**A.2 Proposal 2**

Let us consider the situation in which Alice possesses the qubits 1 and 2, Bob possesses qubits 3, 4, 5 and 6 and Charlie possesses the 7th qubit. Alice has an unknown qubit *α*∣0i þ *β*∣1i which she wants to share with Bob and Charlie. Now Alice can measure in a particular basis. Suppose she measures in the GHZ Basis. Now, Bob can perform a four-qubit measurement and convey his outcome to Charlie through a classical channel. Having known the outcome of both their measurement, Charlie will obtain a certain single qubit quantum state. The outcome of the measurement performed by Bob and the state obtained by Charlie is given as follows: if Bob measures states ∣*x*�i, Charlie obtains states *α*∣0i � *β*∣1i, while if Bob measures states ∣*Y*�i then Charlie obtains the states *β*∣0i � *α*∣1i, where <sup>∣</sup>*x*�i ¼ <sup>1</sup> <sup>4</sup> *α*∣0000i þ *α*∣0111i þ *α*∣1001i þ *α*∣1110i � ð Þ *β*j1001i þ *β*j0000i þ *β*j1110i � *β*j0111i and <sup>∣</sup>*Y*�i ¼ <sup>1</sup> <sup>4</sup> *α*∣0001i þ *α*∣0110i þ *α*∣1000i � *α*∣1111i � ð Þ *β*j1000i þ *β*j0001i � *β*j1111i � *β*j0110i

#### **A.3 Proposal 3**

Let us consider the situation in which Alice possesses the qubits 1, 2, 3 and 4, Bob possesses qubits 5 and 6 and Charlie possesses the 7th qubit. Alice has an unknown qubit *α*∣0i þ *β*∣1i which she wants to share with Bob and Charlie. Based on the state Alice measures ð Þ j*Ai*i∀*i* ∈f g 1, 2, 3, 4, 5, 6, 7, 8 , Bob and Charlie obtain a corresponding state ∣*BCi*i, where

<sup>∣</sup>*A*1i ¼ <sup>1</sup> <sup>4</sup> ð Þ <sup>j</sup>01111i�j01011iþj10010iþj11001iþj11100iþj11101i�j11000<sup>i</sup> <sup>∣</sup>*A*2i ¼ <sup>1</sup> <sup>4</sup> ð Þ <sup>j</sup>01111iþj01011i�j10010i�j11001i�j11100iþj11101i�j11000<sup>i</sup> <sup>∣</sup>*A*3i ¼ <sup>1</sup> <sup>4</sup> ð Þ <sup>j</sup>01111iþj01011iþj10010iþj11001iþj11100i�j11101iþj11000<sup>i</sup> <sup>∣</sup>*A*4i ¼ <sup>1</sup> <sup>4</sup> ð Þ <sup>j</sup>01111i�j01011i�j10010i�j11001i�j11100i�j11101iþj11000<sup>i</sup> <sup>∣</sup>*A*5i ¼ <sup>1</sup> <sup>4</sup> ð Þ <sup>j</sup>11111i�j11011iþj00010iþj01001iþj01100iþj01101i�j01000<sup>i</sup> <sup>∣</sup>*A*6i ¼ <sup>1</sup> <sup>4</sup> ð Þ <sup>j</sup>11111iþj11011i�j00010i�j01001i�j01100iþj01101i�j01000<sup>i</sup> <sup>∣</sup>*A*7i ¼ <sup>1</sup> <sup>4</sup> ð Þ <sup>j</sup>11111iþj11011iþj00010iþj01001iþj01100i�j01101iþj01000<sup>i</sup> <sup>∣</sup>*A*8i ¼ <sup>1</sup> <sup>4</sup> ð Þ <sup>j</sup>11111i�j11011i�j00010i�j01001i�j01100i�j01101iþj01000<sup>i</sup>

