Each normalized state is a member of an orthonormal basis: A simple proof
aa r X i v : . [ qu a n t - ph ] J u l Each normalized state is a member of anorthonormal basis: A simple proof
Iman Sargolzahi and Ehsan Anjidani Department of Physics, University of Neyshabur, P. O. Box 91136-899, Neyshabur, Iran. [email protected] Department of Mathematics, University of Neyshabur, P. O. Box 91136-899, Neyshabur, Iran. [email protected]
Abstract
In a finite dimensional Hilbert space, each normalized vector (state) can bechosen as a member of an orthonormal basis of the space. We give a proof ofthis statement in a manner that seems to be more comprehensible for physicsstudents than the formal abstract one.
Finite dimensional Hilbert spaces, e.g. spin of a particle or polarization of a photon,is of great interest in quantum mechanics, specially in quantum information andcomputation theory [1]. So, studying the mathematical properties of such spaces isnecessary for physics students, specially in graduate levels.One useful result in a finite dimensional Hilbert space is that an arbitrary nor-malized state can be chosen as a member of an orthonormal basis of the space. In thelinear algebra theory, this statement is proven [2]: A set including only one state, con-structs a linearly independent set. But this proof seems rather abstract, at least, forphysics students. So in this paper, instead, we prove this statement by constructing
I. Sargolzahi and E. Anjidani directly a basis including our arbitrarily chosen state. Doing so, in a two dimensionalHilbert space (one-qubit state space), is simple using the Bloch sphere. Then, byinduction, we generalize this result to the higher finite dimensional Hilbert spacestoo. We hope that this method be more comprehensible for physics students.The paper is organized as follows: In the next section, we review the propertiesof vector spaces briefly. In section 3, using the Bloch sphere, it is shown that in theone-qubit case, each state is a member of an orthonormal basis. Our main result isgiven in section 4, in which, by induction, we generalize the result of section 3 toarbitrary finite dimensional spaces. In section 5, we discuss the consequences of thisresult briefly. When our result can be used for the countable infinite dimensional caseis the subject of section 6 and, finally, our paper is ended by a summary in section 7.
A vector space V on complex numbers C is a set of elements, like | a i , | b i , | c i , · · · ,which we call them vectors and satisfy two following properties [3]:1. Sum of each two vectors | a i and | b i of V is also a vector in V like | c i : | a i ∈ V, | b i ∈ V ⇒ | a i + | b i = | c i ∈ V. (1)2. Multiplication of a vector | a i by a complex number (scalar) z , is also a memberof V : | a i ∈ V, z ∈ C ⇒ z | a i ∈ V. (2)Two above properties are obviously the generalization of the properties of real vectorsin the three dimensional space:1. Sum of two vectors A and B is also a vector like C .2. Multiplication of a vector A by a real scalar c is also a vector (parallel to A ).In addition, in a vector space there is a zero vector, which we denote by 0, with theproperty | a i + 0 = | a i for any | a i ∈ V. (3) For more detailed definition of a vector space see e.g. [2, 3, 4] ach normalized state is a member of an orthonormal basis: A simple proof {| a i , · · · , | a n i} are linearly independent if theequality z | a i + · · · + z n | a n i = 0 (4)(with z i ∈ C , i = 1 , · · · , n ) is satisfied only when for all 1 ≤ i ≤ n , z i = 0. Assumethat the maximum number of linearly independent vectors in the vector space V is afinite number n < ∞ . Now, consider the following equality, in which the vector | b i isalso nonzero: z | a i + · · · + z n | a n i + z | b i = 0 . (5)Obviously, the above equation has a trivial solution: z i = 0, ( i = 1 , · · · , n ) and z = 0.If there were no other solution except the trivial one, then {| a i , · · · , | a n i , | b i} wouldbe a linearly independent set with ( n + 1) members, which is in contradiction to ourassumption. So, equation (5) has another solution; i.e. at least one of z i and also z are nonzero. This has an important consequence: For each | b i ∈ V , we have | b i = z − z | a i + · · · + z n − z | a n i = n X i =1 c i | a i i , (6)where c i := z i − z . In other words, any vector in the vector space V can be written as alinear combination of vectors | a i , · · · , | a n i . The set {| a i , · · · , | a n i} is called a basisfor the vector space V . Such a vector space, in which the number of the basis vectorsis finite ( n < ∞ ), in other words, the maximum number of linearly independentvectors is finite, is called a finite dimensional ( n -dimensional) vector space [2, 3].A finite dimensional vector space V equipped with an inner product is called aHilbert space or an inner product space [1]. A function ( ., . ) from V × V to C is aninner product if it satisfies the following properties [1, 3, 4]:1. ( ., . ) is linear in the second argument ( λ i are complex numbers.): | a i , m X i =1 λ i | b i i ! = m X i =1 λ i ( | a i , | b i i ) . (7)2. For any | a i , | b i ∈ V , ( | a i , | b i ) = ( | b i , | a i ) ∗ . (8)3. For any | a i ∈ V , ( | a i , | a i ) ≥ , (9)where the equality occurs if and only if | a i = 0. I. Sargolzahi and E. Anjidani
