*B.1.2 Probabilistic circular teleportation scheme for arbitrary one-qubit states*

Not only is the seven-qubit resource state useful for linear and bidirectional teleportation but can also facilitate the probabilistic teleportation of an arbitrary single-qubit states in a circular manner between three networknodes (users). Let us say we have Alice, Bob and Charlie in the system, with the first qubit used as a control qubit, qubits 1 and 4 given to Alice, qubits 2 and 6 given to Bob and qubits 3 and 7 given to Charlie. Let us say the arbitrary states are ∣*ψA*i ¼ *αA*∣0*A*i þ *βA*∣1*A*i, ∣*ψB*i ¼ *αB*∣0*B*i þ *βB*∣1*B*i and ∣*ψC*i ¼ *αC*∣0*C*i þ *βC*∣1*C*i. Then, the composite state is given by ∣*ψA*i ⊗ ∣*ψB*i ⊗ ∣*ψC*i ⊗ ∣Γ7i*TA*1*B*1*C*1*A*1*B*1*C*<sup>1</sup> , where ∣Γ7i*<sup>T</sup>* is the control qubit. We apply a CNOT gate using the qubits A, B and C of the arbitrary states as the control-qubits and the first qubits of each user as the target-qubit. Let us for simplicity only consider the case where ∣Γ7i*<sup>T</sup>* ¼ ∣0i.

Let us now measure the first qubits of Alice, Bob and Charlie in the Z-basis. Let us say ∣Γ7i*<sup>A</sup>*1*B*1*C*<sup>1</sup> ¼ ∣010i, then we have the state ∣*ψ*<sup>00</sup> i¼� <sup>1</sup> <sup>4</sup> ððj001i j010iÞ*αAαBαC*∣0*A*0*B*0*C*i þ jð 100i j111iÞ*αAαBβC*∣0*A*0*B*1*C*iþ jð 000i þ j011iÞ*αAβBαC*∣0*A*1*B*0*C*iþ jð 101iþ j110iÞ*αAβBβC*∣0*A*1*B*1*C*i þ ð∣100i ∣111iÞ*βAαBαC*∣1*A*0*B*0*C*i þ ð Þ �j001iþj010i *βAαBβC*∣1*A*0*B*1*C*iþ �ð ∣101i ∣110iÞ*βAβBαC*∣1*A*1*B*0*C*i þ ð Þ j000iþj011i *βAβBβC*∣1*A*1*B*1*C*iÞ. We can now measure the control qubits in the X-basis. So, let us say, we have ∣*QAQBQC*i ¼ ∣þ*<sup>A</sup>*�*B*þ*<sup>C</sup>*i, then we obtain the state <sup>∣</sup>*C*1i j*A*1i �j*B*1i þ *<sup>χ</sup>*j*B*3i � *<sup>χ</sup>*�<sup>1</sup> <sup>ð</sup> <sup>ð</sup> <sup>j</sup> *<sup>B</sup>*4i�j*B*2iÞ þ j*A*2i �j ð þ *<sup>B</sup>*1i þ *<sup>χ</sup>*j*B*3<sup>i</sup> *<sup>χ</sup>*�<sup>1</sup>j*B*4iþj*B*2iÞÞ þ <sup>∣</sup>*C*<sup>2</sup> i j*A*1i �j*B*1i � *<sup>χ</sup>*j*B*3i � *<sup>χ</sup>*�<sup>1</sup> <sup>ð</sup> <sup>ð</sup> <sup>j</sup>*B*4iþj*B*2iÞþj *<sup>A</sup>*2i �j*B*1i � *<sup>χ</sup>*j*B*3i þ *<sup>χ</sup>*�<sup>1</sup> <sup>ð</sup> <sup>j</sup>*B*4i�j*B*<sup>2</sup> iÞÞ þ <sup>∣</sup>*C*3i j*A*1i j*B*4i � *<sup>χ</sup>*j*B*2i � *<sup>χ</sup>*�<sup>1</sup> <sup>ð</sup> <sup>ð</sup> <sup>j</sup> *<sup>B</sup>*i�j*B*3iÞ þ j*A*2i j*B*4i � *<sup>χ</sup>*j*B*2i þ *<sup>χ</sup>*�<sup>1</sup> ð j <sup>j</sup> *<sup>B</sup>*1i�*B*3iÞÞ þ <sup>∣</sup>*C*4i j*A*1i j*B*4i þ *<sup>χ</sup>*j*B*2i � *<sup>χ</sup>*�<sup>1</sup> ð j <sup>ð</sup> <sup>j</sup>*B*1i�j*B*3iÞþj*A*2i jð *<sup>B</sup>*4i þ *<sup>χ</sup>*<sup>j</sup> *<sup>B</sup>*2i þ *<sup>χ</sup>*�<sup>1</sup><sup>j</sup> *<sup>B</sup>*1iþ*B*3iÞÞ, where ∣*C*1i ¼ *βC*∣0i þ *α<sup>C</sup>* ∣1i, ∣*C*2i ¼ *βC*∣0i � *αC*∣1i, ∣*C*<sup>3</sup> i ¼ *βC*∣1i þ *αC*∣0i, ∣*C*4i ¼ *βC*∣1i � *α<sup>C</sup>* ∣0i, ∣*B*1i ¼ *αB*∣1i þ *βB*∣0i, ∣*B*2i ¼ *αB*∣1i� *βB*∣0i, ∣*B*3i ¼ *αB*∣0i þ *<sup>β</sup>B*∣1i, <sup>∣</sup>*B*4i ¼ *<sup>α</sup>B*∣0i� *<sup>β</sup>B*∣1i, *<sup>χ</sup>* <sup>¼</sup> *<sup>a</sup>*<sup>2</sup> *<sup>a</sup>*<sup>1</sup> with *a*<sup>1</sup> ¼ *β<sup>A</sup>* þ *αA*, *a*<sup>2</sup> ¼ *α<sup>A</sup>* � *βA*, ∣*A*1i ¼ *a*<sup>1</sup> ∣0i þ *a*2∣1i, ∣*A*1i ¼ *a*1∣0i � *a*2∣1i. Therefore I see that the users can obtain states derived from the original state of the users next to them Alice ð ! Bob ! Charlie ! Alice). However, as you can see, this can be done in a probabilistic manner with one of the users not quite obtaining the original state but rather a derivative-state based on the original.