and

$$\begin{aligned} \vert BC\_1\rangle &= a\vert 1\rangle \vert \Phi\_-\rangle + a\vert 0\rangle \vert \Psi\_-\rangle + \beta\vert 0\rangle \vert \Phi\_+\rangle + \beta\vert 1\rangle \vert \Psi\_+\rangle \\ \vert BC\_2\rangle &= a\vert 1\rangle \vert \Phi\_-\rangle - a\vert 0\rangle \vert \Psi\_-\rangle - \beta\vert 0\rangle \vert \Phi\_+\rangle + \beta\vert 1\rangle \vert \Psi\_+\rangle \\ \vert BC\_3\rangle &= a\vert 1\rangle \vert \Phi\_-\rangle - a\vert 0\rangle \vert \Psi\_-\rangle + \beta\vert 0\rangle \vert \Phi\_+\rangle - \beta\vert 1\rangle \vert \Psi\_+\rangle \\ \vert BC\_4\rangle &= a\vert 1\rangle \vert \Phi\_-\rangle + a\vert 0\rangle \vert \Psi\_-\rangle - \beta\vert 0\rangle \vert \Phi\_+\rangle - \beta\vert 1\rangle \vert \Psi\_+\rangle \\ \vert BC\_5\rangle &= \beta\vert 1\rangle \vert \Phi\_-\rangle + \beta\vert 0\rangle \vert \Psi\_-\rangle + a\vert 0\rangle \vert \Phi\_+\rangle + a\vert 1\rangle \vert \Psi\_+\rangle \\ \vert BC\_6\rangle &= \beta\vert 1\rangle \vert \Phi\_-\rangle - \beta\vert 0\rangle \vert \Psi\_-\rangle - a\vert 0\rangle \vert \Phi\_+\rangle + a\vert 1\rangle \vert \Psi\_+\rangle \\ \vert BC\_7\rangle &= \beta\vert 1\rangle \vert \Phi\_-\rangle - \beta\vert 0\rangle \vert \Psi\_-\rangle + a\vert 0\rangle \vert \Phi\_+\rangle - a\vert 1\rangle \vert \Psi\_+\rangle \\ \vert BC\_8\rangle &= \beta\vert 1\rangle \vert \Phi\_-\rangle + \beta\vert 0\rangle \vert \Psi\_-\rangle - a\vert 0\rangle \vert \Phi\_+\rangle - a\vert 1\rangle \vert \Psi\_+\rangle \end{aligned}$$

Bob can now perform a Bell measurement on his particles, and Charlie can obtain a particular resultant state by applying the appropriate unitary operation.

For example, if the joint-state obtained by Bob and Charlie is *β*∣1i∣Φ�i þ *β*∣0i∣Ψ�i � *α*∣0i∣Φþi � *α*∣1i∣Ψþi, one can see that Charlie will obtain the state ∣*Ci*i, *i* ¼ 1, 2, 3, 4 corresponding to the state measured by Bob ∣*Bi*i, where ∣*B*1i ¼ 1ffiffi 2 <sup>p</sup> <sup>∣</sup>01i, <sup>∣</sup>*B*2i ¼ <sup>1</sup>ffiffi 2 <sup>p</sup> <sup>∣</sup>10i, <sup>∣</sup>*B*3i ¼ <sup>1</sup>ffiffi 2 <sup>p</sup> <sup>∣</sup>11i, <sup>∣</sup>*B*4i ¼ <sup>1</sup>ffiffi 2 <sup>p</sup> ∣00i and ∣*C*1i ¼ *α*∣0i þ *β*∣1i, ∣*C*2i ¼ *α*∣0i � *β*∣1i, ∣*C*3i ¼ *α*∣1i þ *β*∣0i, ∣*C*4i ¼ *α*∣1i � *β*∣0i

