Lecture 13B: Supplementary Notes on Advanced Topics. 1 Inner Products and Outer Products for Single Particle States

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1 Lecture 13B: Supplementary Notes on Advanced Topics Outer Products, Operators, Density Matrices In order to explore the complexity of many particle systems a different way to represent multiparticle states is often introduced. The key quantity, the density matrix, is central to easily representing the nature of a multipartite state, be it eigenstate, superposition, or mixture. 1 Inner Products and Outer Products for Single Particle States Recall the definition of the inner product: Given two vectors, one from a vector space V n (C) (ket space) and the other from the dual Ṽ n (C) (bra space). Inner Product: A B = c, where c is a scalar related to the projection of one vector on the other. Since we are dealing with complex vector spaces, the answer may be complex. The outer product is another operation but this time takes a ket and a bra vector and returns an operator. Given two vectors, one from a vector space V n (C) (ket space) and the other from the dual Ṽ n (C) (bra space). Outer Product: A B = ˆQ, where ˆQ is an operator again related to the projection of one vector on the other but now returning a vector in the direction in question if it operators on a vector (either bra or ket). A few examples will help illuminate the nature of this product. Consider a photon prepared in a polarization state of P (by sending light through a linear polarizer). Now consider this photon impinging on a PBS that transmits H and reflects V polarized photons. The state transforms into the following after passing through such a device, P 1 [ H T + V R ]. Let s construct an operator that affects this transformation. Define an operator ˆQ PBS, ˆQ PBS P = 1 [ H T + V R ] = [ H T H + V R V ] P (1) = 1 [ H T H + V R V ] [ H + V ] = 1 [ H T H H + V R V V ] () = 1 [ H T + V R ] (3) The operator ˆQ PBS = H T H + V R V represents the action of the PBS device orientated in the HV basis. In what follows we will not discuss outer products so much as operators constructed out of ket-bra combinations (these always being operators that act on other vectors). 1.1 Examples of operators represented in terms of the form A B In what follows we will restrict ourselves to two state observers, and to keep the discussion familiar we will concentrate on our three complementary polarization bases (HV, PM, and RL) Identity Operator, ˆ1 This is usually the first operator considered, being the simplest. It takes a vector and returns the same vector. In one particular basis (say RL) all that need be done is to include all of the basis vectors in the operator. Here we require, ˆ1 = R R + L L. If this operator acts on either R, L, R, or even R the original vector is returned. It is important to convince yourself that the identity operator remains the identity operator if it is transformed to a new basis (use the relations between bases and transform to another basis). 1

2 1.1. Operator, ˆρ, corresponding to the state of a particle: Density Matrix Given a particle (photon) in a state ψ we can construct an operator based on the bra and ket of this state, i.e. ˆρ ψ ψ. What is the meaning of this type of operator? Let s examine the effects of this operator acting on various kets. First of all let s see what ˆρ acting on ψ results in, ˆρ ˆψ = ( ψ ψ ) ψ = ψ, (4) ψ ˆρ = ψ ( ψ ψ ) = ψ, (5) The special feature of this operator ρ, which we call the density matrix, is the following: The probability to find the state of the object in state ψ in the state ψ is Prob(measure outcome ψ when in state ψ) = ψ ψ = ψ ˆρ ψ = 1 Consider the probability to measure outcome H when the object is in state ψ, Prob(measure outcome H when in state ψ) = H ψ = H ψ H ψ = H ψ ψ H = H ˆρ H We see that the density matrix is an equivalent way to represent the state of the object, yielding the same probabilities. Why density matrix? In the example considered earlier we had a photon in a state of P polarization. When we represent this state in terms of a density matrix we get many terms if expressed in the HV basis, ˆρ = P P = 1 ( H + V )( H + V ) = 1 ( H H + H V + V H + V V ) This can be written as a matrix where the top row relates to H, the bottom row to V, the left column to H and the right column to V. This is in a form where we are representing our vectors as column and row vectors, ( ) ( ) H =, V =, H = ( ) and, V = (0 1). 0 1 The density matrix takes the following form, ˆρ = 1 ( ). (6) The density matrix has a few constraints that stem from the probabilities, these are reviewed in the appendix. The main feature to note is that the diagonal elements (from upper left to lower right) are related to the probability to measure the object in each state and the off diagonal elements are related to mixing of different basis vectors. These off diagonal elements are indicators of superpositions, if they are zero, then the object is not in a superposition in that basis. Compare the above density matrix to one for a photon in a state of H polarization, ˆρ = ( 0 0 ). (7) In this basis, the photon is not in a superposition but in a definite state. (Definite states, and also superpositions are examples of pure states. They can be represented by vectors in Hilbert space. Mixed states can not (see following).) Eigenstates and Superpositions and Mixtures Oh My! The real power of the density matrix is to classify states as being eigenstates (like our P photon in the PM basis), superpositions (like P in the HV basis), and mixtures. The difference between an ensemble of objects prepared in a mixture as contrasted with a superposition was introduced a few lectures back. Consider a source of photons that can be polarized in the H, V, or P direction and measurements made in the HV basis only. Eigenstate: If every photon emitted from the source is in the H direction, what we have is an eigenstate or pure state. If measured, every single photon will be measured to be H. The density matrix in this case is given by (7) above. Superposition: If every photon is emitted from the source in the P direction, then this photon is in a superposition in the HV basis. In this case the density matrix has off diagonal terms and is represented by (6)

