# Mathematical explanation of bra-ket notation in quantum mechanics

$$\newcommand{\hp}[1]{\hphantom{#1}}$$

We have the entangled state of two pairs of qubits:

$$|\psi \rangle =\frac{1}{2}|0011\rangle-\frac{1}{2}|0110\rangle-\frac{1}{2}|1001\rangle+\frac{1}{2}|1100\rangle \tag{01}\label{01}$$

Then unitary transformations $$A$$ and $$B$$ are applied to it:

\begin{align} A & =\:\:\frac{1}{2}\:\, \begin{bmatrix} \hp{-}i & \hp{-}1 &\hp{-}1 & \hp{-}i\hp{..} \\ -i & \hp{-}1 & -1 & \hp{-}i\hp{..} \\ \hp{-}i & \hp{-}1 & -1 & -i\hp{..} \\ -i & \hp{-}1 &\hp{-}1 & -i\hp{..} \end{bmatrix} \tag{02a}\label{02a}\\ B & =\frac{1}{\sqrt{2}} \begin{bmatrix} \hp{-}1 & \,\hp{.}0 & \hp{-}0 & \hp{-}1\hp{-} \\ -1 & \,\hp{.}0 & \hp{-}0 & \hp{-}1\hp{-} \\ \hp{-} 0 & \,\hp{.}1 & \hp{-}1 & \hp{-}0\hp{-}\\ \hp{-} 0 & \,\hp{.}1 & -1 & \hp{-}0\hp{-} \end{bmatrix} \tag{02b}\label{02b} \end{align}

\begin{align} &(A \otimes B) |\psi \rangle =\\ &\frac{1}{2 \sqrt{2}} \left(|0000\rangle -|0010\rangle -|0101\rangle +|0111\rangle +|1001\rangle +|1011\rangle -|1100\rangle -|1110\rangle\right) \tag{03}\label{03} \end{align}

I am not educated that deep in quantum mechanics so I need explanation how the last expression was achieved.

I can compute $$A \otimes B$$ whether it is tensor product or Kronecker product (I am not sure which of the two).

But then how the result of the product is plied to $$|\psi \rangle$$ is not clear to me. I need to know what math is applied.

You can see the source of the problem in this document Quantum Pseudo-Telepathy on page 22.

Here is computed product $$A \otimes B$$ in "Kronecker product form" and in "tensor product form" if that helps.

$$A \otimes B=\frac{1}{2 \sqrt{2}}\left(\tiny{ \begin{array}{cccccccccccccccc} i & 0 & 0 & i & 1 & 0 & 0 & 1 & 1 & 0 & 0 & 1 & i & 0 & 0 & i \\ -i & 0 & 0 & i & -1 & 0 & 0 & 1 & -1 & 0 & 0 & 1 & -i & 0 & 0 & i \\ 0 & i & i & 0 & 0 & 1 & 1 & 0 & 0 & 1 & 1 & 0 & 0 & i & i & 0 \\ 0 & i & -i & 0 & 0 & 1 & -1 & 0 & 0 & 1 & -1 & 0 & 0 & i & -i & 0 \\ -i & 0 & 0 & -i & 1 & 0 & 0 & 1 & -1 & 0 & 0 & -1 & i & 0 & 0 & i \\ i & 0 & 0 & -i & -1 & 0 & 0 & 1 & 1 & 0 & 0 & -1 & -i & 0 & 0 & i \\ 0 & -i & -i & 0 & 0 & 1 & 1 & 0 & 0 & -1 & -1 & 0 & 0 & i & i & 0 \\ 0 & -i & i & 0 & 0 & 1 & -1 & 0 & 0 & -1 & 1 & 0 & 0 & i & -i & 0 \\ i & 0 & 0 & i & 1 & 0 & 0 & 1 & -1 & 0 & 0 & -1 & -i & 0 & 0 & -i \\ -i & 0 & 0 & i & -1 & 0 & 0 & 1 & 1 & 0 & 0 & -1 & i & 0 & 0 & -i \\ 0 & i & i & 0 & 0 & 1 & 1 & 0 & 0 & -1 & -1 & 0 & 0 & -i & -i & 0 \\ 0 & i & -i & 0 & 0 & 1 & -1 & 0 & 0 & -1 & 1 & 0 & 0 & -i & i & 0 \\ -i & 0 & 0 & -i & 1 & 0 & 0 & 1 & 1 & 0 & 0 & 1 & -i & 0 & 0 & -i \\ i & 0 & 0 & -i & -1 & 0 & 0 & 1 & -1 & 0 & 0 & 1 & i & 0 & 0 & -i \\ 0 & -i & -i & 0 & 0 & 1 & 1 & 0 & 0 & 1 & 1 & 0 & 0 & -i & -i & 0 \\ 0 & -i & i & 0 & 0 & 1 & -1 & 0 & 0 & 1 & -1 & 0 & 0 & -i & i & 0 \\ \end{array}} \right)$$