#### **B.2 Quantum teleportation of arbitrary two-qubit state**

#### *B.2.1 Linear teleportation scheme*

Similarly, an arbitrary two qubit quantum state can be teleported using the resource-state. In this case Alice possesses qubits 1, 2, 3, 4 and 5, and the 6th and 7th particles belong to Bob. Alice wants to transport an arbitrary state <sup>∣</sup>*ψ*ð Þ<sup>2</sup> i ¼ *<sup>α</sup>*∣00i þ *<sup>μ</sup>*∣10i þ *<sup>γ</sup>*∣01i þ *<sup>β</sup>*∣11<sup>i</sup> to Bob. The combined state of the system is <sup>∣</sup>Γð Þ<sup>2</sup> <sup>7</sup> i ¼ <sup>∣</sup>*ψ*ð Þ<sup>2</sup> <sup>i</sup> ⊗ ∣Γ7i,

$$\begin{aligned} \vert \Gamma\_{\mathcal{T}}^{(L)} \rangle &= a(A\_{00} \vert 00 \rangle + A\_{01} \vert 01 \rangle + A\_{10} \vert 10 \rangle + A\_{11} \vert 11 \rangle) \\ &+ \mu(B\_{00} \vert 00 \rangle + B\_{01} \vert 01 \rangle + B\_{10} \vert 10 \rangle + B\_{11} \vert 11 \rangle) \\ &+ \gamma(C\_{00} \vert 00 \rangle + C\_{01} \vert 01 \rangle + C\_{10} \vert 10 \rangle + C\_{11} \vert 11 \rangle) \\ &+ \beta(D\_{00} \vert 00 \rangle + D\_{01} \vert 01 \rangle + D\_{10} \vert 10 \rangle + D\_{11} \vert 11 \rangle) \end{aligned} \tag{A6}$$

