Efficient simulation of quantum computers: the Gottesman ...quantum.phys.cmu.edu/groupth/talk30Jan2009.pdf · E cient simulation of quantum computers: the Gottesman-Knill theorem

Efficient simulation of quantum computers: theGottesman-Knill theorem or an application of group

theory to quantum information (part 2)

Vlad Gheorghiu

Department of PhysicsCarnegie Mellon University

Pittsburgh, PA 15213, U.S.A.

January 30, 2008

Vlad Gheorghiu (CMU) Efficient simulation of quantum computers January 30, 2008 1 / 16

Outline

1 Brief review

2 Stabilizer groupsThe action of the Pauli groupStabilizersConjugation of stabilizer groups under Clifford operations

3 The Gottesman-Knill theorem

4 Simple example

5 References

Both lectures (Wed. Jan 28 and Today, Jan 30) are available onlineat http://quantum.phys.cmu.edu/groupth


Brief review

Classical computers

Quantum computers - evolution is unitary, the group U(2n)

|ψ〉final = U|ψ〉initial = U|0, 0, . . . , 0〉

Evolving a quantum state requires in general O(2n) operations!

The Pauli group on one qudit

P1 = {±1,±i}{I ,X ,Y ,Z}

where the Pauli matrices are

X =

(0 11 0

),Y =

(0 −ii 0

),Z =

(1 00 −1

)with XY = iZ and X 2 = Y 2 = Z 2 = I .

The Pauli group on n qudits

Pn = {±1,±i}{X a1Zb1 ⊗ X a2Zb2 ⊗ . . .⊗ X anZbn}


Brief review

Classical computers





P1 = {±1,±i}{I ,X ,Y ,Z}


X =

(0 11 0

),Y =

(0 −ii 0

),Z =

(1 00 −1





Brief review

Classical computers





P1 = {±1,±i}{I ,X ,Y ,Z}


X =

(0 11 0

),Y =

(0 −ii 0

),Z =

(1 00 −1





Brief review

Classical computers





P1 = {±1,±i}{I ,X ,Y ,Z}


X =

(0 11 0

),Y =

(0 −ii 0

),Z =

(1 00 −1





Brief review

The Clifford group C1 on one qubit

C1P1C†1 = P1

The Clifford group Cn on n qubits

CnPnC†n = Pn

Up to complex phases, the Clifford group Cn is generated by H,S andCNOT , where

H =1√2

(1 11 −1

),S =

(1 00 i

)and CNOT =

1 0 0 00 1 0 00 0 0 10 0 1 0

.

For example, HXH† = Z , CNOT (X ⊗ I )CNOT † = X ⊗ X .


Brief review


C1P1C†1 = P1


CnPnC†n = Pn


H =1√2

(1 11 −1

),S =

(1 00 i

)and CNOT =

1 0 0 00 1 0 00 0 0 10 0 1 0

.



Brief review


C1P1C†1 = P1


CnPnC†n = Pn


H =1√2

(1 11 −1

),S =

(1 00 i

)and CNOT =

1 0 0 00 1 0 00 0 0 10 0 1 0

.



Brief review


C1P1C†1 = P1


CnPnC†n = Pn


H =1√2

(1 11 −1

),S =

(1 00 i

)and CNOT =

1 0 0 00 1 0 00 0 0 10 0 1 0

.



Stabilizer groups The action of the Pauli group

The action of a group G on a set A is defined as a binary function

G × A −→ A

denoted by (g , a) −→ g · a that satisfies the following two axioms

1 (gh) · a = g · (h · a) for all g , h ∈ G and a ∈ A;2 e · a = a for every a ∈ A (here e is the identity of G ).

Now let G be the Pauli group P1 and A the Hilbert space of 1 qudit,C2.

A basis of C2 is formally denoted by {|0〉, |1〉}.We define an action of the Pauli group on the Hilbert space of onequbit by specifying how the elements of G act on the basis

1 I |0〉 = |0〉, I |1〉 = |1〉 (identity)2 X |0〉 = |1〉, X |1〉 = |0〉 (bit flip)3 Z |0〉 = |0〉, Z |1〉 = −1|1〉 (phase flip)4 Y |0〉 =?, Y |1〉 =?.

Now we know how G is acting on any element of C2, by linearity, e.g.X (α|0〉+ β|1〉) = α|1〉+ β|0〉.Check that we have a valid action function.




