Important Theorems in Quantum Mechanics: Hermitian Operators and Eigenfunctions - Prof. Wi, Study notes of Physical Chemistry

Download Important Theorems in Quantum Mechanics: Hermitian Operators and Eigenfunctions - Prof. Wi and more Study notes Physical Chemistry in PDF only on Docsity! Notes on VIT #1,2,3 for the non- degenerate and degenerate cases; and a summary of Quantum Measurment tools. AL The Three Very Imporant Theorems, the case of non- degeneracy. #1 Hermitian operators have real eigenvalues. To prove: A is Hermetian, and has an eigenfucntion fa with eigenvalue "a", then "a" is a real number. Proof: By assumption: Ÿtfa*HtL A faHtL ‚ t = Ÿt HA fa HtLL* faHtL ‚ t, and AfaHtL = afaHtL. The integrations co-ordinates and range of integration, denoted by t, are appropriate to the problem at hand. Then Ÿtfa*HtL a faHtL ‚ t = Ÿt Ha faHtLL* faHtL ‚ t, or, a Ÿtfa*HtL faHtL ‚ t = a* Ÿtfa*HtL faHtL ‚ t, from which a = a*, from which VIT#1 follows. QED #2 Hermitian linear operators have orthogonal eigenfunctions. Let us assume that Hermitian operator A has an eigenfunction fa with non- degenate eigenvalue "a". We need make no assumptions about the possible degeneracy of any of the other eigenfunctions of A. To prove: Ÿtfa*HtL fbHtL ‚ t = Ÿtfb* HtL faHtL ‚ t = 0, where the pair "b, and fbHtL" are any other eigenvalue, eigenfunction of A. By assumption of the non-degeneracy of "a" we have a ≠ b. Proof: As A is Hermitian: Ÿtfb* HtL A faHtL ‚ t = Ÿt HA fb HtLL* faHtL ‚ t, and thus a Ÿtfa*HtL faHtL ‚ t = b Ÿtfa*HtL faHtL ‚ t, as both eigenvalues are real from VIT #1. Subtracting the right side from the left gives: (a - b ) Ÿtfa*HtL faHtL ‚ t = 0. As a ≠b by assumption, it follows that Ÿtfa*HtL faHtL ‚ t = 0. QED #3 Commuting Hermitian linear operaors have simultaneous eigenfunctions. Consider the two distinct operators A,B, such that [A,B] = AB - BA =0. Now assume that one of the operators, say A has an engenfunction fa with non-degenerate eigenvalue a. Note that we need assume nothing about the degeneracy of eigenvalues of B. To prove: fa is automatically an eigenfunction of operator B. Proof: as [A,B] is the "zero operator" it give zero when applied to "any function," specifically: [A,B] fa = 0. Thus, AB fa - BA fa = 0. Using the fact that fa is an eigenfunction of A, this may be re-written as: A (B fa N = a (B fa N , this is equation (*). The non-degeneracy of "a" now implies that these is only a single linearly independent function which satisfies Eqn. (*), and that is fa itself, or a constant multiple of itself. Note that this is the 2nd time in the proof we have used linearity. Thus (B fa N µ fa , where "µ " denotes proportionality. As these funcions (B fa M and fa are proportional there is some constant, b, such that: Bfa = bfa. Thus fa is an eigenfunction of operator B. QED. REMARK: note that the eigenvalues a and b are completely unrelated! They usually have different units, and so can't even be compared. For example one might be an "energy," the other an "angular momentum." B) The Three Very Imporant Theorems, the case of degeneracy. #1 Eigenvalues of Hermitian operators are real. This result has nothing to do with degeneracy, so the proof is the same! PREAMBLE to VIT #2 and #3 in the case of degeneracy. Before discussing VIT #2 and #3, which are actually "changed" by the presence of degeneracy, a few useful remarks about "degenerate" eigenvalues and "degenerate" eigenfuctions may be of use. Suppose operator A has two, and only two, distinct (i. e. linearly independent, which simply means one is not simply "a constant" times the other) eigenfuctions, say fa 1 and fa2 which have the same eigenvalue "a". We then say that "a" is doubly degenerate. If there and N, and only N, linearly independent eigenfunctions which share the same eigenvalue then the degeneracy is N-fold. We can also say that the eigenfuncions belong to an N-fold degenerate set of eigenfuctions. For example the degeneracy of "p-orbitals" is 3-fold, and "d-orbitals" is 5-fold....so these are not ideas irrelavent to CHEMISTRY. If they were we'd happily skip them as mathematical oddities and simply move on. Note also that a set of functions which share an eigenvalue of operator A, most often are non-degenerate with respect to the action of other operators, say B, or even C! An important fact, which