Basics

QuBit, quantum teleportation and entanglement. These terms are essential for understanding quantum technologies such as the quantum computer or quantum networks. Using graphs and small examples, these three terms are discussed in more detail in this article. In the text, as a supplement, there are digression sections that contain more detailed information on the content such as mathematical definitions and derivations.

Excursus: Why quantum computing? The Grover Algorithm

This example uses the Grover algorithm to illustrate why quantum computers are superior to classical computers in terms of computing time.To find an element from an unsorted database with N elements, a computing time is required that is proportional to O(N) (large O notation). Already in 1996 Grover could show that with an algorithm based on quantum technology the number of computational steps can be reduced to an order of O(N0,5).By reducing the number of computational steps, Grover’s algorithm reduces the time required to find an item in a database. This algorithm is one of several that exploit the properties of quantum phenomena to achieve an increase in computational speed [3].

further information about the Grover algorithm: https://qiskit.org/textbook/ch-algorithms/grover.html

The QuBit

Computers use switching states of transistors that can assume the classical state 0 or 1 to store information or perform computing operations. In a quantum computer, the smallest unit of information is called a QuBit. Unlike the bit, which assumes either one or the other state (0 or 1), a QuBit Q consists of an arbitrary superposition of these two states:


Ket{Q}=alphaKet{0}+betaKet{1}qquad(1)

α and β are (in general) complex numbers indicating the fraction for 0 and 1, respectively. The state 0 resp. 1 and the state Q are represented as vectors on the surface of a Bloch sphere [1]:



Ket{0}


Ket{1}


Ket{Q}=alphaKet{0}+betaKet{1}

The notation ⟨ | 〉 is called Bra-Ket notation. This notation is a simplified representation for column vectors | 〉 (Bra) and ⟨ | for row vectors (Ket), respectively..

Excursus: Vector representation and braket notation


Ket{0}=begin{bmatrix}
1 \
0 \
end{bmatrix}
;
Ket{1}=begin{bmatrix}
0 \
1 \
end{bmatrix}qquad(2)


Bra{0}=begin{bmatrix}
1 & 0
end{bmatrix}
;
Bra{1}=begin{bmatrix}
0 & 1
end{bmatrix}qquad(3)
The first row in the column vectors or first column in the row vectors refers to the value |0〉 and the second row or column refers to the value |1〉 (see also representation on the Bloch sphere).

Die Werte für α und β sind dabei so normiert, dass ihr Quadrat den Wert 1 ergibt:


|alpha|^{2} + |beta|^{2}=1qquad(4)

In this case, |α|2 and |β|2 indicate the probabilities of which state the QuBit is currently in. A permissible state for a QuBit is shown in the next figure, where the probability of encountering 0 or 1 is 50% to 50% [1].:


Ket{Q}=frac{1}{sqrt{2}}Ket{0} + frac{1}{sqrt{2}}Ket{1}qquad (5)



Ket{+}

=


=


frac{1}{sqrt{2}}


frac{1}{sqrt{2}}Ket{0}

+

+

frac{1}{sqrt{2}}


frac{1}{sqrt{2}}Ket{1}

This state is abbreviated as |+〉.

Question: When does the QuBit “decide” for one of the two states?

Measurement of a QuBit state

Once the state of a QuBit is determined by a measurement, it transitions to one of the two classical states 0 or 1 with probability |α|2 or |β|2, respectively.

Quantum entanglement

 

Excursus: Representation of two QuBits via tensor product

For the explanation of the entanglement of two quanta or QuBits, it is necessary to explain the representation of the state of two QuBits, as well as the Hadamard and CNOT gates.
Two non-entangled QuBits can be represented by the following state vector:

Ket{a}=a_{00}Ket{00}+a_{01}Ket{01} +a_{10}Ket{10} +a_{11}Ket{11}=begin{bmatrix}
a_{00}\
a_{01}\
a_{10}\
a_{11}
end{bmatrix}qquad(6)
For the probabilities, again:

|a_{00}|^2+|a_{01}|^2+|a_{10}|^2+|a_{11}|^2=1qquad(7)
In general, the state or states of two (or more QuBits) can be described by the tensor product (dyadic product):