3 above. A superpositions (even entangled states) are pure states and can be represented by a vector in a Hilbert space. Mixture: Now consider the source to be emitting H and V photons randomly (let s say 50/50). In terms of our original ket vectors, we can not represent such a state easily. Is it the sum of H and V? No, because that is a superposition. To mathematically represent such a state necessitates a density matrix. In this case there is equal probability to measure the object in H or V and it is not in a superposition - there are no off diagonal terms. Thus the density matrix for the mixture is, ˆρ mixture = 1 ( ). (8) 0 1 This represents a state of classical ignorance, the photon is in one of the states but is not in a superposition. Thus diagonal matrices (meaning there are no off diagonal terms) with more than one non-zero element are mixtures. The example of flipping a coin and its state before revealing the result is such an example, it is in a 50/50 mixture but not in a superposition. Once revealed, the matrix does collapse to one or the other result Decoherence Theory Thus, the whole question of decoherence theory is how does interaction with the environment drive a system from a superposition to a mixture? All agree that decoherence is an important fact for yielding macroscopic (classical) phenomena from microscopic (quantum) phenomena. However, there is still debate as to whether this solves the measurement problem. For it is true that decoherence severely limits the existence of macroscopic superpositions however it does not determine which of the outcomes will result from a measurement (alive or dead cat). Schrödinger s cat is no longer in a superposition of being alive and dead but quantum theory still does not provide an answer to which state is revealed when the door is opened (collapses to one of the results). To try to understand how this works, consider the case of the decaying atom and a detector. Before a measurement is done the system is in a superposition, [ n + y ] D. The detector is in a state prior to a measurement (D). After the measurement is done the two subsystems become correlated, [ n 1 + y ]. Where the two correlated states are either that the atom has not decayed (n) and the detector is in position 1, or the atom has decayed (y) and the detector is in position. The density matrix in this case, without interactions of the environment, is, ˆρ noe = 1 ( ). (9) Including the interactions with the environment gives a larger superposition upon measurement (since the environment would clearly be different in the two cases). Before measurement we have, ψ 0 = 1 [ n + y ] D E. Note that the environment E does not become entangled with the atom since it is microscopic - it only interacts with the detector, assumed macroscopic. After the measurement the state becomes, [ n 1 E 1 + y E ]. Since we never measure the environment, this state can be reduced to involve only the states that can be measured, the atom and the detector. This is mathematically represented by a partial trace, you sum over the subsystems Hilbert spaces that are not measured. This is same as when finding correlations between two observers in a three observer system. In order to get the correct expression, you need to include both of the possibilities of the third observer, i.e. sum over the possibilities of the third observer. In this case, taking a partial trace over the environment leaves a density matrix only representing the atom and the detector, but the result is that 3