we then prove: any linear combination of degenerate functions of operator A is also an eigenfunction of operator A. Proof: Consider A( ⁄i=1N cifai ) = ⁄i=1N ciAfai = ⁄i=1N ciafai = a(⁄i=1N cifai ). QED. Note that linearity of A was used "several times" in this simple demonsteation. Note also that if any one of the funcions fi in the sum didn't have eigenvalue "a", then the a would not factor outsidet he sum, and the result would be false. Said specifically: "linear combinations of non-degenerate eigenfuctions" of an operator are NOT eigenfunctions of that opertor. As the above result has nothing to do with "what linear combination" has been taken, it applies equally to all linear combinations. Can we then generate an infintie number of eigenfuctions of operator A? Of course not: For a double degenerate eigenvalue only two linearly independent eigenfuctions exist, and taking linear combinations does not expand the "space" of degenerate functions beyond two, it just gives different representaions of the same space. An analog in 2D Cartesian space is that any two linearly independent vectors can serve to represent any other vector: but this may be done in an infinite number of way, simply by "rotating" the x,y axies on the page....but nothing new is actually produced by doing that. VIT #2 in the case of degeneracy: "We can always choose to take all of the eigenfunctions of an Hermetian operator to be orthonormal, but we might have to do a bit of work to find the linear combinations of degenerate eigenfunctions that makes this work." Discussion: Where do things not happen "automatically?" as in the case of a non- degenerate eigenvalue of A? Supose "a" is a multiply degenerate eigenvalue, and we examine fa1 and fa2 (note that these superscrips are NOT powers!). Following the Proof of VIT #2 in part A) we find that the key result: (a-a) Ÿt I fa1 HtLM * fa 2HtL ‚ t = 0 does not tell us whether Ÿt I fa1 HtLM * fa 2HtL ‚ t is zero or not, it might be that fa1 and fa2 are orthogonal, but we haven't proven it. It might also be the case that they are NOT orthogonal. BUT, and we state this without detailed proof: if fa1 and fa2 are NOT orthogonal it is always possible to find two linearly independent and orthonormal linear combinations of fa1 and fa2 which ARE orthogonal. We will see examples of both cases as Chem 455A develops. VIT#3 in the case of degeneracy: "We can always choose appropraite linear combinations of degenerate eigenfunctions to obtain simultaneous eigenfunctions of two commuting observables, but we might have to do a bit of work to find those linear combinations." Looking back to part A) the key step was: (B fa L µ fa , which implied that fa was an eigenfunction of B. In the case of degeneracy, we have to replace this by (B fa L µ "ANY linear combination of the degenerate eigenfucntions of A with eigenvalue a". Thus (B fa L is NOT simply a multiple of fa . What do we do? We need to then FIND appropriate linear combinations of the degenerate eigenfucntions of A which ARE also eigenfunctions of B. This can, and we don't give the proof, always be done. We will see examples of this, also, as Chem. 455A proceeds. C) The Measurement Postulate stated in three different ways Statement #1: Given an observable "A" the only results of measurments are the eigenvalues of A. Commentary: this gives us a set (often infinite) of eigenvalues {ai <, which will include anything we measure regarding propert A. However, this doen't tell us how what we might see in a given experimental situation depends on Y(t), the wavefunction. What we can know about a system is contained the wavefunction (Postulate #1) so there should be a relationship between Y and "which of the possible eigenvalues {ai} we might actually expect to find," if the system is in state Y. So a somewhat more useful statement is: Statement #2: The average value of a large number of independent measurements of the property A, when the system is always prepared with wave function Y, is <A> ≡ Ÿt Y* Aop Y ‚ t, where Y is the normalized wavefunction. The uncertainty DA is then defined as < A2 > - H < A >L2+ , and gives a measure of the "spread" in expected results. Statement #3: However, we can go further, and actaully predict which eigenvalues will actually be observed, and with what probability. This is a lot more