G × A −→ A

denoted by (g , a) −→ g · a that satisfies the following two axioms1 (gh) · a = g · (h · a) for all g , h ∈ G and a ∈ A;

2 e · a = a for every a ∈ A (here e is the identity of G ).








G × A −→ A

denoted by (g , a) −→ g · a that satisfies the following two axioms1 (gh) · a = g · (h · a) for all g , h ∈ G and a ∈ A;2 e · a = a for every a ∈ A (here e is the identity of G ).








G × A −→ A









G × A −→ A



A basis of C2 is formally denoted by {|0〉, |1〉}.

We define an action of the Pauli group on the Hilbert space of onequbit by specifying how the elements of G act on the basis






G × A −→ A




1 I |0〉 = |0〉, I |1〉 = |1〉 (identity)

2 X |0〉 = |1〉, X |1〉 = |0〉 (bit flip)3 Z |0〉 = |0〉, Z |1〉 = −1|1〉 (phase flip)4 Y |0〉 =?, Y |1〉 =?.





G × A −→ A




1 I |0〉 = |0〉, I |1〉 = |1〉 (identity)2 X |0〉 = |1〉, X |1〉 = |0〉 (bit flip)

3 Z |0〉 = |0〉, Z |1〉 = −1|1〉 (phase flip)4 Y |0〉 =?, Y |1〉 =?.





G × A −→ A




1 I |0〉 = |0〉, I |1〉 = |1〉 (identity)2 X |0〉 = |1〉, X |1〉 = |0〉 (bit flip)3 Z |0〉 = |0〉, Z |1〉 = −1|1〉 (phase flip)

4 Y |0〉 =?, Y |1〉 =?.





G × A −→ A









G × A −→ A





Now we know how G is acting on any element of C2, by linearity, e.g.X (α|0〉+ β|1〉) = α|1〉+ β|0〉.

Check that we have a valid action function.




G × A −→ A








Similarly, we define the action of the Pauli group on n qudits Pn onthe Hilbert space of n qudits Cn

2

First note that a basis of Cn2 is formally specified as

{|0, 0, . . . , 0〉, |0, 0, . . . , 1〉, . . . , |1, 1, . . . , 1〉}Recall that an element g of Pn consists of a phase times a tensorproduct of n Pauli matrices

Each component of g acts individually on the correspondingcomponent of the basis element

Example: consider 3 qubits, let g = X ⊗ I ⊗ Z be an element of P3

and let us see how it acts on |1, 0, 1〉; thenX ⊗ I ⊗ Z |1, 0, 1〉 = −|0, 0, 1〉. Extend the action by linearity.

So now we know how Pn acts on the Hilbert space Cn2 of n qubits

What is (Z ⊗ X ⊗ X )(α|0, 0, 1〉+ β|1, 1, 0〉)? The result isα|0, 1, 0〉 − β|1, 0, 1〉.




2


{|0, 0, . . . , 0〉, |0, 0, . . . , 1〉, . . . , |1, 1, . . . , 1〉}

Recall that an element g of Pn consists of a phase times a tensorproduct of n Pauli matrices









2











2











2











2











2







What is (Z ⊗ X ⊗ X )(α|0, 0, 1〉+ β|1, 1, 0〉)?

The result isα|0, 1, 0〉 − β|1, 0, 1〉.




2









Stabilizer groups Stabilizers

We now have an action of Pn on the Hilbert space Cn2

Suppose S is a subgroup of Pn

Define VS to be the set of n qubit states which are fixed by everyelement of S , i.e.

VS = {|ψ〉 ∈ Cn2 : s|ψ〉 = (+1)|ψ〉,∀s ∈ S}

VS is the vector space stabilized by S , and S is called the stabilizer ofVS

Convince yourself that VS is a vector space (show that an arbitrarylinear combination of elements from VS also belongs to VS )

Show that VS is the intersection of the subspaces fixed by eachoperator in S (that is, the eigenvalues one eigenspaces of elements ofS)






VS = {|ψ〉 ∈ Cn2 : s|ψ〉 = (+1)|ψ〉,∀s ∈ S}









VS = {|ψ〉 ∈ Cn2 : s|ψ〉 = (+1)|ψ〉,∀s ∈ S}









VS = {|ψ〉 ∈ Cn2 : s|ψ〉 = (+1)|ψ〉,∀s ∈ S}









VS = {|ψ〉 ∈ Cn2 : s|ψ〉 = (+1)|ψ〉,∀s ∈ S}









VS = {|ψ〉 ∈ Cn2 : s|ψ〉 = (+1)|ψ〉,∀s ∈ S}






Example: what is the vector space stabilized by {I ,X} ⊂ P1?