### **B. Quantum teleportation**

#### **B.1 Quantum teleportation of arbitrary one-qubit state**

#### *B.1.1 Linear teleportation scheme*

To begin with, an arbitrary single qubit state can be teleported using the resource state ∣Γ7i will be considered. In this case Alice possesses qubits 1, 2, 3, 4, 5, 6 and the 7th particle belongs to Bob. Alice wants to transport an arbitrary state <sup>∣</sup>*ψ*ð Þ<sup>1</sup> i ¼ *<sup>α</sup>*∣0i þ *<sup>β</sup>*∣1<sup>i</sup> to Bob. The combined state of the system is <sup>∣</sup>Γð Þ<sup>1</sup> <sup>7</sup> i ¼ <sup>∣</sup>*ψ*ð Þ<sup>1</sup> <sup>i</sup> ⊗ ∣Γ7i. Alice measures the seven qubits in her possession via the seven qubit orthonormal states:

<sup>∣</sup>*ξ*�i ¼ <sup>∣</sup>0000i∣Ψ<sup>0</sup> *GHZ*i � <sup>∣</sup>0001i∣Ψ<sup>3</sup> *GHZ*i þ <sup>∣</sup>0010i∣Ψ<sup>7</sup> *GHZ*i <sup>þ</sup>∣0011i∣Ψ<sup>4</sup> *GHZ*i � <sup>∣</sup>0100i∣Ψ<sup>5</sup> *GHZ*i � <sup>∣</sup>0101i∣Ψ<sup>6</sup> *GHZ*i <sup>þ</sup>∣0110i∣Ψ<sup>2</sup> *GHZ*i þ <sup>∣</sup>0111i∣Ψ<sup>1</sup> *GHZ*i�ð∣1000i∣Ψ<sup>2</sup> *GHZ*i �∣1001i∣Ψ<sup>1</sup> *GHZ*i � <sup>∣</sup>1010i∣Ψ<sup>5</sup> *GHZ*i � <sup>∣</sup>1011i∣Ψ<sup>6</sup> *GHZ*i <sup>þ</sup>∣1100i∣Ψ<sup>7</sup> *GHZ*i þ <sup>∣</sup>1101i∣Ψ<sup>4</sup> *GHZ*i þ <sup>∣</sup>1110i∣Ψ<sup>0</sup> *GHZ*i �∣1111i∣Ψ<sup>3</sup> *GHZ*iÞ (A4) <sup>∣</sup>*ν*�i ¼ <sup>∣</sup>1000i∣Ψ<sup>0</sup> *GHZ*i � <sup>∣</sup>1001i∣Ψ<sup>3</sup> *GHZ*i þ <sup>∣</sup>1010i∣Ψ<sup>7</sup> *GHZ*i <sup>þ</sup>∣1011i∣Ψ<sup>4</sup> *GHZ*i � <sup>∣</sup>1100i∣Ψ<sup>5</sup> *GHZ*i � <sup>∣</sup>1101i∣Ψ<sup>6</sup> *GHZ*i <sup>þ</sup>∣1110i∣Ψ<sup>2</sup> *GHZ*i þ <sup>∣</sup>1111i∣Ψ<sup>1</sup> *GHZ*i�ð∣0000i∣Ψ<sup>2</sup> *GHZ*i �∣0001i∣Ψ<sup>1</sup> *GHZ*i � <sup>∣</sup>0010i∣Ψ<sup>5</sup> *GHZ*i � <sup>∣</sup>0011i∣Ψ<sup>6</sup> *GHZ*i <sup>þ</sup>∣0100i∣Ψ<sup>7</sup> *GHZ*i þ <sup>∣</sup>0101i∣Ψ<sup>4</sup> *GHZ*i þ <sup>∣</sup>0110i∣Ψ<sup>0</sup> *GHZ*i �∣0111i∣Ψ<sup>3</sup> *GHZ*iÞ (A5)

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 .

Alice then conveys the outcome of the measurement results to Bob via two classical bits. Bob then applies a suitable unitary operation from the set *I*, *σx*, *iσy*, *σ<sup>z</sup>* to recover the original state, sent by Alice. In this way, one can teleport an arbitrary single-qubit state using the state ∣Γ7i.