where

*<sup>A</sup>*<sup>00</sup> <sup>¼</sup> <sup>∣</sup>0000i∣Ψ<sup>0</sup> *GHZ*i þ <sup>∣</sup>0001i∣Ψ<sup>4</sup> *GHZ*i � <sup>∣</sup>0010i∣Ψ<sup>5</sup> *GHZ*i � <sup>∣</sup>0011i∣Ψ<sup>7</sup> *GHZ*i, *<sup>A</sup>*<sup>11</sup> <sup>¼</sup> <sup>∣</sup>0000i∣Ψ<sup>1</sup> *GHZ*i þ <sup>∣</sup>0001i∣Ψ<sup>5</sup> *GHZ*i∣0010i∣Ψ<sup>2</sup> *GHZ*i∣0011i∣Ψ<sup>7</sup> *GHZ*i, *<sup>A</sup>*<sup>01</sup> <sup>¼</sup> <sup>∣</sup>0000i∣Ψ<sup>6</sup> *GHZ*i � <sup>∣</sup>0001i∣Ψ<sup>2</sup> *GHZ*i þ <sup>∣</sup>0010i∣Ψ<sup>4</sup> *GHZ*i þ <sup>∣</sup>0011i∣Ψ<sup>0</sup> *GHZ*i, *<sup>A</sup>*<sup>10</sup> ¼ �∣0000i∣Ψ<sup>7</sup> *GHZ*i � <sup>∣</sup>0001i∣Ψ<sup>3</sup> *GHZ*i þ <sup>∣</sup>0010i∣Ψ<sup>5</sup> *GHZ*i þ <sup>∣</sup>0011i∣Ψ<sup>1</sup> *GHZ*i, *<sup>B</sup>*<sup>00</sup> <sup>¼</sup> <sup>∣</sup>1000i∣Ψ<sup>0</sup> *GHZ*i þ <sup>∣</sup>1001i∣Ψ<sup>4</sup> *GHZ*i∣1010i∣Ψ<sup>2</sup> *GHZ*i þ <sup>∣</sup>1011i∣Ψ<sup>6</sup> *GHZ*i, *<sup>B</sup>*<sup>11</sup> <sup>¼</sup> <sup>∣</sup>1000i∣Ψ<sup>0</sup> *GHZ*i þ <sup>∣</sup>1001i∣Ψ<sup>5</sup> *GHZ*i � <sup>∣</sup>1010i∣Ψ<sup>3</sup> *GHZ*i � <sup>∣</sup>1011i∣Ψ<sup>7</sup> *GHZ*i, *<sup>B</sup>*<sup>01</sup> <sup>¼</sup> <sup>∣</sup>1000i∣Ψ<sup>6</sup> *GHZ*i � <sup>∣</sup>1001i∣Ψ<sup>2</sup> *GHZ*i þ <sup>∣</sup>1010i∣Ψ<sup>4</sup> *GHZ*i þ <sup>∣</sup>1011i∣Ψ<sup>0</sup> *GHZ*i, *<sup>B</sup>*<sup>10</sup> ¼ �∣1000i∣Ψ<sup>7</sup> *GHZ*i � <sup>∣</sup>1001i∣Ψ<sup>3</sup> *GHZ*i þ <sup>∣</sup>1010i∣Ψ<sup>5</sup> *GHZ*i þ <sup>∣</sup>1011i∣Ψ<sup>1</sup> *GHZ*i, *<sup>C</sup>*<sup>00</sup> <sup>¼</sup> <sup>∣</sup>0100i∣Ψ<sup>0</sup> *GHZ*i þ <sup>∣</sup>0101i∣Ψ<sup>4</sup> *GHZ*i � <sup>∣</sup>0110i∣Ψ<sup>3</sup> *GHZ*i þ <sup>∣</sup>0111i∣Ψ<sup>6</sup> *GHZ*i, *<sup>C</sup>*<sup>11</sup> <sup>¼</sup> <sup>∣</sup>0100i∣Ψ<sup>1</sup> *GHZ*i þ <sup>∣</sup>0101i∣Ψ<sup>5</sup> *GHZ*i � <sup>∣</sup>0110i∣Ψ<sup>3</sup> *GHZ*i � <sup>∣</sup>0111i∣Ψ<sup>7</sup> *GHZ*i, *<sup>C</sup>*<sup>01</sup> <sup>¼</sup> <sup>∣</sup>0100i∣Ψ<sup>6</sup> *GHZ*i � <sup>∣</sup>0101i∣Ψ<sup>2</sup> *GHZ*i∣0110i∣Ψ<sup>4</sup> *GHZ*i þ <sup>∣</sup>0111i∣Ψ<sup>0</sup> *GHZ*i, *<sup>C</sup>*<sup>10</sup> <sup>¼</sup> <sup>∣</sup>0100i∣Ψ<sup>7</sup> *GHZ*i � <sup>∣</sup>0101i∣Ψ<sup>3</sup> *GHZ*i þ <sup>∣</sup>0110i∣Ψ<sup>5</sup> *GHZ*i þ <sup>∣</sup>0111i∣Ψ<sup>1</sup> *GHZ*i, *<sup>D</sup>*<sup>00</sup> <sup>¼</sup> <sup>∣</sup>1100i∣Ψ<sup>0</sup> *GHZ*i þ <sup>∣</sup>1101i∣Ψ<sup>4</sup> *GHZ*i � <sup>∣</sup>1110i∣Ψ<sup>2</sup> *GHZ*i þ <sup>∣</sup>1111i∣Ψ<sup>6</sup> *GHZ*i, *<sup>D</sup>*<sup>11</sup> <sup>¼</sup> <sup>∣</sup>1100i∣Ψ<sup>1</sup> *GHZ*i þ <sup>∣</sup>1101i∣Ψ<sup>5</sup> *GHZ*i∣1110i∣Ψ<sup>3</sup> *GHZ*i∣1111i∣Ψ<sup>7</sup> *GHZ*i, *<sup>D</sup>*<sup>01</sup> <sup>¼</sup> <sup>∣</sup>1100i∣Ψ<sup>6</sup> *GHZ*i � <sup>∣</sup>1101i∣Ψ<sup>2</sup> *GHZ*i þ <sup>∣</sup>1110i∣Ψ<sup>4</sup> *GHZ*i þ <sup>∣</sup>1111i∣Ψ<sup>0</sup> *GHZ*i, *<sup>D</sup>*<sup>10</sup> <sup>¼</sup> <sup>∣</sup>1100i∣Ψ<sup>6</sup> *GHZ*i � <sup>∣</sup>1101i∣Ψ<sup>3</sup> *GHZ*i þ <sup>∣</sup>1110i∣Ψ<sup>5</sup> *GHZ*i þ <sup>∣</sup>1111i∣Ψ<sup>1</sup> *GHZ*i,

where ∣Ψ0,1 *GHZ*i ¼ <sup>1</sup>ffiffi 2 <sup>p</sup> ½ � <sup>j</sup>000i�j111<sup>i</sup> , <sup>∣</sup>Ψ2,3 *GHZ*i ¼ <sup>1</sup>ffiffi 2 <sup>p</sup> ½ � <sup>j</sup>001i�j110<sup>i</sup> , <sup>∣</sup>Ψ4,5 *GHZ*i ¼ 1ffiffi 2 <sup>p</sup> ½ � <sup>j</sup>010i�j101<sup>i</sup> and <sup>∣</sup>Ψ6,7 *GHZ*i ¼ <sup>1</sup>ffiffi 2 <sup>p</sup> ½ � j100i�j011i . Now, Bob can carry out a combination of unitary operations, according to the given table, to obtain the original state teleported by Alice.



#### *B.2.2 Bidirectional teleportation of arbitrary two-qubit states*

The resource-state can also be used for bidirectional quantum teleportation. Bidirectional Controlled Quantum Teleportation (BCQT) protocols have been proposed for multi-qubit resource states, such as five-qubit [81], six-qubit [82, 83], seven-qubit [84–86] and eight-qubit states [87]. Bidirectional Controlled Quantum Teleportation can teleport arbitrary states between two users under the supervision of a third party. Zha et al proposed the first scheme for BCQT of single qubit states using a maximally entangled seve-qubit quantum state [85]. There have been schemes proposed for BCQT that utilise states with the same number of qubits as the quantum channel being used, and thereby realise bidirectional teleportation of arbitrary single-and two-qubit states under the controller Charlie [84, 86].