Answer: {|0〉}.Another example: what about {I ⊗ I ,Z ⊗ Z} ⊂ Pn? Answer:Span{|0, 0〉, |1, 1〉}.The last one... {I ,Z1Z2,Z2Z3,Z1Z3} stabilizesSpan{|0, 0, 0〉, |1, 1, 1〉}S and VS are dual to each other, in the sense that one uniquelydetermines the other and vice versa.

Clever idea: one can describe (some) subspaces (VS ) by subgroups(S). What’s the advantage?



Example: what is the vector space stabilized by {I ,X} ⊂ P1?Answer: {|0〉}.

Another example: what about {I ⊗ I ,Z ⊗ Z} ⊂ Pn? Answer:Span{|0, 0〉, |1, 1〉}.The last one... {I ,Z1Z2,Z2Z3,Z1Z3} stabilizesSpan{|0, 0, 0〉, |1, 1, 1〉}S and VS are dual to each other, in the sense that one uniquelydetermines the other and vice versa.




Example: what is the vector space stabilized by {I ,X} ⊂ P1?Answer: {|0〉}.Another example: what about {I ⊗ I ,Z ⊗ Z} ⊂ Pn?

Answer:Span{|0, 0〉, |1, 1〉}.The last one... {I ,Z1Z2,Z2Z3,Z1Z3} stabilizesSpan{|0, 0, 0〉, |1, 1, 1〉}S and VS are dual to each other, in the sense that one uniquelydetermines the other and vice versa.




Example: what is the vector space stabilized by {I ,X} ⊂ P1?Answer: {|0〉}.Another example: what about {I ⊗ I ,Z ⊗ Z} ⊂ Pn? Answer:Span{|0, 0〉, |1, 1〉}.

The last one... {I ,Z1Z2,Z2Z3,Z1Z3} stabilizesSpan{|0, 0, 0〉, |1, 1, 1〉}S and VS are dual to each other, in the sense that one uniquelydetermines the other and vice versa.




Example: what is the vector space stabilized by {I ,X} ⊂ P1?Answer: {|0〉}.Another example: what about {I ⊗ I ,Z ⊗ Z} ⊂ Pn? Answer:Span{|0, 0〉, |1, 1〉}.The last one... {I ,Z1Z2,Z2Z3,Z1Z3} stabilizesSpan{|0, 0, 0〉, |1, 1, 1〉}

S and VS are dual to each other, in the sense that one uniquelydetermines the other and vice versa.




Example: what is the vector space stabilized by {I ,X} ⊂ P1?Answer: {|0〉}.Another example: what about {I ⊗ I ,Z ⊗ Z} ⊂ Pn? Answer:Span{|0, 0〉, |1, 1〉}.The last one... {I ,Z1Z2,Z2Z3,Z1Z3} stabilizesSpan{|0, 0, 0〉, |1, 1, 1〉}S and VS are dual to each other, in the sense that one uniquelydetermines the other and vice versa.




Example: what is the vector space stabilized by {I ,X} ⊂ P1?Answer: {|0〉}.Another example: what about {I ⊗ I ,Z ⊗ Z} ⊂ Pn? Answer:Span{|0, 0〉, |1, 1〉}.The last one... {I ,Z1Z2,Z2Z3,Z1Z3} stabilizesSpan{|0, 0, 0〉, |1, 1, 1〉}S and VS are dual to each other, in the sense that one uniquelydetermines the other and vice versa.




The necessary and sufficient conditions that S must satisfy in order tostabilize a nontrivial vector space VS are

1 S must be Abelian

2 −I does not belong to S

The necessity is easy...

Describe the stabilizer group S of some subspace VS by specifying itsgenerators

A group of size K has at most log(K ) generators!

So a stabilizer subspace of n qubits can be specified by a set of kindependent generators, and it turns out that k is always smaller thann

In the last example from the previous slide, S was generated by〈Z1Z2,Z2Z3〉 = 〈Z ⊗ Z ⊗ I , I ⊗ Z ⊗ Z 〉




1 S must be Abelian2 −I does not belong to S





















































Theorem

Let S = 〈g1, g2, . . . , gk〉 be a stabilizer group generated by k independentand commuting generators from Pn, such that −I /∈ S. Then the vectorspace stabilized by S has dimension 2n−k .

Enlarging S reduces the dimension of VS

What if k = n?

