\ast$. Thus $T:\mathbb{C}^n\otimes \mathbb{C}^n\rightarrow \mathbb{C}^n\otimes (\mathbb{C}^n)^\ast$ via \begin{align} T\left( \begin{pmatrix} \langle 1\mid \varphi\rangle\\ \langle 2 \mid \varphi\rangle\\ \vdots\\ \langle n\mid \varphi\rangle \end{pmatrix}\otimes_o\begin{pmatrix} \langle 1\mid \psi\rangle\\ \langle 2 \mid \psi\rangle\\ \vdots\\ \langle n\mid \psi\rangle \end{pmatrix}\right) =\begin{pmatrix} \langle 1\mid \varphi\rangle\\ \langle 2 \mid \varphi\rangle\\ \vdots\\ \langle n\mid \varphi\rangle \end{pmatrix}\otimes_o\begin{pmatrix} \overline{\langle 1\mid \psi\rangle}\\ \overline{\langle 2 \mid \psi\rangle}\\ \vdots\\ \overline{\langle n\mid \psi\rangle} \end{pmatrix}. \end{align}