# Prove that $\text{Tr}(M|ψ\rangle\langleϕ|)=\langleϕ|M|ψ\rangle$

### Question:

I am studying alone, and I found p.76 of the book quantum computation and quantum information of nielsen &c huang that: $$\text{Tr}(M |\psi\rangle \langle\psi)=\langle\psi| M |\psi\rangle\,.$$
I want to be sure that my understanding of the formula is correct by proving a more general expression: $$\text{Tr}(M |\psi\rangle\langle\phi|)=\langle\phi| M |\psi\rangle\,,$$ with $$M$$ an operator is correct.

1- We know that $$\text{Tr}(A) = \sum_i \lambda_i\,,$$ with $$\lambda_i$$ all the eigenvalues of $$A$$. Hence with $$|i\rangle$$ the eigenvector associated to the eigenvalue $$\lambda_i$$ we can write that $$\text{Tr}(A) = \sum_i \langle i|A|i\rangle \text{ as }\,\,\, A|i\rangle=\lambda_i |i\rangle\,.$$
Rem: To simplify the scripture I suppose that to each eigenvalue match only one eigenvector (I do not think that it will change a lot my demonstration in the case this is not the case).

2- By identifying $$A=M |\psi\rangle\langle\phi|$$ we can then write (with $$|i\rangle$$ the eigenvectors of $$M |\psi\rangle \langle\phi|$$), \begin{align} \text{Tr}(M |\psi\rangle \langle\phi|) &= \sum_i \langle i|M |\psi\rangle \langle\phi|i\rangle\\ &= \sum_i \langle\phi|i\rangle \langle i|M |\psi\rangle\\ &= \langle \phi| \bigg(\sum_i |i\rangle \langle i| \bigg) M |\psi\rangle \,. \end{align} The term $$\sum_i ∣i⟩⟨i∣ = I$$ is the identity operator since it represents a sum over a complete set of projectors onto an orthonormal basis.

3- Hence we get $$\text{Tr}(M |\psi \rangle \langle\phi|)=\langle\phi| M |\psi \rangle\,.$$

Is this correct? Did I forgot to write some conditions in order to make my prove more precise?

• – glS
Feb 11 at 10:38
• You don't actually need the spectral decomposition of $A$. You can let $|i\rangle$ be any orthonormal basis, and just define the trace to be $\text{Tr}(A)=\sum_i\langle i|A|i\rangle$. Feb 12 at 16:26

Your proof is not general, it assumes implicitly that the operator $$M|\psi\rangle\langle \phi|$$ is diagonalizable (there is a basis of eigenvectors).

You can instead just use basic fact of Trace, namely $$\mathrm{Tr}(AB)=\mathrm{Tr}(BA)$$.

So $$\mathrm{Tr}(M|\psi\rangle\langle \phi|)=\mathrm{Tr}(\langle \phi|M|\psi\rangle)$$

But $$\langle \phi|M|\psi\rangle \in \mathbb{C}$$ is just a scalar so

$$\mathrm{Tr}(\langle \phi|M|\psi\rangle)=\langle \phi|M|\psi\rangle$$

Edit. More insight

Here is a visual proof for those who are unsure about the formula:

$$M|\psi\rangle$$ is an operator applied to vector $$|\psi\rangle$$, the result is a vector which lives in the same state space as $$|\psi\rangle$$. Let us write the matrix representation in a given basis :

\begin{alignat*}{1} M|\psi\rangle&=\begin{bmatrix} \alpha_1 \\ \alpha_2\\ \vdots\\ \alpha_l\\ \vdots\\ \end{bmatrix} \end{alignat*}

$$|\phi\rangle$$ is a vector which lives in same state space as $$|\psi\rangle$$, let its representation be:

\begin{alignat*}{1} |\phi\rangle&=\begin{bmatrix} \beta_1 \\ \beta_2\\ \vdots\\ \beta_l\\ \vdots\\ \end{bmatrix} \end{alignat*}

Its dual vector is represented by the conjugate transpose of the previous column:

\begin{alignat*}{1} \langle\phi|&=\begin{bmatrix} \overline{\beta_1} & \overline{\beta_2} & \dots & \overline{\beta_l} & \dots\\ \end{bmatrix} \end{alignat*}

So now think about what happens when we make the matrix multiplication of a column by a row: we get a square (maybe infinite) matrix

\begin{alignat*}{1} M|\psi\rangle \langle\phi|&=\begin{bmatrix} \alpha_1\overline{\beta_1} & \alpha_1\overline{\beta_1}& \dots &\alpha_1\overline{\beta_i}&\dots&&\\ \alpha_2\overline{\beta_1} & \alpha_2\overline{\beta_2}& \dots &\dots&\dots&&\\ \vdots & & \ddots &&&&\\ \alpha_i\overline{\beta_1} & & &\alpha_i\overline{\beta_i}&&&\\ \vdots & & & &\ddots&&\\ \end{bmatrix} \end{alignat*}

You can check that the trace of the latter is indeed the inner product $$\langle\phi|M|\psi\rangle$$.