Stabilizer states

A stabilizer group generated by n independent generators stabilizes aone-dimensional vector space. The latter is called a stabilizer state.

Example: Let S = 〈Z ⊗ I , I ⊗ Z 〉. So n = 2, k = 2. What is the statestabilized by S? Answer: |00〉.So the state |0, 0, . . . , 0〉 is a stabilizer state with stabilizer groupgenerated by ... ... ... .......... 〈Z1,Z2, . . . ,Zn〉



Theorem



What if k = n?

Stabilizer states





Theorem



What if k = n?

Stabilizer states





Theorem



What if k = n?

Stabilizer states





Theorem



What if k = n?

Stabilizer states


Example: Let S = 〈Z ⊗ I , I ⊗ Z 〉. So n = 2, k = 2. What is the statestabilized by S?

Answer: |00〉.So the state |0, 0, . . . , 0〉 is a stabilizer state with stabilizer groupgenerated by ... ... ... .......... 〈Z1,Z2, . . . ,Zn〉



Theorem



What if k = n?

Stabilizer states


Example: Let S = 〈Z ⊗ I , I ⊗ Z 〉. So n = 2, k = 2. What is the statestabilized by S? Answer: |00〉.

So the state |0, 0, . . . , 0〉 is a stabilizer state with stabilizer groupgenerated by ... ... ... .......... 〈Z1,Z2, . . . ,Zn〉



Theorem



What if k = n?

Stabilizer states


Example: Let S = 〈Z ⊗ I , I ⊗ Z 〉. So n = 2, k = 2. What is the statestabilized by S? Answer: |00〉.So the state |0, 0, . . . , 0〉 is a stabilizer state with stabilizer groupgenerated by ...

... ... .......... 〈Z1,Z2, . . . ,Zn〉



Theorem



What if k = n?

Stabilizer states


Example: Let S = 〈Z ⊗ I , I ⊗ Z 〉. So n = 2, k = 2. What is the statestabilized by S? Answer: |00〉.So the state |0, 0, . . . , 0〉 is a stabilizer state with stabilizer groupgenerated by ... ...

... .......... 〈Z1,Z2, . . . ,Zn〉



Theorem



What if k = n?

Stabilizer states


Example: Let S = 〈Z ⊗ I , I ⊗ Z 〉. So n = 2, k = 2. What is the statestabilized by S? Answer: |00〉.So the state |0, 0, . . . , 0〉 is a stabilizer state with stabilizer groupgenerated by ... ... ...

.......... 〈Z1,Z2, . . . ,Zn〉



Theorem



What if k = n?

Stabilizer states


Example: Let S = 〈Z ⊗ I , I ⊗ Z 〉. So n = 2, k = 2. What is the statestabilized by S? Answer: |00〉.So the state |0, 0, . . . , 0〉 is a stabilizer state with stabilizer groupgenerated by ... ... ... ..........

〈Z1,Z2, . . . ,Zn〉



Theorem



What if k = n?

Stabilizer states




Stabilizer groups Conjugation of stabilizer groups under Clifford operations

Remember that the Clifford group maps Pauli operators to Paulioperators

Let c ∈ Cn. Note that S ′ = cSc† is also a stabilizer group, and thedimension of VS ′ equals the dimension of VS . Why?

Convince yourself that if S stabilizes a stabilizer state |ψ〉, thenS ′ = cSc† stabilizes |ψ′〉 =c |ψ〉.If we know how c conjugates the generators of S , we then know howc conjugates the whole S . Why?










Convince yourself that if S stabilizes a stabilizer state |ψ〉, thenS ′ = cSc† stabilizes |ψ′〉 =

c |ψ〉.If we know how c conjugates the generators of S , we then know howc conjugates the whole S . Why?





Convince yourself that if S stabilizes a stabilizer state |ψ〉, thenS ′ = cSc† stabilizes |ψ′〉 =c |ψ〉.

If we know how c conjugates the generators of S , we then know howc conjugates the whole S . Why?












The Gottesman-Knill theorem

We can now state and prove the following theorem


A quantum unitary evolution that uses only the following elements can besimulated efficiently on a classical computer:

1 preparation of qubits in computational basis states (w.l.o.g. can be taken tobe |0, 0, . . . , 0〉 )

2 evolution U from the Clifford group

3 measurements in the computational basis.

The proof is now quite simple...