$$A \otimes B=\frac{1}{2 \sqrt{2}}\left(\tiny{ \begin{array}{cccc} \left( \begin{array}{cccc} i & 0 & 0 & i \\ -i & 0 & 0 & i \\ 0 & i & i & 0 \\ 0 & i & -i & 0 \\ \end{array} \right) & \left( \begin{array}{cccc} 1 & 0 & 0 & 1 \\ -1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 0 & 1 & -1 & 0 \\ \end{array} \right) & \left( \begin{array}{cccc} 1 & 0 & 0 & 1 \\ -1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 0 & 1 & -1 & 0 \\ \end{array} \right) & \left( \begin{array}{cccc} i & 0 & 0 & i \\ -i & 0 & 0 & i \\ 0 & i & i & 0 \\ 0 & i & -i & 0 \\ \end{array} \right) \\ \left( \begin{array}{cccc} -i & 0 & 0 & -i \\ i & 0 & 0 & -i \\ 0 & -i & -i & 0 \\ 0 & -i & i & 0 \\ \end{array} \right) & \left( \begin{array}{cccc} 1 & 0 & 0 & 1 \\ -1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 0 & 1 & -1 & 0 \\ \end{array} \right) & \left( \begin{array}{cccc} -1 & 0 & 0 & -1 \\ 1 & 0 & 0 & -1 \\ 0 & -1 & -1 & 0 \\ 0 & -1 & 1 & 0 \\ \end{array} \right) & \left( \begin{array}{cccc} i & 0 & 0 & i \\ -i & 0 & 0 & i \\ 0 & i & i & 0 \\ 0 & i & -i & 0 \\ \end{array} \right) \\ \left( \begin{array}{cccc} i & 0 & 0 & i \\ -i & 0 & 0 & i \\ 0 & i & i & 0 \\ 0 & i & -i & 0 \\ \end{array} \right) & \left( \begin{array}{cccc} 1 & 0 & 0 & 1 \\ -1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 0 & 1 & -1 & 0 \\ \end{array} \right) & \left( \begin{array}{cccc} -1 & 0 & 0 & -1 \\ 1 & 0 & 0 & -1 \\ 0 & -1 & -1 & 0 \\ 0 & -1 & 1 & 0 \\ \end{array} \right) & \left( \begin{array}{cccc} -i & 0 & 0 & -i \\ i & 0 & 0 & -i \\ 0 & -i & -i & 0 \\ 0 & -i & i & 0 \\ \end{array} \right) \\ \left( \begin{array}{cccc} -i & 0 & 0 & -i \\ i & 0 & 0 & -i \\ 0 & -i & -i & 0 \\ 0 & -i & i & 0 \\ \end{array} \right) & \left( \begin{array}{cccc} 1 & 0 & 0 & 1 \\ -1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 0 & 1 & -1 & 0 \\ \end{array} \right) & \left( \begin{array}{cccc} 1 & 0 & 0 & 1 \\ -1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 0 & 1 & -1 & 0 \\ \end{array} \right) & \left( \begin{array}{cccc} -i & 0 & 0 & -i \\ i & 0 & 0 & -i \\ 0 & -i & -i & 0 \\ 0 & -i & i & 0 \\ \end{array} \right) \\ \end{array}} \right)$$

• Can you be more specific what is unclear about this? $\lvert \psi\rangle$ is a 4-qubit state, so it is 16-dimensional, and your $A\otimes B$ are also 16-dimensional. What problem exactly do you have in applying the latter to the former? Nov 24, 2022 at 19:41
• Yes $A\otimes B$ is 16x16 but how to transform $\lvert \psi\rangle$ in 16x16 matrix? Then what operation to use? Dot product or something else? Nov 24, 2022 at 19:48
• Since equation (03) (to be proved) is missing the imaginary unit, I suspect that the authors of the linked article give the result $\:\left(A_3\otimes B_3\right)|\psi\rangle\:$ and not the result $\:\left(A_2\otimes B_3\right)|\psi\rangle$. Note that between the 9 pairs of matrices $\:\left(A_i,B_j\right)-i,j=1,2,3\:$ the pair $\:\left(A_3,B_3\right)\:$ is the only pair with real matrices. I'll check it in a couple of days and if so I'll post an ADDENDUM in my answer. See Page 22 from ''Quantum Pseudo-Telepathy''. Nov 28, 2022 at 8:26

The bra vector can be treated as a row vector, while the ket vector is a column vector. So, if a vector space has $$N$$ dimensions, the row vectors for a basis would be $$\langle 0|, \langle 1|, ⋯, \langle N-1|$$, while the corresponding column vectors for the basis would be $$|0\rangle , |1\rangle , ⋯, |N-1\rangle$$.