Ket{a}=begin{bmatrix}
a_{0} \
a_{1} \
end{bmatrix}
;
Ket{b}=begin{bmatrix}
b_{0} \
b_{1} \
end{bmatrix}qquad(8)


Ket{ab}=Ket{a} otimes Ket{b}=begin{bmatrix}
a_{0} times begin{bmatrix}
b_{0}\
b_{1}
end{bmatrix} & \
a_{1} times begin{bmatrix}
b_{0}\
b_{1}
end{bmatrix} &
end{bmatrix}=begin{bmatrix}
a_{0}b_{0}\
a_{0}b_{1}\
a_{1}b_{0}
\
a_{1}b_{1}
end{bmatrix}qquad(9)

 

Together, the QuBits can assume four different states, like two ordinary bits.
Entangled QuBits consist of a superposition of two certain states and decay into one of the two states after a measurement. For two entangled QuBits QA, QB the following four states exist:


Ket{Q_{A},Q_{B}}=frac{1}{sqrt{2}}(Ket{00}pmKet{11})qquad(10)


Ket{Q_{A},Q_{B}}=frac{1}{sqrt{2}}(Ket{01}pmKet{10})qquad(11)

The state (10) is also called Bell state or Bell pair.

Question: How are 2 QuBits connected so that they are interleaved?

Two so-called (circuit) gates are needed for entanglement: The Hadamard and CNOT gates.

The Hadamard gate

The Hadamard gate has the task of transferring the (absolute) state of a QuBit (|0〉 or |1〉) into the superposition state [1]:


Ket{0}


Ket{+}

Ket{-}


Ket{1}

 

This corresponds on the Bloch sphere to a projection in the equatorial plane. The Hadamard gate transforms the state
|0〉 to |+〉 (see equation (13)) or |1〉 to |-〉 (see equation (14)). If the QuBit would again be measured into this state (|+〉 or |-〉), it will transition to the |1〉 or |0〉 state with a probability of 50%.

Excursus: Mathematical representation of a Hadamard gate

Mathematisch gesehen, handelt es sich bei dem Hadamard-Gatter (Hadmard-Operator) um eine 2×2 Matrix, die die Vektoren (Zustände) |0〉 bzw. |1〉 in den Vektor (Zustand) |+〉 bzw. |-〉 transformiert:


H=frac{1}{sqrt{2}}begin{bmatrix}
1 &1 \
1 & -1
end{bmatrix}qquad(12)\
HKet{0}=Ket{+}=frac{1}{sqrt{2}}(Ket{0}+Ket{1})qquad(13)\
HKet{1}=Ket{-}=frac{1}{sqrt{2}}(Ket{0}-Ket{1})qquad(14)

 

The CNOT gate

The second gate required for entanglement is the CNOT gate. Two QuBits are required for this gate QA and QB. The QuBit QA is called the control bit, because the state of QB depends on its state [1]:

The CNOT gate inverts the value of QuBit QB only if QA has the value 1. Otherwise the value of QB remains unchanged.

Exkurs: Mathematische Darstellung des CNOT-Gatters

The CNOT gate (CNOT operator) can be mathematically expressed by a 4×4 matrix, which multiplied by the status vector of two QuBits, gives the following state after the CNOT operation.:
CNOT=begin{bmatrix}
1 & 0 & 0 & 0 \
0 & 0 & 0 & 1 \
0 & 0 & 1 & 0 \
0 & 0 & 1 & 0
end{bmatrix}qquad(15)


Ket{a}=begin{bmatrix}
a_{00}\
a_{01}\
a_{10}\
a_{11}
end{bmatrix}


CNOTKet{a}=begin{bmatrix}
a_{00}\
a_{11}\
a_{10}\
a_{01}
end{bmatrix}qquad(16)
The CNOT gate thus swaps the amplitudes of a01 and a11.