Let us say Alice and Bob would like to teleport two-qubit states to each other by utilizing the seven-qubit genuinely entangled resource state. We assume the form of the two-qubit states to be

$$|\phi\rangle\_{A\_1A\_2} = a\_0|00\rangle + a\_1|01\rangle + a\_2|10\rangle + a\_3|11\rangle\tag{A7}$$

$$|\phi\rangle\_{B\_1B\_2} = \beta\_0|\mathbf{0}\mathbf{0}\rangle + \beta\_1|\mathbf{0}\mathbf{1}\rangle + \beta\_2|\mathbf{1}\mathbf{0}\rangle + \beta\_3|\mathbf{1}\mathbf{1}\rangle\tag{A8}$$

For the resource-state, let Alice have the qubits 1,4 and 7, while Bob has the qubits 2, 3 and 6 and Charlie has the qubit 5. The steps for the scheme are as follows:


We will now be looking more closely at these steps with a specific one instance to illustrate each step.

*Step 1*: Alice measures qubit 7 of the resource state and *A*<sup>1</sup> in the bell basis. If Alice measures ∣*ψ*þi, the remainder state is

1 4 ffiffi 2 <sup>p</sup> <sup>ð</sup>∣000i∣*ψ*þi *<sup>α</sup>*0j0i6j0i*A*<sup>2</sup> <sup>þ</sup> *<sup>α</sup>*1j0i6j1i*A*<sup>2</sup> <sup>þ</sup> *<sup>α</sup>*2j1i6j0i*A*<sup>2</sup> <sup>þ</sup> *<sup>α</sup>*3j1i6j1i*A*<sup>2</sup> ð Þ ffiffi 2 p þ <sup>∣</sup>001i∣*ϕ*�i *<sup>α</sup>*0j1i6j0i*A*<sup>2</sup> <sup>þ</sup> *<sup>α</sup>*1j1i6j1i*A*<sup>2</sup> <sup>þ</sup> *<sup>α</sup>*2j0i6j0i*A*<sup>2</sup> <sup>þ</sup> *<sup>α</sup>*3j0i6j1i*A*<sup>2</sup> ð Þ ffiffi 2 p þ <sup>∣</sup>010i∣*ψ*�i �*α*0j1i6j0i*A*<sup>2</sup> � *<sup>α</sup>*1j1i6j1i*A*<sup>2</sup> � *<sup>α</sup>*2j0i6j0i*A*<sup>2</sup> � *<sup>α</sup>*3j0i6j1i*A*<sup>2</sup> ð Þ ffiffi 2 p þ <sup>∣</sup>011i∣*ϕ*þi *<sup>α</sup>*0j0i6j0i*A*<sup>2</sup> <sup>þ</sup> *<sup>α</sup>*1j0i6j1i*A*<sup>2</sup> � *<sup>α</sup>*2j1i6j0i*A*<sup>2</sup> � *<sup>α</sup>*3j1i6j1i*A*<sup>2</sup> ð Þ ffiffi 2 p þ <sup>∣</sup>100i∣*ϕ*þi *<sup>α</sup>*0j1i6j0i*A*<sup>2</sup> <sup>þ</sup> *<sup>α</sup>*1j1i6j1i*A*<sup>2</sup> <sup>þ</sup> *<sup>α</sup>*2j0i6j0i*A*<sup>2</sup> <sup>þ</sup> *<sup>α</sup>*3j0i6j1i*A*<sup>2</sup> ð Þ ffiffi 2 p þ <sup>∣</sup>101i∣*ψ*�i �*α*0j0i6j0i*A*<sup>2</sup> � *<sup>α</sup>*1j0i6j1i*A*<sup>2</sup> � *<sup>α</sup>*2j1i6j0i*A*<sup>2</sup> � *<sup>α</sup>*3j1i6j1i*A*<sup>2</sup> ð Þ ffiffi 2 p þ <sup>∣</sup>110i∣*ϕ*�i �*α*0j0i6j0i*<sup>A</sup>*<sup>2</sup> � *<sup>α</sup>*1j0i6j1i*<sup>A</sup>*<sup>2</sup> <sup>þ</sup> *<sup>α</sup>*2j1i6j0i*<sup>A</sup>*<sup>2</sup> <sup>þ</sup> *<sup>α</sup>*3j1i6j1i*<sup>A</sup>*<sup>2</sup> ð Þ ffiffi 2 p þ <sup>∣</sup>111i∣*ψ*þi �*α*0j1i6j0i*<sup>A</sup>*<sup>2</sup> � *<sup>α</sup>*1j1i6j1i*<sup>A</sup>*<sup>2</sup> <sup>þ</sup> *<sup>α</sup>*2j0i6j0i*<sup>A</sup>*<sup>2</sup> <sup>þ</sup> *<sup>α</sup>*3j0i6j1i*<sup>A</sup>*<sup>2</sup> ð Þ ffiffi 2 p Þ∣*ϕ*i*<sup>B</sup>*

*:*

Alice communicates her result to Bob using a classical channel.