Vectors in the tensor product $$H⊗H'$$ of two vector spaces $$H$$ and $$H'$$ can be punned as vectors in a larger vector space, such that if $$u ∈ H$$ and $$u' ∈ H'$$, then the corresponding column vectors would be denoted by concatenation as $$|u\rangle ⊗|u'\rangle = |uu'\rangle$$ and the dual row vector as $$\langle u'|⊗\langle u| = \langle u'u|$$. The reversal of order extends the algebra consistently such that the adjoint $$A^†$$ remains an involution operation (i.e. $$A^{††} = A$$ and $$(AB)^† = B^† A^†$$) - applicable now to all matrices, both square and non-square. Correspondingly, $$|u\rangle ^† = \langle u|$$, $$\langle u|^† = |u\rangle$$, similarly for $$\langle u'|$$ and $$|u'\rangle$$, and $$|uu'\rangle ^† = (|u\rangle ⊗|u'\rangle )^† = |u'\rangle ^†⊗|u\rangle ^† = \langle u'|⊗\langle u| = \langle u'u|.$$

Matrices, algebraically, are all sums of ket-bra combinations, e.g. $$|0\rangle \langle 0|$$ is the unit matrix whose only non-zero element is 1 at the upper left corner. So, the tensor product of matrices can be defined in such a way that $$(A⊗A')|u\rangle ⊗|u'\rangle = A|u\rangle ⊗A'|u'\rangle = |(Au)(A'u')\rangle .$$ Thus, if $$A = |v\rangle \langle w|$$ and $$A' = |v'\rangle \langle w'|$$, then $$(A⊗A')|u\rangle ⊗|u'\rangle = (|v\rangle \langle w||u\rangle )⊗(|v'\rangle \langle w'||u'\rangle ) = |vv'\rangle \langle w|u\rangle \langle w'|u'\rangle .$$ Algebraically, if we can treat the product sequentially, then $$\langle w'w| |uu'\rangle = \langle w'|\langle w||u\rangle |u'\rangle = \langle w|u\rangle \langle w'||u'\rangle = \langle w|u\rangle \langle w'|u'\rangle .$$ Hence, with that convention, we have: $$(A⊗A')|uu'\rangle = |vv'\rangle \langle w|u\rangle \langle w'|u'\rangle = |v\rangle |v'\rangle \langle w'|\langle w| |uu'\rangle = |v\rangle A'\langle w| |uu'\rangle .$$ Thus, the tensor product for matrices can be defined in terms of the bras and kets by: $$|v\rangle \langle w|⊗A' = |v\rangle A'\langle w|.$$

For qubits, the vector space has 2 basis elements. For brevity, we will define the following: $$b = \langle 0|, \hspace 1em p = \langle 1|, \hspace 1em d = |0\rangle , \hspace 1em q = |1\rangle ,$$ as well as the corresponding tensor products - with their renumbering as elements $$0, 1, 2, 3$$: $$bb = \langle 00| = \langle 0|, \hspace 1em pb = \langle 10| = \langle 1|, \hspace 1em bp = \langle 01| = \langle 2|, \hspace 1em pp = \langle 11| = \langle 3|,\\ dd = |00\rangle = |0\rangle , \hspace 1em dq = |01\rangle = |1\rangle , \hspace 1em qd = |10\rangle = |2\rangle , \hspace 1em qq = |11\rangle = |3\rangle .$$ So, the tensor product of two 1-qubit spaces serves as a 4-dimensional vector space with 4 basis elements.

If we adopt the axioms: $$bd = 1 = pq, \hspace 1em bq = 0 = pd, \hspace 1em db + qp = 1$$ then all the matrix operations can be represented. In fact, the associative linear algebra generated from the 4 elements $$\{b,d,p,q\}$$, given, in the abstract, by these relations, is universal. It contains every finite dimensional real (and complex) vector and matrix algebra within it! So, it's not just for qubits, it can be used for everything.

Footnote: this applies generally to semi-rings, as well. In particular, in Computer Science, if you attach this algebra to the algebra of regular expressions, the result is an algebra for context-free expressions molded almost directly on the Chomsky-Schützenberger Theorem. So, in effect, Chomsky and Schützenberger are the Faraday to our Maxwell for this new paradigm, though we haven't gotten around (quite yet) to writing our "Maxwell's Treatise" for context-free expressions. Consider this advanced notice.

From these relations, in fact, you can write $$\langle i| |j\rangle = δ^i_j, \hspace 1em (i, j = 0, 1, 2, 3), \\ |0\rangle \langle 0| + |1\rangle \langle 1| + |2\rangle \langle 2| + |3\rangle \langle 3| = 1$$ where the Kroenecker delta is defined by $$δ^i_j = 1$$ if $$i = j$$ and $$δ^i_j = 0$$ if $$i ≠ j$$, for $$i, j = 0, 1, 2, 3$$.

The last of these relations demonstrated as follows: $$ddbb + dqpb + qdbp + qqpp = d(db + qp)b + q(db + qp)p = d1b + q1p = db + qp = 1.$$

With these conventions in place, we can write: \begin{align} ψ &= ½ (ddqq - dqqd - qddq + qqdd),\\ A &= ½ i(ddbb + ddpp - dqbb + dqpp + qdbb - qdpp - qqbb - qqpp) \\ &+ ½(ddpb + ddbp + dqpb - dqbp + qdpb - qdbp + qqpb + qqbp),\\ B &= √½ (ddbb + ddpp - dqbb + dqpp + qdpb + qdbp + qqpb - qqbp). \end{align} According to your account, which is inherited from the account given in your reference, we should have: \begin{align} A⊗B ψ &= √⅛ (dddd - ddqd - dqdq + dqqq + qddq + qdqq - qqdd - qqqd) \\ &= √⅛ ((dd - qq)dd - (dd + qq)qd + (qd - dq)dq + (dq + qd)qq). \end{align} We won't, by the way. It's wrong. They made a mistake, and you can already see that clearly: there are no factors of $$i$$ in their product, even though they appear in the matrix $$A$$.