4 it is now a mixture! The atom and detector are no longer in a superposition but a mixture. Schrödinger s cat is not in a superposition of alive and dead, but in a mixture - a statement of classical ignorance. Some dispute that this completely solves the measurement problem as a) which of the two states is finally observed is still inherently random and b) there still is a superposition with the environment. Most agree that decoherence is a major advance in quantum theory, an explanation why we do not observe macroscopic quantum superpositions, but there are still unanswered questions related to measurement. 1. Decoherence Theory A model of a two state system (A) in a superposition (cat state) interacting with the environment (E) can be written as Ψ = α 0 A 0 E + β 1 A 1 E. (10) The density matrix for this state is ˆρ A&E = 1 α 0 0 α β αβ 0 0 β. (11) Since the state of the environment is never measured (FAPP 1 ) we want to find the density matrix for only subsystem A. To do this, one performs a partial trace over the ignored subsystem (here E). This is done by summing over all the possible outcomes for system E (just as with probability), resulting in a reduced density matrix for system A alone, ˆρ A = Tr E ˆρ A&E = E i Ψ Ψ i E = E 0 Ψ Ψ 0 E + E 1 Ψ Ψ 1 E iofe = E 0 [ α 0 0 E 0 E 0 + α β 0 0 E 1 E 1 + ] 0 E + E 1 [ α 0 0 E 0 E 0 + β 1 1 E 1 E 1 + ] 1 E (1) In the full expression notice that only the first and last terms survive if 0 E is orthogonal to 1 E. Assuming they are orthogonal we have only the diagonal terms left. ( ) ˆρ A = α β α = 0 β (13) If the two states of the environment are orthogonal then the density matrix, the quantum state of system A has become a mixture when we ignore the environment. Of course the whole system is still in a superposition but we only measure system A. This process is called decoherence. We have lost the phase information (phase of α and β ) when we trace out the environment. The whole study of decoherence is more complex than this but this is the main idea. Often one needs to examine the states of the environment and determine whether they are orthogonal or nearly orthogonal etc. Appendix: Mathematical Properties of the Density Matrix The main constraint on any density matrix is that its trace must be 1. The trace is simply the sum of the diagonal elements. The reason stems from the fact that if measured in a basis, the probabilities to measure it in any state must add up to 1. To show why, consider a non-physical state ˆρ bad = R L. Why is this non-physical? No matter which outcome you try to observe this particle in the probability vanishes, Prob(measure outcome R when in state bad) = R ˆρ bad R = R R L R = 0. Thus the diagonal terms are directly related to the probability to measure the object in the various states in that basis. Thus the diagonal entries must be positive and constrained to [0, 1]. The density matrix is an example of what is called a positive operator (more formally its determinant must be zero or positive). In short, the density matrix must be a positive, trace 1 operator. 1 For All Practical Purposes 4

5 .1 Multiparticle density matrices If we consider the case of two entangled particles, recall that a possible state vector could be, ( H V V H ) where the first ket refers to particle 1 and the second to particle. To represent the density matrix for this system we must increase the size of our Hilbert space from to 4 dimensions. Now, the basis can be represented by the four basis kets: H H, H V, V H, and H H. In this expanded basis the density matrix is, ˆρ entangled = (14) 0 0 Recall that, in terms of vectors, the criteria to have entanglement was that the vector could not be factorized into a product of one on the other, ψ ψ 1 ψ. The same criterion applies to density matrices, if they cannot be expressed as a direct product: ρ total = ρ 1 ρ then they are entangled. A direct product of matrices is simply each element of one matrix multiplied by the other matrix, thus expanding the size to N M where N is the size of the first matrix and M is the size of the second. An example will make it more clear. Consider two matrices, ( ) ( ) a11 a A = 1 b11 b, B = 1. a 1 a b 1 b The direct product of these two matrices is, ( ) ( ) b11 b a 1 b11 b 11 a 1 a 11 b 11 a 11 b 1 a 1 b 11 a 1 b 1 A B = b 1 b 1 ( ) ( b 1 b ) b11 b a 1 b11 b 1 a 1 = a 11 b 1 a 11 b a 1 b 1 a 1 b a 1 b 11 a 1 b 1 a b 11 a b 1 b 1 b b 1 b a 1 b 1 a 1 b a b 1 a b 5