The first assumption guarantees that we start with a stabilizer state|ψ〉, with stabilizer group Sψ generated by n independent generators〈Z1,Z2, . . . ,Zn〉Now let’s pick a generator of the Clifford group, call it u






























The first assumption guarantees that we start with a stabilizer state|ψ〉, with stabilizer group Sψ generated by n independent generators〈Z1,Z2, . . . ,Zn〉

Now let’s pick a generator of the Clifford group, call it u













Under the action of u, the initial stabilizer state |ψ〉 stabilized bysome Sψ will evolve to u|ψ〉

Remember the duality between a stabilizer state and a stabilizer group

Now |ψ′〉 = u|ψ〉 can be as well described by its stabilizer groupSψ′ = uSψu†

To specify Sψ′ , it is enough to see how the n generators of Stransform under conjugation by u, so we need to finduZ1u

†, . . . , uZnu†

To summarize, implementing a generator u of the Clifford groupimplies updating the n generators that describe the initial quantumstate

This step requires O(n2) operations on a classical computer

In conclusion, if our unitary evolution U is a product of m terms, eacha generator of the Clifford group, then the computation can besimulated by a classical computer in time O(mn2)



Under the action of u, the initial stabilizer state |ψ〉 stabilized bysome Sψ will evolve to u|ψ〉Remember the duality between a stabilizer state and a stabilizer group



†, . . . , uZnu†









†, . . . , uZnu†









†, . . . , uZnu†









†, . . . , uZnu†









†, . . . , uZnu†









†, . . . , uZnu†






Intuitively, the way the classical computer performs the simulation issimply to keep track of the generators of the stabilizer as the variousClifford elementary operations are being performed in the computation

At the end, we are left with some stabilizer group that stabilizes somestate

Knowing the stabilizer is equivalent to knowing the state (rememberthe duality)

End of story...

It is clear that one cannot perform universal quantum computationwith only Clifford operations and measurements in the computationalbasis

It is not at all obvious that Clifford operations together with only onequbit unitary that does not belong to the Clifford group form a denseset in U(2n), that is, one can implement universal quantumcomputing using this gates!






End of story...








End of story...








End of story...








End of story...








End of story...




Simple example

Simulate the evolution of the state |ψinitial〉 = |00〉 throughU = CNOT12H1

The following table is all we need for simulating any Clifford evolution1 Conjugation by CNOT : X1 → X1X2,X2 → X2,Z1 → Z1,Z2 → Z1Z2

2 Conjugation by H: X → Z ,Z → X3 Conjugation by S : X → Y ,Z → Z4 Conjugation by X : X → X ,Z → −Z5 Conjugation by Y : X → −X ,Z → −Z6 Conjugation by Z : X → −X ,Z → Z

|00〉 ⇔ 〈Z ⊗ I , I ⊗ Z 〉After H, 〈X ⊗ I , I ⊗ Z 〉After CNOT , 〈X ⊗ X ,Z ⊗ Z 〉The final stabilizer defines |ψfinal〉 = |00〉+|11〉√

2, so we simulated the

evolution without computing U|ψinitial〉


Simple example


The following table is all we need for simulating any Clifford evolution

1 Conjugation by CNOT : X1 → X1X2,X2 → X2,Z1 → Z1,Z2 → Z1Z2






Simple example








Simple example




|00〉 ⇔ 〈Z ⊗ I , I ⊗ Z 〉

After H, 〈X ⊗ I , I ⊗ Z 〉After CNOT , 〈X ⊗ X ,Z ⊗ Z 〉The final stabilizer defines |ψfinal〉 = |00〉+|11〉√




Simple example




|00〉 ⇔ 〈Z ⊗ I , I ⊗ Z 〉After H, 〈X ⊗ I , I ⊗ Z 〉

After CNOT , 〈X ⊗ X ,Z ⊗ Z 〉The final stabilizer defines |ψfinal〉 = |00〉+|11〉√




Simple example




|00〉 ⇔ 〈Z ⊗ I , I ⊗ Z 〉After H, 〈X ⊗ I , I ⊗ Z 〉After CNOT , 〈X ⊗ X ,Z ⊗ Z 〉

The final stabilizer defines |ψfinal〉 = |00〉+|11〉√2

, so we simulated the



Simple example








References

1 Michael A. Nielsen and Isaac L. Chuang, Quantum Computationand Quantum Information, Cambridge University Press (2000)

2 Daniel Gottesman, PhD Thesis, arXiv:quant-ph/9705052, preprint


Efficient simulation of quantum computers: the Gottesman ...quantum.phys.cmu.edu/groupth/talk30Jan2009.pdf · E cient simulation of quantum computers: the Gottesman-Knill theorem

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