We have the following reductions: \begin{align} A &= ½ (i(d + q)(d - q)bb + (d + q)(d + q)pb + i(d - q)(d + q)pp + (d - q)(d - q)bp), \\ B &= √½ (d(d - q)bb + q(d + q)pb + d(d + q)pp + q(d - q)bp). \end{align} We can multiply the matrices one at a time, by using the conventions $$Aψ = A⊗Iψ, \hspace 1em Bψ = I⊗Bψ$$ noting that the general identity applies to matrix tensor products: $$(A⊗B)(A'⊗B') = AA'⊗BB'$$ so that we can actually write the product $$A⊗B$$ in either of two orders: $$(A⊗I)(I⊗B) = A⊗B = (I⊗B)(A⊗I).$$

Thus, applying $$B$$ first, we get: $$Bψ = ½√½ (ddd(d + q) - dqq(d - q) - qdq(d + q) + qqd(d - q))$$ For instance, showing how this applies to a given term, we have: $$B ddqq = I⊗B dd⊗qq = (Idd)⊗(Bqq), \\ Bqq = √½ (d(d - q)bbqq + q(d + q)pbqq + d(d + q)ppqq + q(d - q)bpqq) = √½ d(d + q),$$ since $$bbqq = 0$$, $$pbqq = 0$$, $$bpqq = 0$$ all cancel, while $$ppqq = 1$$.

Applying $$A$$ next, to this result, we get, with a little algebra: \begin{align} A⊗Bψ &= ¼√½ i(d + q)(d - q)d(d + q) - ¼√½ (d + q)(d + q)q(d - q) \\ & - ¼√½ (d - q)(d - q)q(d + q) + ¼√½ i(d - q)(d + q)d(d - q) \\ & = ¼√½ i((d + q)(d - q) + (d - q)(d + q))dd \\ & - ¼√½ ((d + q)(d + q) + (d - q)(d - q))qd \\ & + ¼√½ i((d + q)(d - q) - (d - q)(d + q))dq \\ & + ¼√½ ((d + q)(d + q) - (d - q)(d - q))qq \\ & = ½√½ (i(dd - qq)dd - (dd + qq)qd + i(qd - dq)dq + (dq + qd)qq). \end{align} You and your reference are both missing the factors of $$i$$ on the terms containing the kets $$dddd = |0000\rangle$$, $$qqdd = |1100\rangle$$, $$qdqq = |1011\rangle$$ and $$dqdq = |0101\rangle$$.

I suspect that that incongruity was what you were really asking about, wasn't it? Remember: the reference is only a pre-print. The answer is that they made a mistake.

• You can use rangle and langle for $\langle$ and $\rangle$ Nov 25, 2022 at 10:05
• Yes, that was the source of my problem - missing $i$ factor in some of the terms. But I never doubted about correctness of the article rather I thought I am maybe doing something wrong since I am new to this topic so how could I doubt any article. The answers that was posted before your answer did not mention any mistakes in the article so this just confirmed I might be doing something wrong. Nov 25, 2022 at 12:03
• I was going to get around to it, after seeing the layout, but J. Murray got to it first and fixed it up (thanks). You can also use the characters directly: 〈 (U+2329) and 〉 (U+232a) ... or ⟨ (U+27e8) and ⟩ (U+27e9), which is what I would normally use. Nov 28, 2022 at 2:33

$$(A \otimes B) |\psi \rangle$$ equals another vector which can be calculated by the matrix multiplication of $$A\otimes B$$ and $$|\psi \rangle$$

Assuming the components of $$(A\otimes B)$$ are in the matrix $$m^{i}_{j}$$ where $$0\leq i, j \leq 15$$ ($$i$$ is the row index and $$j$$ is the column index) , and the components of $$|\psi \rangle$$ are in the column matrix $$v^j$$ where $$0\leq j \leq 15$$, then the matrix multiplication is:

$$\sum _{j=0}^{15} m^{i}_j v^{j}$$

The result is another column matrix $$u^i$$, which represents the components of the vector $$(A\otimes B) |\psi \rangle$$ in the basis we did the computation in.

To motivate the matrix multiplication definition, you have to learn that $$u^i$$ is a linear transformation of $$v^j$$

$$A\otimes B$$ is a $$16\times 16$$ matrix and $$|\psi\rangle$$ is a 16-dimensional vector. You can write $$|\psi\rangle$$ as a linear combination of basis vectors: \begin{align} |0000\rangle = |0\rangle\otimes|0\rangle\otimes|0\rangle\otimes|0\rangle = \begin{bmatrix}1\\0\\\vdots\\0\end{bmatrix}, \\ |0001\rangle = |0\rangle\otimes|0\rangle\otimes|0\rangle\otimes|1\rangle = \begin{bmatrix}0\\1\\\vdots\\0\end{bmatrix}, \\ \end{align} and so on... To get $$|\psi\rangle$$ you just take the linear combination of the basis vectors above.