• Thank a lot for yout help ! Indeed it is easier and from far better! Feb 10 at 11:43
• Sorry I have question. You use the property that $Tr(AB)=Tr(BA)$ but $A,B$ are matrixes while it is not the case for $|\phi>$ that you ve moved to left. How did you justify this? Feb 10 at 15:47
• $|\phi\rangle$ is a matrix: it is a vector and a vector is matrix with 1 column. The formula is valid for any format; of course $AB$ has to be a square matrix. In quantum physics we may have "infinite" matrix, so this is a kind of generalized matrix multiplication. Feb 10 at 18:50
• Thk a lot for your editing that add useful information Feb 13 at 10:23

$$\newcommand{\bra}[1]{\left<#1\right|}\newcommand{\ket}[1]{\left|#1\right>}\newcommand{\bk}[2]{\left<#1\middle|#2\right>}\newcommand{\bke}[3]{\left<#1\middle|#2\middle|#3\right>}$$ If you want to be more rigorous in proving that $$\text{Tr}(M |\psi\rangle\langle\phi|)=\langle\phi| M |\psi\rangle\,,\tag{1}$$ from scratch, without using any identity/formula, a good approach would be to consider everything in terms of the vector and matrix elements as a summation with respect to the basis and then do some algebra, and everything should work out automatically.

Let $$\{ \ket{i}\}$$ be the basis we are working with. Then, we can write

$$\ket{\psi} = \sum_i a_i \ket{i}\,,\tag2$$ $$\ket{\phi} = \sum_i b_i \ket{i}\,,\tag3$$ $$M = \sum_{ij} m_{ij} \ket{i}\bra{j}\,.\tag4$$

Now, let's first compute RHS of Eq. $$(1)$$.

\begin{align} \langle\phi| M |\psi\rangle &= \bigg(\sum_i b^*_i \bra{i}\bigg)\cdot\bigg(\sum_{jk} m_{jk} \ket{j}\bra{k}\bigg)\cdot\bigg(\sum_l a_l \ket{l}\bigg)\,,\tag{5.1}\\ &= \sum_{ijkl} b_i^* m_{jk}a_l \langle i|j\rangle \langle k|l\rangle\,,\tag{5.2}\\ &= \sum_{ijkl} b_i^* m_{jk}a_l \cdot \delta_{ij} \cdot \delta_{kl}\,,\tag{5.3}\\ &= \sum_{ik} b_i^* m_{ik}a_k\,.\tag{5.4} \end{align}

Now, conmputing the part inside the trace in the LHS of Eq.$$(1)$$,

\begin{align} M \ket{\psi} \bra{\phi} &= \bigg(\sum_{jk} m_{jk} \ket{j}\bra{k}\bigg) \cdot \bigg(\sum_l a_l \ket{l}\bigg) \cdot \bigg(\sum_i b^*_i \bra{i}\bigg)\,,\tag{6.1}\\ &= \sum_{ijkl} m_{jk}a_l b^*_i \ket{j} \langle k|l\rangle\bra{i}\,,\tag{6.2}\\ &= \sum_{ijkl} m_{jk}a_l b^*_i \ket{j}\cdot \delta_{kl} \cdot\bra{i}\,,\tag{6.3}\\ &= \sum_{ijk} m_{jk} a_k b^*_i \ket{j} \bra{i}\,.\tag{6.4} \end{align}

Now, taking trace, \begin{align} \text{Tr} (M \ket{\psi} \bra{\phi}) &= \text{Tr}\bigg( \sum_{ijk} m_{jk} a_k b^*_i \ket{j} \bra{i} \bigg)\,,\tag{7.1}\\ &= \sum_{x} \bra{x} \bigg( \sum_{ijk} m_{jk} a_k b^*_i \ket{j} \bra{i} \bigg) \ket{x}\,,\tag{7.2}\\ &= \sum_{xijk} m_{jk} a_k b^*_i \langle x | j \rangle \langle i | x \rangle\,,\tag{7.3}\\ &= \sum_{xijk} m_{jk} a_k b^*_i \cdot \delta_{xj} \cdot\delta_{ix}\,,\tag{7.4}\\ &= \sum_{ik} m_{ik} a_k b^*_i \tag{7.5}\,. \end{align}

Comparing Eq. $$(5.4)$$ and $$(7.5)$$, we can conclude that we have proved Eq. $$(1)\,.$$