We define the norm of a vector | a i by k | a i k = p ( | a i , | a i ) . (10)If k | a i k = 1, then we say that | a i is a normalized vector. In addition, if( | a i , | b i ) = 0 , (11)then we say that the two vectors | a i and | b i are orthogonal.Following the Gram-Schmidt procedure [3, 4], one can construct an orthonormalbasis {| e i , · · · , | e n i} from the basis {| a i , · · · , | a n i} in an inner product space. Theorthonormality of vectors | e i i means that ( | e i i , | e j i ) = δ ij , where δ ij is the Kroneckerdelta function ( δ ij = 0 for i = j and δ ij = 1 for i = j ). The Gram-Schmidtprocedure is as follows: | e i := | a ik | a i k , (12)and for each 2 ≤ i ≤ n , | e i i := | e ′ i ik | e ′ i i k , (13)where | e ′ i i := | a i i − i − X j =1 ( | e j i , | a i i ) | e j i . (14)It is easy to see that the set {| e i , · · · , | e n i} is orthonormal. In addition, sinceaccording to (12), (13) and (14) we can write each | a i i in terms of vectors | e j i , foreach | b i ∈ V we have | b i = n X i =1 c i | a i i = n X i =1 d i | e i i , (15)where the coefficients d i are complex numbers. We may use the column matrixnotation for the orthonormal basis vectors | e i i : | e i = n × , · · · , | e n i = n × . (16)Hence, for each | b i ∈ V we have | b i = n X i =1 d i | e i i = d + · · · + d n = d ... d n . (17) ach normalized state is a member of an orthonormal basis: A simple proof | v i , | w i ∈ V , where | v i = v ... v n , | w i = w ... w n , we have [1] ( | v i , | w i ) = (cid:16) v ∗ · · · v ∗ n (cid:17) w ... w n = n X i =1 v ∗ i w i . (18)We can continue the above discussion and introduce linear operators on V , theirmatrix representation and ... [1, 5].Let’s come back to our question: whether an arbitrary normalized vector | b i ∈ V can be chosen as a member of an orthonormal basis. As we mentioned, the answeris yes, since the set {| b i} is a linearly independent set. Therefore | b i is the basisvector of a one dimensional subspace of V and so a basis vector of the V itself [2].In addition, one can find exactly n − | w i i in such a waythat {| b i , | w i , · · · , | w n − i} constructs a linearly independent set and so a basis for V (see [2, 3]). Equipped with the linearly independent set {| b i , | w i , · · · , | w n − i} and using the Gram-Schmidt procedure, we can construct the orthonormal basis {| v i (= | b i ) , | v i , · · · , | v n i} for V , and therefore the proof is completed.The above discussion, though is satisfying for a mathematician, may be ratherabstract for a physics student. So, instead, we prove that each normalized vector | b i is a member of an orthonormal basis, by constructing such a basis directly from theorthonormal computational basis {| e i , · · · , | e n i} . This can be done simply for thetwo dimensional case, i.e. qubits, using the Bloch sphere. For the completeness of ourdiscussion, we give this result in the next section and then in section 4, we generalizeit to the higher dimensional cases. In a two dimensional Hilbert space, according to (15), each (normalized) state can bewritten as: | ψ i = α | i + β | i , (19) I. Sargolzahi and E. Anjidani where we use the notation | i and | i for the orthonormal computational basis, insteadof | e i and | e i . Decomposition coefficients α and β are (in general) complex numbers.Since | ψ i is normalized, from (10) and (18) we have | α | + | β | = 1 . (20)So, we can write (19) as follows | ψ i = e iγ (cid:18) cos θ | i + e iϕ sin θ | i (cid:19) , (21)where i = √− γ, θ, ϕ are real numbers. Comparing (19) and (21), we have α = e iγ cos θ , β = e i ( γ + ϕ ) sin θ | ψ i as[1]: | ψ i = cos θ | i + e iϕ sin θ | i (0 ≤ ϕ ≤ π ) . (22)Obviously, choosing ϕ out of the above interval does not give us any new state. Also,it can be shown that choosing 0 ≤ θ ≤ π covers all possible states; i.e. if we choose θ out of this interval, then, up to a phase factor, it coincides with a state with 0 ≤ θ ≤ π .So, all possible one-qubit states can be written as follows: | ψ i = cos θ | i + e iϕ sin θ | i (0 ≤ ϕ ≤ π, ≤ θ ≤ π ) . (23)Therefore, we can relate each state in a two dimensional Hilbert space to a point ona unit three dimensional sphere, called the Bloch sphere [1].Now, consider an arbitrary state | ψ i (with θ = θ and ϕ = ϕ ): | ψ i = cos θ | i + e iϕ sin θ | i . (24)It is easy to show that | ψ i is a member of an orthonormal basis for our one-qubitstate space. Consider the state | φ i (with θ = π − θ and ϕ = ϕ + π ): | φ i = cos π − θ | i + e iϕ sin π − θ | i = sin θ | i − e iϕ cos θ | i . (25)Using (18), it is easy to show that | ψ i and | φ i are orthonormal. In addition, wehave | i = cos θ | ψ i + sin θ | φ i , | i = e − iϕ (cid:18) sin θ | ψ i − cos θ | φ i (cid:19) . (26) ach normalized state is a member of an orthonormal basis: A simple proof | ψ i , in this two dimensional state space, in terms of | ψ i and | φ i . This can be done by inserting | i and | i from (26) in (19). Therefore, {| ψ i , | φ i} is an orthonormal basis for one-qubit state space, which includes ourarbitrarily chosen state | ψ i . Now we want to generalize the result of the previous section to an arbitrary finitedimensional Hilbert space. We begin with the three dimensional case: one-qutritstate space.An arbitrary one-qutrit state can be decomposed in terms of the computationalbasis {| i , | i , | i} as follows | ω i = α ′ | i + β ′ | i + γ ′ | i , (27)where α ′ , β ′ , γ ′ are complex numbers. The normality condition is as follows | α ′ | + | β ′ | + | γ ′ | = 1 . (28)We want to show that the arbitrary state | ω i is a member of an orthonormalbasis. We do this by constructing such a basis explicitly. (We restrict ourselves tothe case α ′ , β ′ , γ ′ = 0, otherwise, the problem reduces to the lower dimensional cases.)Consider the projection of | ω i into the subspace spanned by {| i , | i} : | ψ ′ i = α ′ | i + β ′ | i . (29)So | ψ i = α ′ ψ ′ | i + β ′ ψ ′ | i := α | i + β | i ( ψ ′ := k | ψ ′ i k ) (30)is a normalized vector in this two dimensional subspace. | ψ i is obviously in the formof (19). So, up to phase factor, it can be written as (24) and we can find an orthogonalstate to | ψ i as (25): | ϕ i = β ∗ | i − α ∗ | i . (31)(Note that | ϕ i is also orthogonal to | ω i .) Now, we can write (27) as: | ω i = ψ ′ | ψ i + γ ′ | i , (32) I. Sargolzahi and E. Anjidani where | ψ i and | i are orthonormal. So, | ω i in (32) is also in the form of (19) whichagain can be written as (24) and we can find an orthogonal state to it as (25): | ν i = γ ′∗ | ψ i − ψ ′ | i ; (33)note that ψ ′ = k | ψ ′ i k is a positive number. Now it is obvious that the set {| ϕ i , | ω i , | ν i} is an orthonormal set. Also an arbitrary state | a i = α | i + α | i + α | i in this one-qutrit state space, can be decomposed in terms of {| ϕ i , | ω i , | ν i} : similarto (26), we can write | i and | i in terms of | ψ i in (30) and | ϕ i in (31) and then wecan write | ψ i and | i in terms of | ω i in (32) and | ν i in (33). So, we can write | a i interms of | ϕ i , | ω i and | ν i .In summary, {| ϕ i , | ω i , | ν i} , including our arbitrarily chosen state | ω i , constructsan orthonormal basis for three dimensional Hilbert space.Also note that | ϕ i and | ν i span a two dimensional subspace, like | i and | i in(19). So, similar to the previous section, instead of | ϕ i and | ν i , we can choose anyother two orthonormal states | ϕ ′ i and | ν ′ i which also span this subspace. Since | ϕ ′ i and | ν ′ i are written in terms of | ϕ i and | ν i , the vectors | ϕ ′ i and | ν ′ i are orthogonalto | ω i . Therefore, the set {| ω i , | ϕ ′ i , | ν ′ i} is also an orthonormal basis for one-qutritstate space.In other words, in a three dimensional Hilbert space (and also in higher dimen-sional cases) in contrast with the two dimensional one, we can find an infinite numberof orthonormal bases including our chosen state | ω i .In the two dimensional Hilbert space, for each | ψ i in (24), up to a phase factor,there is only one | φ i in (25) such that the set {| ψ i , | φ i} constructs an orthonormalbasis including | ψ i .Generalization of our result to the higher dimensions can be done by induction:Assume that in an n -dimensional Hilbert space, each arbitrary state is a member ofan orthonormal basis. We want to show that it is also true for ( n + 1)-dimensionalHilbert space.Let’s show the computational basis as {| i , | i , · · · } . So, in ( n + 1)-dimensionalstate space, we can decompose an arbitrary state | ω i as | ω i = n +1 X i =1 c ′ i | i i , (34) ach normalized state is a member of an orthonormal basis: A simple