In this case, you have written $$|\psi\rangle$$ in a basis with notation $$|b_N b_{N-1}\ldots b_2 b_1\rangle$$ with where $$b_i \in \{0,1\}$$. Note that $$|\psi\rangle$$ lives in a Hilbert space of $$2^N$$ dimensions.

You could equivalently label each basis state as $$|n\rangle= \sum_{k=1}^N 2^{k\cdot b_k}$$. It may be most convenient to write the number $$n$$ in the familiar decimal form: $$|\psi\rangle = \frac{1}{2} \big( |3\rangle + |6\rangle + |9\rangle + |12\rangle \big).$$

Then, the basis state $$|n\rangle$$ when written in the Kronecker representation is the vector with zero in all positions except for the $$n$$-th spot, which should have a $$1$$.

Another way to figure out the representation is to recursively applying the Kronecker product. Remember that $$|b_N b_{N-1}\ldots b_2 b_1\rangle= |b_N\rangle\otimes |b_{N-1}\rangle\otimes \ldots \otimes |b_2 \rangle\otimes |b_1\rangle$$, and the representation of a single qubit.

$$\newcommand{\bl}[1]{\boldsymbol{#1}} \newcommand{\e}{\bl=} \newcommand{\p}{\bl+} \newcommand{\m}{\bl-} \newcommand{\mb}[1]{\mathbf {#1}} \newcommand{\mc}[1]{\mathcal {#1}} \newcommand{\mr}[1]{\mathrm {#1}} \newcommand{\mf}[1]{\mathfrak{#1}} \newcommand{\gr}{\bl>} \newcommand{\les}{\bl<} \newcommand{\greq}{\bl\ge} \newcommand{\leseq}{\bl\le} \newcommand{\il}[1]{\:#1\:} \newcommand{\plr}[1]{\left(#1\right)} \newcommand{\blr}[1]{\left[#1\right]} \newcommand{\vlr}[1]{\left\vert#1\right\vert} \newcommand{\Vlr}[1]{\left\Vert#1\right\Vert} \newcommand{\lara}[1]{\left\langle#1\right\rangle} \newcommand{\lav}[1]{\left\langle#1\right|} \newcommand{\vra}[1]{\left|#1\right\rangle} \newcommand{\lavra}[2]{\left\langle#1|#2\right\rangle} \newcommand{\lavvra}[3]{\left\langle#1\right|#2\left|#3\right\rangle} \newcommand{\vp}{\vphantom{\dfrac{a}{b}}} \newcommand{\Vp}[1]{\vphantom{#1}} \newcommand{\hp}[1]{\hphantom{#1}} \newcommand{\x}{\bl\times} \newcommand{\ox}{\bl\otimes} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\qqlraqq}{\qquad\bl{-\!\!\!-\!\!\!-\!\!\!\longrightarrow}\qquad} \newcommand{\qqLraqq}{\qquad\boldsymbol{\e\!\e\!\e\!\e\!\Longrightarrow}\qquad} \newcommand{\tl}[1]{\tag{#1}\label{#1}}$$ At first your effort to solve your problem via the expression of the product $$\:\plr{\mr A \bl\otimes \mr B}\:$$ as a $$\:16\times 16\:$$ matrix and the kets $$\:\vra{0011},\vra{0110},\vra{1001},\vra{1100}\:$$ as $$\:16\times 1\:$$ one-column matrices is correct. But I think that a better way is to keep working at most with $$\:4\times 4\:$$ and $$\:4\times 1\:$$ matrices. For example, to solve our problem we must find the kets \begin{align} \vra{\xi_1}&\e\plr{\mr A \bl\otimes \mr B}\vra{0011}\e\plr{\mr A \bl\otimes \mr B\vp}\plr{\vra{00}\bl\otimes \vra{11}\vp}\e\plr{\mr A\vra{00}\bl\otimes\mr B\vra{11}\vp} \tl{04.1}\\ \vra{\xi_2}&\e\plr{\mr A \bl\otimes \mr B}\vra{0110}\e\plr{\mr A \bl\otimes \mr B\vp}\plr{\vra{01}\bl\otimes \vra{10}\vp}\e\plr{\mr A\vra{01}\bl\otimes\mr B\vra{10}\vp} \tl{04.2}\\ \vra{\xi_3}&\e\plr{\mr A \bl\otimes \mr B}\vra{1001}\e\plr{\mr A \bl\otimes \mr B\vp}\plr{\vra{10}\bl\otimes \vra{01}\vp}\e\plr{\mr A\vra{10}\bl\otimes\mr B\vra{01}\vp} \tl{04.3}\\ \vra{\xi_4}&\e\plr{\mr A \bl\otimes \mr B}\vra{1100}\e\plr{\mr A \bl\otimes \mr B\vp}\plr{\vra{11}\bl\otimes \vra{00}\vp}\e\plr{\mr A\vra{11}\bl\otimes\mr B\vra{00}\vp} \tl{04.4} \end{align} So the problem is reduced to : first $$\textbf{(a)}$$ to find the 4-dimensional kets produced by the application of the operators $$\il{\mr A, \mr B}$$ on the basis kets $$\il{\mc E_4\bl\equiv\{\vra{00},\vra{01},\vra{10},\vra{11}\}}$$, that is to find \begin{align} &\mr A\vra{00}\quad \mr A\vra{01}\quad\mr A\vra{10}\quad\mr A\vra{11} \tl{05a}\\ &\mr B\vra{00}\quad \mr B\vra{01}\quad\mr B\vra{10}\quad\mr B\vra{11} \tl{05b} \end{align} and second $$\textbf{(b)}$$ to obtain the 16-dimensional kets $$\il{\vra{\xi_i}}$$ from equations \eqref{04.1} to \eqref{04.4} and products of a member of \eqref{05a} by a member of \eqref{05b}.