*Step 2*: Bob measures qubit 2 of the resource state and *B*<sup>1</sup> in the bell basis. If Bob Measures ∣*ψ*þi, the remainder state is

1 2 ffiffi 2 <sup>p</sup> <sup>ð</sup>∣000<sup>i</sup> *<sup>α</sup>*0j0i6j0i*<sup>A</sup>*<sup>2</sup> <sup>þ</sup> *<sup>α</sup>*1j0i6j1i*<sup>A</sup>*<sup>2</sup> <sup>þ</sup> *<sup>α</sup>*2j1i6j0i*<sup>A</sup>*<sup>2</sup> <sup>þ</sup> *<sup>α</sup>*3j1i6j1i*<sup>A</sup>*<sup>2</sup> ð Þ ffiffi 2 p � 1 ffiffi 2 p *β*0j00i þ *β*1j01i þ *β*2j10i þ *β*<sup>3</sup> ð Þ j11i 4,*B*<sup>2</sup> þ ∣001i � *<sup>α</sup>*0j1i6j0i*<sup>A</sup>*<sup>2</sup> <sup>þ</sup> *<sup>α</sup>*1j1i6j1i*<sup>A</sup>*<sup>2</sup> <sup>þ</sup> *<sup>α</sup>*2j0i6j0i*<sup>A</sup>*<sup>2</sup> <sup>þ</sup> *<sup>α</sup>*3j0i6j1i*<sup>A</sup>*<sup>2</sup> ð Þ ffiffi 2 p � 1 ffiffi 2 p *β*0j00i þ *β*1j01i þ *β*2j10i þ *β*<sup>3</sup> ð Þ j11i 4,*B*<sup>2</sup> þ ∣010i � *<sup>α</sup>*0j0i6j0i*<sup>A</sup>*<sup>2</sup> <sup>þ</sup> *<sup>α</sup>*1j0i6j1i*<sup>A</sup>*<sup>2</sup> � *<sup>α</sup>*2j1i6j0i*<sup>A</sup>*<sup>2</sup> � *<sup>α</sup>*3j1i6j1i*<sup>A</sup>*<sup>2</sup> ð Þ ffiffi 2 p � 1 ffiffi 2 p *β*0j00i þ *β*1j01i *β*2j10i *β*<sup>3</sup> ð Þ j11i 4,*B*<sup>2</sup> þ ∣011i∣ � *<sup>α</sup>*0j0i6j0i*<sup>A</sup>*<sup>2</sup> <sup>þ</sup> *<sup>α</sup>*1j0i6j1i*<sup>A</sup>*<sup>2</sup> � *<sup>α</sup>*2j1i6j0i*<sup>A</sup>*<sup>2</sup> � *<sup>α</sup>*3j1i6j1i*<sup>A</sup>*<sup>2</sup> ð Þ ffiffi 2 p þ 1 ffiffi 2 p *β*0j10i þ *β*1j11i *β*2j00i *β*<sup>3</sup> ð Þ j01i 4,*B*<sup>2</sup> þ ∣100i � *<sup>α</sup>*0j1i6j0i*<sup>A</sup>*<sup>2</sup> <sup>þ</sup> *<sup>α</sup>*1j1i6j1i*<sup>A</sup>*<sup>2</sup> <sup>þ</sup> *<sup>α</sup>*2j0i6j0i*<sup>A</sup>*<sup>2</sup> <sup>þ</sup> *<sup>α</sup>*3j0i6j1i*<sup>A</sup>*<sup>2</sup> ð Þ ffiffi 2 p � 1 ffiffi 2 p *β*0j10i þ *β*1j11i þ *β*2j00i þ *β*<sup>3</sup> ð Þ j01i 4,*B*<sup>2</sup> þ ∣101i � �*α*0j0i6j0i*<sup>A</sup>*<sup>2</sup> � *<sup>α</sup>*1j0i6j1i*<sup>A</sup>*<sup>2</sup> � *<sup>α</sup>*2j1i6j0i*<sup>A</sup>*<sup>2</sup> � *<sup>α</sup>*3j1i6j1i*<sup>A</sup>*<sup>2</sup> ð Þ ffiffi 2 p

$$\begin{split} \mathbb{X} \times & \frac{1}{\sqrt{2}} (\beta\_0 |\mathbf{0}\mathbf{0}\rangle + \beta\_1 |\mathbf{0}\mathbf{1}\rangle \quad -\beta\_2 |\mathbf{1}\mathbf{0}\rangle - \beta\_3 |\mathbf{1}\mathbf{1}\rangle)\_{4,B\_2} + |\mathbf{1}\mathbf{1}\mathbf{0}\rangle \\ \times & \frac{(-a\_0 |\mathbf{0}\rangle\_6 |\mathbf{0}\rangle\_{A\_2} - a\_1 |\mathbf{0}\rangle\_6 |\mathbf{1}\rangle\_{A\_2} + a\_2 |\mathbf{1}\rangle\_6 |\mathbf{0}\rangle\_{A\_2} + a\_3 |\mathbf{1}\rangle\_6 |\mathbf{1}\rangle\_{A\_2})}{\sqrt{2}} \\ \times & \frac{1}{\sqrt{2}} (\beta\_0 |\mathbf{1}\mathbf{0}\rangle + \beta\_1 |\mathbf{1}\mathbf{1}\rangle - \beta\_2 |\mathbf{1}\mathbf{0}\rangle - \beta\_3 |\mathbf{0}\mathbf{1}\rangle)\_{4,B\_2} + |\mathbf{1}\mathbf{1}\mathbf{1}\rangle \\ \times & \frac{(-a\_0 |\mathbf{1}\rangle\_6 |\mathbf{0}\rangle\_{A\_2} - a\_1 |\mathbf{1}\rangle\_6 |\mathbf{1}\rangle\_{A\_2} + a\_2 |\mathbf{0}\rangle\_6 |\mathbf{0}\rangle\_{A\_2} + a\_3 |\mathbf{0}\rangle\_6 |\mathbf{1}\rangle\_{A\_2})}{\sqrt{2}} \\ \times & \frac{1}{\sqrt{2}} (\beta\_0 |\mathbf{0}\mathbf{0}\rangle + \beta\_1 |\mathbf{0}\mathbf{1}\rangle + \beta\_2 |\mathbf{1}\mathbf{0}\rangle + \beta\_3 |\mathbf{1}\mathbf{1}\rangle)\_{4,B\_2}) \end{split}$$

Bob communicates his result via a classical channel to Alice.