proof c ′ i are complex numbers and the normality condition is as follows n +1 X i =1 | c ′ i | = 1 . (35)We suppose that all c ′ i are nonzero, otherwise the problem reduces to the lower di-mensional cases.Now, consider the projection of | ω i into the n -dimensional subspace spanned by {| i , · · · , | n i} : | ψ ′ i = n X i =1 c ′ i | i i . (36)Hence | ψ i = | ψ ′ i ψ ′ = n X i =1 c ′ i ψ ′ | i i := n X i =1 c i | i i , ( ψ ′ := k | ψ ′ i k ) , (37)is a normalized vector in this n -dimensional subspace. Since in the n -dimensionalHilbert space each state is a member of an orthonormal basis, we can find n − | ϕ i i such that {| ϕ i , · · · , | ϕ n − i , | ψ i} constructs an orthonormal basis of the n -dimensional subspace spanned by {| i , · · · , | n i} . Thus, {| ϕ i , · · · , | ϕ n − i , | ψ i , | n + 1 i} is an orthonormal basis of the ( n + 1)-dimensional Hilbert space.In addition, we have | ω i = ψ ′ | ψ i + c ′ n +1 | n + 1 i . (38)States | ψ i and | n + 1 i span a two dimensional subspace which also can be spannedby orthonormal states | ω i and | ν i = c ′∗ n +1 | ψ i − ψ ′ | n + 1 i . (39)Therefore the set {| ϕ i , · · · , | ϕ n − i , | ω i , | ν i} , including the arbitrary state | ω i , is anorthonormal basis of ( n + 1)-dimensional state space and the proof is completed. In the previous section, we proved that in a finite dimensional state space, an arbitrarystate | ω i can be chosen as a member of an orthonormal basis. It seems valuable todiscuss briefly the consequences of this result.The first consequence of the above result is that, in addition to the computationalbasis, there exist other orthonormal bases for a finite dimensional state space. It has0 I. Sargolzahi and E. Anjidani an important consequence: existence of unitary operators. A unitary operator U isan operator satisfying the following relations [5]: U U † = I , U † U = I, (40)where U † is the adjoint of U and I is the identity operator. Now if we consider twoorthonormal bases {| i , · · · , | n i} and {| ϕ i , · · · , | ϕ n i} for an n -dimensional Hilbertspace, then it can be shown simply that U = n X i =1 | ϕ i i h i | (41)is a unitary operator [5].Also, using this fact that each state can be chosen as a member of an orthonormalbasis, one can prove the Cauchy-Schwarz inequality, as given in [1]. Till now, our discussion was restricted to the finite dimensional Hilbert spaces, whichare used widely in quantum information and computation theory [1]. However, count-able infinite dimensional Hilbert spaces are also of interest in quantum theory. Stan-dard examples in this context are the harmonic oscillator and the infinite well [6].Proving, in the general case, that an arbitrary normalized state | ω i , in a countableinfinite dimensional Hilbert space V , is a member of an orthonormal basis of V , isout of the scope of this paper. But, there is an important special case for which wecan prove the above statement simply, using our result in section 4.Assume that the set {| i , | i , · · · } is the orthonormal computational basis of ourcountable infinite dimensional Hilbert space V . So we can write each | ψ i ∈ V as | ψ i = ∞ X i =1 c i | i i , (42)where c i are complex numbers. Now, consider the special case that our arbitrarilychosen state | ω i can be written as | ω i = N X i =1 α i | i i , (43)where N is a finite positive integer. Therefore, | ω i belongs to a finite dimensionalsubspace of V , spanned by {| i , | i , · · · , | N i} . Let’s denote this subspace as V ′ . ach normalized state is a member of an orthonormal basis: A simple proof V ′ whichincludes | ω i : {| ω i = | e i , | e i , · · · , | e N i} . So, the whole V can be spanned by theorthonormal basis {| ω i = | e i , | e i , · · · , | e N i , | N + 1 i , | N + 2 i , · · · } , which includesour chosen state | ω i . In a finite dimensional Hilbert space, each normalized state can be chosen as a memberof an orthonormal basis of the space. Instead of the formal proof of this statement,we give a simple proof by constructing such a basis directly. In two dimensional case,this can be done by using the Bloch sphere. We generalize the two dimensional caseto the higher dimensional state spaces by induction. We hope that our proof be moreclear for physics students than the formal one.In addition, even when the Hilbert space is countable infinite dimensional, we canuse the above result, for an important special case, given in section 6.
Acknowledgments
We would like to thank an anonymous referee for his (her) suggestion to add a sectionabout the infinite dimensional case.
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