For $$\textbf{(a)}$$ note that the matrix representation of the basis kets $$\il{\mc E_4\bl\equiv\{\vra{00},\vra{01},\vra{10},\vra{11}\}}$$ are $$\il{4\times 1}$$ one-column matrices with 3 zeros and one unit since \begin{align} \vra{00} & \e \vra{0}\bl\otimes\vra{0}\e \begin{bmatrix} 1\\ 0 \end{bmatrix}\bl\otimes \begin{bmatrix} 1\\ 0 \end{bmatrix}\e \begin{bmatrix} 1 \\ 0 \\ 0\\ 0 \end{bmatrix} \tl{06.1}\\ \vra{01} & \e \vra{0}\bl\otimes\vra{1}\e \begin{bmatrix} 1\\ 0 \end{bmatrix}\bl\otimes \begin{bmatrix} 0\\ 1 \end{bmatrix}\e \begin{bmatrix} 0 \\ 1 \\ 0\\ 0 \end{bmatrix} \tl{06.2}\\ \vra{10} & \e \vra{1}\bl\otimes\vra{0}\e \begin{bmatrix} 0\\ 1 \end{bmatrix}\bl\otimes \begin{bmatrix} 1\\ 0 \end{bmatrix}\e \begin{bmatrix} 0 \\ 0 \\ 1\\ 0 \end{bmatrix} \tl{06.3}\\ \vra{11} & \e \vra{1}\bl\otimes\vra{1}\e \begin{bmatrix} 0\\ 1 \end{bmatrix}\bl\otimes \begin{bmatrix} 0\\ 1 \end{bmatrix}\e \begin{bmatrix} 0 \\ 0 \\ 0\\ 1 \end{bmatrix} \tl{06.4} \end{align} By above expressions the kets \eqref{05a} and \eqref{05b} are columns of the matrices $$\il{\mr A, \mr B}$$ respectively \begin{align} \mr A\vra{00} & \e \begin{bmatrix} a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34} \\ a_{41} & a_{42} & a_{43} & a_{44} \end{bmatrix} \begin{bmatrix} 1 \\ 0 \\ 0\\ 0 \end{bmatrix}\e \begin{bmatrix} a_{11} \\ a_{21} \\ a_{31}\\ a_{41} \end{bmatrix}\e\tfrac12 \begin{bmatrix} \hp{-}i \\ -i \\ \hp{-}i \\ -i \end{bmatrix} \bl\implies \nonumber\\ \mr A\vra{00} & \e \tfrac12 i\plr{\vra{00}\m\vra{01}\p\vra{10}\m\vra{11}\vp} \tl{07.1}\\ \mr A\vra{01} & \e \begin{bmatrix} a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34} \\ a_{41} & a_{42} & a_{43} & a_{44} \end{bmatrix} \begin{bmatrix} 0 \\ 1 \\ 0\\ 0 \end{bmatrix}\e \begin{bmatrix} a_{12} \\ a_{22} \\ a_{32}\\ a_{42} \end{bmatrix}\e\tfrac12 \begin{bmatrix} \:\:1\:\:\\ \:\:1\:\:\\ \:\:1\:\:\\ \:\:1\:\: \end{bmatrix} \bl\implies \nonumber\\ \mr A\vra{01} & \e \tfrac12 \plr{\vra{00}\p\vra{01}\p\vra{10}\p\vra{11}\vp} \tl{07.2}\\ \mr A\vra{10} & \e \begin{bmatrix} a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34} \\ a_{41} & a_{42} & a_{43} & a_{44} \end{bmatrix} \begin{bmatrix} 0 \\ 0 \\ 1\\ 0 \end{bmatrix}\e \begin{bmatrix} a_{13} \\ a_{23} \\ a_{33}\\ a_{43} \end{bmatrix}\e\tfrac12 \begin{bmatrix} \hp{-}1 \\ -1 \\ -1 \\ \hp{-}1 \end{bmatrix} \bl\implies \nonumber\\ \mr A\vra{10} & \e \tfrac12 \plr{\vra{00}\m\vra{01}\m\vra{10}\p\vra{11}\vp} \tl{07.3}\\ \mr A\vra{11} & \e \begin{bmatrix} a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34} \\ a_{41} & a_{42} & a_{43} & a_{44} \end{bmatrix} \begin{bmatrix} 0 \\ 0 \\ 0\\ 1 \end{bmatrix}\e \begin{bmatrix} a_{14} \\ a_{24} \\ a_{34}\\ a_{44} \end{bmatrix}\e\tfrac12 \begin{bmatrix} \hp{-}i \\ \hp{-}i\\ -i\\ -i \end{bmatrix} \bl\implies \nonumber\\ \mr A\vra{11} & \e \tfrac12 i\plr{\vra{00}\p\vra{01}\m\vra{10}\m\vra{11}\vp} \tl{07.4} \end{align} and \begin{align} \mr B\vra{00} & \e \begin{bmatrix} b_{11} & b_{12} & b_{13} & b_{14} \\ b_{21} & b_{22} & b_{23} & b_{24} \\ b_{31} & b_{32} & b_{33} & b_{34} \\ b_{41} & b_{42} & b_{43} & b_{44} \end{bmatrix} \begin{bmatrix} 1 \\ 0 \\ 0\\ 0 \end{bmatrix}\e \begin{bmatrix} b_{11} \\ b_{21} \\ b_{31}\\ b_{41} \end{bmatrix}\e\tfrac{1}{\sqrt 2} \begin{bmatrix} \hp{-}1 \\ -1 \\ \hp{-}0 \\ \hp{-}0 \end{bmatrix} \bl\implies \nonumber\\ \mr B\vra{00} & \e \tfrac{1}{\sqrt 2}\plr{\vra{00}\m\vra{01}\vp} \tl{08.1}\\ \mr B\vra{01} & \e \begin{bmatrix} b_{11} & b_{12} & b_{13} & b_{14} \\ b_{21} & b_{22} & b_{23} & b_{24} \\ b_{31} & b_{32} & b_{33} & b_{34} \\ b_{41} & b_{42} & b_{43} & b_{44} \end{bmatrix} \begin{bmatrix} 0 \\ 1 \\ 0\\ 0 \end{bmatrix}\e \begin{bmatrix} b_{12} \\ b_{22} \\ b_{32}\\ b_{42} \end{bmatrix}\e\tfrac{1}{\sqrt 2} \begin{bmatrix} \:\:0\:\:\\ \:\:0\:\:\\ \:\:1\:\:\\ \:\:1\:\: \end{bmatrix} \bl\implies \nonumber\\ \mr B\vra{01} & \e \tfrac{1}{\sqrt 2} \plr{\vra{10}\p\vra{11}\vp} \tl{08.2}\\ \mr B\vra{10} & \e \begin{bmatrix} b_{11} & b_{12} & b_{13} & b_{14} \\ b_{21} & b_{22} & b_{23} & b_{24} \\ b_{31} & b_{32} & b_{33} & b_{34} \\ b_{41} & b_{42} & b_{43} & b_{44} \end{bmatrix} \begin{bmatrix} 0 \\ 0 \\ 1\\ 0 \end{bmatrix}\e \begin{bmatrix} b_{13} \\ b_{23} \\ b_{33}\\ b_{43} \end{bmatrix}\e\tfrac{1}{\sqrt 2} \begin{bmatrix} \hp{-}0 \\ \hp{-}0 \\ \hp{-}1 \\ -1 \end{bmatrix} \bl\implies \nonumber\\ \mr B\vra{10} & \e \tfrac{1}{\sqrt 2} \plr{\vra{10}\m\vra{11}\vp} \tl{08.3}\\ \mr B\vra{11} & \e \begin{bmatrix} b_{11} & b_{12} & b_{13} & b_{14} \\ b_{21} & b_{22} & b_{23} & b_{24} \\ b_{31} & b_{32} & b_{33} & b_{34} \\ b_{41} & b_{42} & b_{43} & b_{44} \end{bmatrix} \begin{bmatrix} 0 \\ 0 \\ 0\\ 1 \end{bmatrix}\e \begin{bmatrix} b_{14} \\ b_{24} \\ b_{34}\\ b_{44} \end{bmatrix}\e\tfrac{1}{\sqrt 2} \begin{bmatrix} \:\:1\:\:\\ \:\:1\:\:\\ \:\:0\:\:\\ \:\:0\:\: \end{bmatrix} \bl\implies \nonumber\\ \mr B\vra{11} & \e \tfrac{1}{\sqrt 2}\plr{\vra{00}\p\vra{01}\vp} \tl{08.4} \end{align}