*Step 3*: Charlie, Alice and Bob measure their qubits in the Z-basis. Let us say they all measure 0, we have the

$$\begin{split} \frac{1}{2}(|000\rangle\frac{(a\_0|0\rangle\langle 0|\rangle\_{A\_2} + a\_1|0\rangle\langle 1|\rangle\_{A\_2} + a\_2|1\rangle\_6|0\rangle\_{A\_2} + a\_3|1\rangle\_6|1\rangle\_{A\_2})}{\sqrt{2}} \\ \times \frac{1}{\sqrt{2}}(\beta\_0|00\rangle + \beta\_1|01\rangle + \beta\_2|10\rangle + \beta\_3|11\rangle)\_{4,B\_2}) \end{split} \tag{A9}$$

*Step 4:* Let Alice apply a CNOT with *A*<sup>2</sup> as control and qubit 1 as target, and let Bob apply a CNOT with and *B*<sup>2</sup> as control and qubit 3 as target, to get

$$\begin{split} \frac{1}{2\sqrt{2}} (|0\rangle \frac{(a\_0|000\rangle + a\_1|101\rangle\_{A\_1} + a\_2|010\rangle + a\_3|111\rangle)\_{1,6,A\_2}}{\sqrt{2}} \\ \times \frac{1}{\sqrt{2}} (\beta\_0|000\rangle + \beta\_1|101\rangle + \beta\_2|010\rangle + \beta\_3|111\rangle)\_{3,4,B\_1} \end{split} \tag{A10}$$

*Step 4:* Alice and Bob measure their qubits *A*<sup>2</sup> and *B*<sup>2</sup> in the X-basis. Let us say they obtain the state <sup>∣</sup>þi ¼ <sup>1</sup>ffiffi 2 <sup>p</sup> ð Þ j0iþj1i , then the composite state is given by

$$\begin{split} \frac{1}{2}(|\mathbf{0}\rangle \frac{(a\_0|\mathbf{0}\rangle + a\_1|\mathbf{1}\mathbf{0}\rangle\_{A\_1} + a\_2|\mathbf{0}\mathbf{1}\rangle + a\_3|\mathbf{1}\mathbf{1}\rangle\_{1,6}}{\sqrt{2}} \\ \times \frac{1}{\sqrt{2}}(\beta\_0|\mathbf{0}\mathbf{0}\rangle + \beta\_1|\mathbf{1}\mathbf{0}\rangle + \beta\_2|\mathbf{0}\mathbf{1}\rangle + \beta\_3|\mathbf{1}\mathbf{1}\rangle)\_{3,4} \end{split} \tag{A11}$$

*Step 5*: We apply unitary transformations to the composite state to now get Alice's initial arbitrary state in Bob's terminal and Bob's initial arbitrary state in Alice's terminal. In this instance, the unitary transformation is simply *I* ⊗ *I* ⊗ *I* ⊗ *I* with *I* being the identity matrix.

#### **B.3 Quantum teleportation of arbitrary three-qubit state**

The seven-qubit resource state can be used for the perfect linear teleportation of an arbitrary three qubit state. In this case, Alice possesses qubits 1, 2, 3, 4 and 5, and the 6th and 7th particles belong to Bob. Alice wants to transport an arbitrary state <sup>∣</sup>*ψ*ð Þ<sup>3</sup> i ¼ *<sup>a</sup>*∣000i þ *<sup>b</sup>*∣001i þ *<sup>c</sup>*∣010i þ *<sup>d</sup>*∣011i þ *<sup>e</sup>*∣100i þ *<sup>f</sup>*∣101i þ *<sup>g</sup>*∣110i þ *<sup>h</sup>*∣111<sup>i</sup> to Bob. Using the decomposition given in *Supplementary Material*, the states possessed by, and the unitary transforms to be performed by, Bob have been recorded, to

accomplish the teleportation of an arbitrary three-qubit state. A point to note here is that we get the GHZ state for *<sup>a</sup>* <sup>¼</sup> *<sup>h</sup>* <sup>¼</sup> <sup>1</sup>ffiffi 2 <sup>p</sup> , *b* ¼ *c* ¼ *d* ¼ *e* ¼ *f* ¼ *g* ¼ 0 and the W state for *<sup>b</sup>* <sup>¼</sup> *<sup>c</sup>* <sup>¼</sup> *<sup>e</sup>* <sup>¼</sup> <sup>1</sup>ffiffi 3 <sup>p</sup> , *a* ¼ *d* ¼ *f* ¼ *g* ¼ *h* ¼ 0.