Now a Bell pair or entanglement is to be generated with these two gates (operators). This circuit entangles the two qubits QA and QB with the initial values 0 [1].:


Ket{Q_A,Q_B}=frac{1}{sqrt{2}}(Ket{0,0}+Ket{1,1})

 

Depending on the measurement result of the control QuBit QA, the QuBit QB is changed or not. After a measurement, the state of both qubits is |0.0〉 or |1.1〉 with a probability of 50%.

Quantum teleportation

The quantum teleportation is shown schematically and as a circuit diagram consisting of CNOT, Hadamard and XZ gates in the following two sketches [1,2].:

The goal of quantum teleportation is to transfer the state of QuBit C to another QuBit B. Due to the No-Cloning Theorem, it is not possible to transfer the state of one QuBit C to another without changing the state of QuBit C. The state of QuBit C is initially QC:


Ket{Q_C}=alphaKet{0}+betaKet{1}qquad(17)

Three QuBits are required for execution: One QuBit has the state to be transferred, the other two interleaved QuBits (A and B) are used to transfer the state from QuBit C to QuBit B.

Sequence of a quantum teleportation

  1. Two QuBits A and B are interleaved with each other. Bob keeps the QuBit B and sends the second QuBit to Alice.
  2. Alice concatenates her QuBit C with the received QuBit A via a CNOT gate and performs a Hadamard transformation
  3. Alice measures the state of QuBit C and A
  4. Alice communicates the result of the measurements- encoded by two bits- via a classical communication channel
  5. Depending on the measurement result, Bob performs transformations on his QuBit B (called XZ or Pauli transformation) so that his QuBit B has the same state as QuBit C.

First, qubit A and B are interleaved with each other. Then QuBit A (Alice) is connected to QuBit C via a CNOT gate. QuBit C is then brought into the superposition state with a Hadamard gate (see circuit diagram).

Excursus: Entanglement and superposition of qubits A, B and C

The interleaving of QuBit A and B results in:


Ket{e}=frac{1}{sqrt{2}}(Ket{00}+Ket{11})=frac{1}{sqrt{2}}(Ket{0}_{A}Ket{0}_{B}+Ket{1}_{A}Ket{1}_{B})qquad(18)

Thus, with the QuBit QC whose one-particle state is to be transferred, the total states are as follows (expressed by tensor product):


Ket{Q_C}otimesKet{e}=frac{1}{sqrt{2}}(alphaKet{000}+alphaKet{011}+betaKet{100}+betaKet{111})qquad(19)

Excursus: Matrix representation of two non-entangled qubits

The question arises how QuBit B, which is not connected to QuBit C via CNOT, is represented mathematically. This will be explained using two QuBits, where no operation is performed on one QuBit and a Hadamard transformation is performed on the second QuBit [3]:


To be able to describe the state of this system, the identity matrix I is written down for the QuBit q0. This matrix leaves the state of a QuBit unchanged:

I=begin{bmatrix}
1 & 0\
0 & 1
end{bmatrix}qquad(20)

The overall state G of this system can thus be described with the matrix representation of H (see equation (12)) as follows:

G=Hotimes I=begin{bmatrix}
H & I\
I & H
end{bmatrix}=begin{bmatrix}
frac{1}{sqrt{2}} & frac{1}{sqrt{2}} & 1 & 0\
frac{1}{sqrt{2}} &-frac{1}{sqrt{2}} & 0 &1 \
1& 0 & frac{1}{sqrt{2}} &frac{1}{sqrt{2}} \
0 & 1 & frac{1}{sqrt{2}} &-frac{1}{sqrt{2}}
end{bmatrix}qquad(21)

Exkurs: Mathematische Beschreibung der Quantenverschränkung

From the matrix representation of two non-entangled qubits (see above representation for q0 and q1) it follows that in the CNOT operation for the uninvolved QuBit B must be multiplied by the identity matrix I (see equation(20)). In the subsequent Hadamard transformation, two identity matrices are required because QuBit A and B are not involved in the transformation:

(Hotimes I otimes I)(CNOT otimes I)(Ket{Q_C}otimes Ket{e})=\
(Hotimes I otimes I)(CNOT otimes I)frac{1}{sqrt{2}}(alphaKet{000}+alphaKet{011}+betaKet{100}+betaKet{111})=\
(Hotimes I otimes I)frac{1}{sqrt{2}}(alphaKet{000}+alphaKet{011}+betaKet{110}+betaKet{101})=\
frac{1}{2}(alpha(Ket{000}+Ket{011}+Ket{100}+Ket{111})+beta(Ket{010}+Ket{010}-Ket{110}-Ket{101}))=\
frac{1}{2}(color{green}{Ket{00}}color{black}{(}color{blue}{alphaKet{0}+betaKet{1}}color{black}{)+} color{green}{Ket{01}}color{black}{(}color{blue}{alphaKet{1}+betaKet{0}}color{black}{)+}color{green}{Ket{10}}color{black}{(}color{blue}{alphaKet{0}-betaKet{1}}color{black}{)+}color{green}{Ket{11}}color{black}{(}color{blue}{alphaKet{1}-betaKet{0}}color{black}{))}quad(22)

Alice performs a measurement on her two QuBits (A and C). As a measurement result she receives with equal probability (25%) one of the four green marked states in the equation above. Alice sends this measurement result encoded in two classical bits to Bob. Depending on the measurement result, Bob’s QuBit B assumes the state marked in blue.

 

Alice performs a measurement on her two QuBits (A and C). She obtains one of the four measurement results in the following table with equal probability (25%):

Measurement result (Alice) QuBit B state (Bob)
|00〉 α|0〉+β|1〉
|01〉 α|1〉+β|0〉
|10〉 α|0〉-β|1〉
|11〉 α|1〉-β|0〉

Note: the states in which Bob’s QuBit is located result from the mathematical descriptions of the Hadamard and CNOT gates and their connections with the three QuBits (see circuit diagram and digressions on quantum teleportation).

X and Z gates

From the above table it can be seen that for the measurement state |00〉 Bob’s QuBit QB is already in the state of the QuBit QC to be transmitted (compare equation (17)). For the three other measurement states, an adjustment of QB is required so that QB has the same state as the QuBit QC. So-called X and Z gates are required for this.

The X gate inverts the state from |0〉 to |1〉 and vice versa [1]:

The Z gate results in a change of state |+〉 to |-〉 or |-〉 to |+〉 [1]:

By using these gates Bob can bring his QuBit QB into the state of QC.
The following table shows, depending on the measurement result, the gates to be used, which Bob must apply to his qubit QB:

Measurement result (Alice) required gates
|00〉 keine
|01〉 X
|10〉 Z
|11〉 ZX

 

The transformation of a QuBit with X and Z gates is also called Pauli transformation.

Excursus: Mathematical representation of X and Z gates

In this example, the X gate changes the QuBit state |0〉 to |1〉:

Ket{0}=begin{bmatrix}
1\
0
end{bmatrix};
X=begin{pmatrix}
0 & 1\
1 & 0
end{pmatrix}qquad(23)


XKet{0}=begin{pmatrix}
0 & 1\
1 & 0
end{pmatrix}begin{bmatrix}
1\
0
end{bmatrix}=begin{bmatrix}
0\
1
end{bmatrix}

The Z gate is also a 2×2 matrix:

Ket{+}=frac{1}{sqrt{2}}begin{bmatrix}
1\
1
end{bmatrix};
Z=begin{pmatrix}
1 & 0\
0 & -1
end{pmatrix}qquad(24)


ZKet{+}=Z=begin{pmatrix}
1 & 0\
0 & -1
end{pmatrix}frac{1}{sqrt{2}}begin{bmatrix}
1\
1
end{bmatrix}=frac{1}{sqrt{2}}begin{bmatrix}
1\
-1end{bmatrix}=Ket{-}

Quellen

[1]: M. Ellerhoff, Mit Quanten Rechnen. Wiesbaden: Springer Spektrum , ISBN 978-3-658-31221-3
[2]: P. Kaufmann, S. Naegele-Jackson, II. Quantenrevolution – die Welt der Qubits: DFN-Mitteilungen Ausgabe 99 Juni 20/21 Seite 22
[3] https://qiskit.org/textbook/preface.html

Stand: 28.06.21