From \eqref{04.1} to \eqref{04.4} using \eqref{07.1} to \eqref{07.4} and \eqref{08.1} to \eqref{08.4} we obtain \begin{align} \vra{\xi_1}&\e\plr{\mr A \bl\otimes \mr B}\vra{0011}\e\plr{\mr A\vra{00}\bl\otimes\mr B\vra{11}\vp} \nonumber\\ &\e\frac{i}{2\sqrt 2}\plr{\vra{00}\m\vra{01}\p\vra{10}\m\vra{11}\vp}\bl\otimes\plr{\vra{00}\p\vra{01}\vp}\bl\implies \nonumber\\ \vra{\xi_1}&\e\frac{i}{2\sqrt 2}\plr{\vra{0000}\m\vra{0100}\p\vra{1000}\m\vra{1100}\p\vra{0001}\m\vra{0101}\p\vra{1001}\m\vra{1101}\vp} \tl{09.1}\\ \vra{\xi_2}&\e\plr{\mr A \bl\otimes \mr B}\vra{0110}\e\plr{\mr A\vra{01}\bl\otimes\mr B\vra{10}\vp} \nonumber\\ &\e\frac{1}{2\sqrt 2}\plr{\vra{00}\p\vra{01}\p\vra{10}\p\vra{11}\vp}\bl\otimes\plr{\vra{10}\m\vra{11}\vp}\bl\implies \nonumber\\ \vra{\xi_2}&\e\frac{1}{2\sqrt 2}\plr{\vra{0010}\p\vra{0110}\p\vra{1010}\p\vra{1110}\m\vra{0011}\m\vra{0111}\m\vra{1011}\m\vra{1111}\vp} \tl{09.2}\\ \vra{\xi_3}&\e\plr{\mr A \bl\otimes \mr B}\vra{1001}\e\plr{\mr A\vra{10}\bl\otimes\mr B\vra{01}\vp} \nonumber\\ &\e\frac{1}{2\sqrt 2}\plr{\vra{00}\m\vra{01}\m\vra{10}\p\vra{11}\vp}\bl\otimes\plr{\vra{10}\p\vra{11}\vp}\bl\implies \nonumber\\ \vra{\xi_3}&\e\frac{1}{2\sqrt 2}\plr{\vra{0010}\m\vra{0110}\m\vra{1010}\p\vra{1110}\p\vra{0011}\m\vra{0111}\m\vra{1011}\p\vra{1111}\vp} \tl{09.3}\\ \vra{\xi_4}&\e\plr{\mr A \bl\otimes \mr B}\vra{1100}\e\plr{\mr A\vra{11}\bl\otimes\mr B\vra{00}\vp} \nonumber\\ &\e\frac{i}{2\sqrt 2}\plr{\vra{00}\p\vra{01}\m\vra{10}\m\vra{11}\vp}\bl\otimes\plr{\vra{00}\m\vra{01}\vp}\bl\implies \nonumber\\ \vra{\xi_4}&\e\frac{i}{2\sqrt 2}\plr{\vra{0000}\p\vra{0100}\m\vra{1000}\m\vra{1100}\m\vra{0001}\m\vra{0101}\p\vra{1001}\p\vra{1101}\vp} \tl{09.4} \end{align} From (01) and \eqref{09.1} to \eqref{09.4} we have \begin{align} &\plr{\mr A \bl\otimes \mr B} \vra\psi \e \frac12 \plr{\vra{\xi_1}\m\vra{\xi_2}\m\vra{\xi_3}\p\vra{\xi_4}\vp}\e \frac12\plr{\vra{\xi_1}\p\vra{\xi_4}\vp}\m\frac12\plr{\vra{\xi_2}\p\vra{\xi_3}\vp} \nonumber\\ &\e \frac{i}{2\sqrt 2}\plr{\vra{0000}\m\vra{1100}\m\vra{0101}\p\vra{1001}\vp}\m\frac{1}{2\sqrt 2}\plr{\vra{0010}\p\vra{1110}\m\vra{0111}\m\vra{1011}\vp} \nonumber\\ &\e \frac{1}{2\sqrt 2}\blr{i\plr{\vra{0000}\m\vra{1100}\m\vra{0101}\p\vra{1001}\vp}\m \plr{\vra{0010}\p\vra{1110}\m\vra{0111}\m\vra{1011}\vp}\Vp{\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}}}} \tl{10} \end{align}