The teleportation of an arbitrary three-qubit state using our resource-state has as the initial composite state,

$$\begin{aligned} \left|\hat{r}\_{\gamma}^{(3)}\right\rangle &= \left|\psi^{(3)}\right\rangle \otimes \left|\hat{r}\_{\gamma}^{(1)}\right\rangle \\ &= \left|aA\_{000}|0000\right\rangle + aA\_{001}|0010\rangle + aA\_{010}|010\rangle + aA\_{011}|011\rangle \\ &+ aA\_{100}|000\rangle + aA\_{101}|001\rangle + aA\_{110}|100\rangle + aA\_{111}|111\rangle \\ &+ bB\_{000}|000\rangle + bB\_{001}|001\rangle + bB\_{000}|001\rangle + bB\_{001}|010\rangle \\ &+ bB\_{001}|100\rangle + bB\_{010}|100\rangle + bB\_{011}|110\rangle + bB\_{011}|111\rangle \\ &+ cC\_{000}|000\rangle + cC\_{001}|001\rangle + cC\_{010}|001\rangle + cC\_{011}|011\rangle \\ &+ dD\_{000}|100\rangle + cD\_{001}|001\rangle + cD\_{100}|100\rangle + cD\_{011}|111\rangle \\ &+ dD\_{000}|000\rangle + dD\_{001}|001\rangle + dD\_{000}|001\rangle + dD\_{011}|011\rangle \\ &+ eD\_{000}|100\rangle + eD\_{001}|001\rangle + eD\_{100}|100\rangle + eD\_{011}|111\rangle \\ &+ eE\_{000}|000\rangle + eE\_{010}|100\rangle + eE\_{100}|101\rangle + eE\_{011}|111\rangle \\ &+ fF\_{000}|100\rangle + fF\_{001}|101\rangle + fE\_{101}|100\rangle + fE\_{011}|111\rangle \\ &+ fF\_{00$$

with

$$|A\_{000}\rangle = |00000000\rangle + |00000101\rangle - |0001011\rangle + |00001100\rangle$$

$$|A\_{001}\rangle = |0000010\rangle + |0001001\rangle + |0001100\rangle - |0000011\rangle$$

$$|A\_{010}\rangle = |0000111\rangle - |0000010\rangle + |0001001\rangle + |0001100\rangle$$

$$|A\_{011}\rangle = |0000000\rangle + |0000101\rangle + |0001011\rangle - |0001110\rangle$$

$$|A\_{100}\rangle = |0000011\rangle + |0000100\rangle - |0001000\rangle + |0001101\rangle$$

$$|A\_{101}\rangle = |0000001\rangle - |0000100\rangle + |0001010\rangle + |0001111\rangle$$

$$|A\_{110}\rangle = |0000001\rangle - |0000100\rangle - |0001010\rangle - |0001111\rangle$$

$$|A\_{111}\rangle = |0001101\rangle - |0000011\rangle - |0000110\rangle - |0001000\rangle$$

∣*B*000i ¼ ∣0010000i þ ∣0010101i � ∣0011011i þ ∣0011110i ∣*B*001i ¼ ∣0010010i þ ∣0011001i þ ∣0011100i � ∣0010111i ∣*B*010i ¼ ∣0010111i � ∣0010010i þ ∣0011001i þ ∣0011100i ∣*B*011i ¼ ∣0010000i þ ∣0010101i þ ∣0011011i � ∣0011110i ∣*B*100i ¼ ∣0010011i þ ∣0010110i � ∣0011000i þ ∣0011101i ∣*B*101i ¼ ∣0010001i � ∣0010100i þ ∣0011010i þ ∣0011111i ∣*B*110i ¼ ∣0010001i � ∣0010100i � ∣0011010i � ∣0011111i ∣*B*111i ¼ ∣0011101i � ∣0010011i � ∣0010110i � ∣0011000i ∣*C*000i ¼ ∣0100000i þ ∣0100101i � ∣0101011i þ ∣0101110i ∣*C*001i ¼ ∣0100010i þ ∣0101001i þ ∣0101100i � ∣0100111i ∣*C*010i ¼ ∣0100111i � ∣0100010i þ ∣0101001i þ ∣0101100i ∣*C*011i ¼ ∣0100000i þ ∣0100101i þ ∣0101011i � ∣0101110i ∣*C*100i ¼ ∣0100011i þ ∣0100110i � ∣0101000i þ ∣0101101i ∣*C*101i ¼ ∣0100001i � ∣0100100i þ ∣0101010i þ ∣0101111i ∣*C*110i ¼ ∣0100001i � ∣0100100i � ∣0101010i � ∣0101111i ∣*C*111i ¼ ∣0101101i � ∣0100011i � ∣0100110i � ∣0101000i ∣*D*000i ¼ ∣0110000i þ ∣0110101i � ∣0111011i þ ∣0111110i ∣*D*001i ¼ ∣0110010i þ ∣0111001i þ ∣0111100i � ∣0110111i ∣*D*010i ¼ ∣0110111i � ∣0110010i þ ∣0111001i þ ∣0111100i ∣*D*011i ¼ ∣0110000i þ ∣0110101i þ ∣0111011i � ∣0111110i ∣*D*100i ¼ ∣0110011i þ ∣0110110i � ∣0111000i þ ∣0111101i ∣*D*101i ¼ ∣0110001i � ∣0110100i þ ∣0111010i þ ∣0111111i ∣*D*110i ¼ ∣0110001i � ∣0110100i � ∣0111010i � ∣0111111i ∣*D*111i ¼ ∣0111101i � ∣0110011i � ∣0110110i � ∣0111000i ∣*E*000i ¼ ∣1000000i þ ∣1000101i � ∣1001011i þ ∣1001110i ∣*E*001i ¼ ∣1000010i þ ∣1001001i þ ∣1001100i � ∣1000111i ∣*E*010i ¼ ∣1000111i � ∣1000010i þ ∣0001001i þ ∣0001100i ∣*E*011i ¼ ∣1000000i þ ∣1000101i þ ∣1001011i � ∣1001110i ∣*E*100i ¼ ∣1000011i þ ∣1000110i � ∣1001000i þ ∣1001101i ∣*E*101i ¼ ∣1000001i � ∣1000100i þ ∣1001010i þ ∣1001111i ∣*E*110i ¼ ∣1000001i � ∣1000100i � ∣1001010i � ∣1001111i ∣*E*111i ¼ ∣1001101i � ∣1000011i � ∣1000110i � ∣1001000i ∣*F*000i ¼ ∣1010000i þ ∣1010101i � ∣1011011i þ ∣1011110i ∣*F*001i ¼ ∣1010010i þ ∣1011001i þ ∣1011100i � ∣1010111i ∣*F*010i ¼ ∣1010111i � ∣1010010i þ ∣1011001i þ ∣1011100i ∣*F*011i ¼ ∣1010000i þ ∣1010101i þ ∣1011011i � ∣1011110i ∣*F*100i ¼ ∣1010011i þ ∣1010110i � ∣1011000i þ ∣1011101i ∣*F*101i ¼ ∣1010001i � ∣1010100i þ ∣1011010i þ ∣1011111i ∣*F*110i ¼ ∣1010001i � ∣1010100i � ∣1011010i � ∣1011111i ∣*F*111i ¼ ∣1011101i � ∣1010011i � ∣1010110i � ∣1011000i