So equation (03) of the post is wrong.

• Yes this is the result I got too. The equation $(03)$ is not mine but from the article (look the link in OP) which I did not suspect to have mistakes. I admit that your way and explanation is better and more systematic than using just one large multiplication in which the result may be correct but the details to better understand quantum mechanics are not all revealed in it. Nov 26, 2022 at 21:09
• Since equation (03) (to be proved) is missing the imaginary unit, I suspect that the authors of the linked article give the result $\:\left(A_3\otimes B_3\right)|\psi\rangle\:$ and not the result $\:\left(A_2\otimes B_3\right)|\psi\rangle$. Note that between the 9 pairs of matrices $\:\left(A_i,B_j\right)-i,j=1,2,3\:$ the pair $\:\left(A_3,B_3\right)\:$ is the only pair with real matrices. I'll check it in a couple of days and if so I'll post an ADDENDUM in my answer. See Page 22 from ''Quantum Pseudo-Telepathy''. Nov 28, 2022 at 8:28
• While it is true that $\left(A_3\otimes B_3\right)|\psi\rangle$ would contain no imaginary unit but on the other hand it contains different components than $\left(A_2\otimes B_3\right)|\psi\rangle$. Nov 28, 2022 at 13:15
• @azerbajdzan : As in my previous comment we must check this. Nov 28, 2022 at 13:18
• OK :-) but I already checked it. $\left(A_2\otimes B_3\right)|\psi\rangle$ contains ${0000,0010,0101,0111,1001,1011,1100,1110}$ while $\left(A_3\otimes B_3\right)|\psi\rangle$ contains ${0001,0010,0100,0111,1000,1011,1101,1110}$. (with some factors, of course, which I omitted) Nov 28, 2022 at 13:22