∣*G*000i ¼ ∣1100000i þ ∣1100101i � ∣1101011i þ ∣1101110i ∣*G*001i ¼ ∣1100010i þ ∣1101001i þ ∣1101100i � ∣1100111i ∣*G*010i ¼ ∣1100111i � ∣1100010i þ ∣1101001i þ ∣1101100i ∣*G*011i ¼ ∣1100000i þ ∣1100101i þ ∣1101011i � ∣1101110i ∣*G*100i ¼ ∣1100011i þ ∣1100110i � ∣1101000i þ ∣1101101i ∣*G*101i ¼ ∣1100001i � ∣1100100i þ ∣1101010i þ ∣1101111i ∣*G*110i ¼ ∣1100001i � ∣1100100i � ∣1101010i � ∣1101111i ∣*G*111i ¼ ∣1101101i � ∣1100011i � ∣0000110i � ∣0001000i ∣*H*000i ¼ ∣1110000i þ ∣1110101i � ∣1111011i þ ∣1111110i ∣*H*001i ¼ ∣1110010i þ ∣1111001i þ ∣1111100i � ∣1110111i ∣*H*010i ¼ ∣1110111i � ∣1110010i þ ∣1111001i þ ∣1111100i ∣*H*011i ¼ ∣1110000i þ ∣1110101i þ ∣1111011i � ∣1111110i ∣*H*100i ¼ ∣1110011i þ ∣1110110i � ∣1111000i þ ∣1111101i ∣*H*101i ¼ ∣1110001i � ∣1110100i þ ∣1111010i þ ∣1111111i ∣*H*110i ¼ ∣1110001i � ∣1110100i � ∣1111010i � ∣1111111i ∣*H*111i ¼ ∣1111101i � ∣1110011i � ∣1110110i � ∣1111000i

An arbitrary three qubit state can be decomposed in terms of these basis-vectors,

$$\begin{aligned} & \left( a|000\rangle + b|001\rangle + c|010\rangle + d|011\rangle + e|100\rangle \right) \\ &+ f|101\rangle + g|110\rangle + h|111\rangle)|\Psi\_7\rangle \\ &= \sum\_{\text{permutations}} ((-1)^{l\_1} A\_{d1i23\delta\_3} + (-1)^{l\_2} B\_{b\_1b\_2b\_3} \\ &+ (-1)^{l\_3} C\_{e\_1e\_3\delta\_3} + (-1)^{l\_4} D\_{d1i2b\_3} \\ &+ (-1)^{l\_5} E\_{e\_1e\_2\epsilon\_3} + (-1)^{l\_6} F\_{f\_1f\_2f\_3} \\ &+ (-1)^{l\_7} G\_{\mathcal{S}\mathcal{S}\mathcal{S}\mathcal{S}^8} + (-1)^{l\_8} H\_{h\_1h\_2b\_3} \\ & ((-1)^{l\_1} d|a\_1a\_2a\_3\rangle + (-1)^{l\_2} b|b\_1b\_2b\_3\rangle \\ &+ (-1)^{l\_5} c|c\_1c\_2c\_3\rangle + (-1)^{l\_4} d|d\_1d\_2b\_3\rangle \\ &+ (-1)^{l\_5} e|c\_1c\_2c\_3\rangle + (-1)^{l\_6} f|f\_1f\_2f\_3\rangle \\ &+ (-1)^{l\_7} g|g\_1g\_2g\_3\rangle + (-1)^{l\_8} h|b\_1h\_2h\_3\rangle) \end{aligned} \tag{A13}$$

where *Ii*ð Þ i ¼ 1, 2, 3, 4, 5, 6, 7, 8 can take values 0 or 1 independently, and *L <sup>j</sup>*ð Þ L ¼ a, b, c, d, e, f, g, h; j ¼ 1, 2, 3 can take values 0 or 1 independently. The summation is over all possible permutation states obtained.

The relevant transformations for the three-qubit teleportation are given in terms of the following basic operations:

Projection of *i* th component *Pi*:

$$P\_1 = \begin{pmatrix} 10 \\ 00 \end{pmatrix} P\_2 = \begin{pmatrix} 00 \\ 01 \end{pmatrix} \tag{A14}$$

Flip and Projection of *i* th component *Fi*:

$$F\_1 = \begin{pmatrix} \mathbf{01} \\ \mathbf{00} \end{pmatrix} \\ F\_2 = \begin{pmatrix} \mathbf{00} \\ \mathbf{10} \end{pmatrix} \\ \tag{A15}$$