infomation that just the average + standard deviation (aka quantum uncertainty.) This is done by analogy with finding the "components of a vector in 3D Euclidean Space by projecting the a vector of interest onto the unit vectors chosen to be the "basis." Suppose we have an ortho-normal set of unit vectors i,j,k: The components of a vector V are given by (v1, v2, v3), and the vector itself may be writen as V = v1 i” + v2 j+ v3k where v1= i”.V ,v2= j .V , and v3 = k .V . Note that we have found the component by "projecting the unit vector onto the vector V". In QM the analog is: The components of a wavefunction Y(t), along the directions defined by the orthonormal eigenfunctions, fi HtL, of an obserable are: ci = Ÿtfi*HtL Y HtL ‚ t, and then Y(t) = ⁄i=1N ci fi HtL. If Y(t) is normalized, and the fi HtL are eigenfunctions of observable "A", with eigenvalue ai , then the probability of observing any one of the eigenvalues, say the j th, is given by Pj= I … cj …M2. If an expansion coefficient, say c14 ,is zero, for Y, we will never observe eigenvalue #14, as there is no "component of Y" in that "direction" in the function space. Notes on VIT #1,2,3 for the non- degenerate and degenerate cases; and a summary of Quantum Measurment tools. AL The Three Very Imporant Theorems, the case of non- degeneracy. #1 Hermitian operators have real eigenvalues. To prove: A is Hermetian, and has an eigenfucntion fa with eigenvalue "a", then "a" is a real number. Proof: By assumption: Ÿtfa*HtL A faHtL ‚ t = Ÿt HA fa HtLL* faHtL ‚ t, and AfaHtL = afaHtL. The integrations co-ordinates and range of integration, denoted by t, are appropriate to the problem at hand. Then Ÿtfa*HtL a faHtL ‚ t = Ÿt Ha faHtLL* faHtL ‚ t, or, a Ÿtfa*HtL faHtL ‚ t = a* Ÿtfa*HtL faHtL ‚ t, from which a = a*, from which VIT#1 follows. QED #2 Hermitian linear operators have orthogonal eigenfunctions. Let us assume that Hermitian operator A has an eigenfunction fa with non- degenate eigenvalue "a". We need make no assumptions about the possible degeneracy of any of the other eigenfunctions of A. To prove: Ÿtfa*HtL fbHtL ‚ t = Ÿtfb* HtL faHtL ‚ t = 0, where the pair "b, and fbHtL" are any other eigenvalue, eigenfunction of A. By assumption of the non-degeneracy of "a" we have a ≠ b. Proof: As A is Hermitian: Ÿtfb* HtL A faHtL ‚ t = Ÿt HA fb HtLL* faHtL ‚ t, and thus a Ÿtfa*HtL faHtL ‚ t = b Ÿtfa*HtL faHtL ‚ t, as both eigenvalues are real from VIT #1. Subtracting the right side from the left gives: (a - b ) Ÿtfa*HtL faHtL ‚ t = 0. As a ≠b by assumption, it follows that Ÿtfa*HtL faHtL ‚ t = 0. QED #3 Commuting Hermitian linear operaors have simultaneous eigenfunctions. Consider the two distinct operators A,B, such that [A,B] = AB - BA =0. Now assume that one of the operators, say A has an engenfunction fa with non-degenerate eigenvalue a. Note that we need assume nothing about the degeneracy of eigenvalues of B. To prove: fa is automatically an eigenfunction of operator B. Proof: as [A,B] is the "zero operator" it give zero when applied to "any function," specifically: [A,B] fa = 0. Thus, AB fa - BA fa = 0. Using the fact that fa is an eigenfunction of A, this may be re-written as: A (B fa N = a (B fa N , this is equation (*). The non-degeneracy of "a" now implies that these is only a single linearly independent function which satisfies Eqn. (*), and that is fa itself, or a constant multiple of itself. Note that this is the 2nd time in the proof we have used linearity. Thus (B fa N µ fa , where "µ " denotes proportionality. As these funcions (B fa M and fa are proportional there is some constant, b, such that: Bfa = bfa. Thus fa is an eigenfunction of operator B. QED. REMARK: note that the eigenvalues a and b are completely unrelated! They usually have different units, and so can't even be compared. For example one might be an "energy," the other an "angular momentum." B) The Three Very Imporant Theorems, the case of degeneracy. #1 Eigenvalues of Hermitian operators are real. This result has nothing to do with degeneracy, so the proof is the same! PREAMBLE to VIT #2 and #3 in the case of degeneracy. Before discussing VIT #2 and #3, which are actually "changed" by the presence of degeneracy, a few useful remarks about "degenerate" eigenvalues and "degenerate" eigenfuctions may be of use. Suppose operator A has two, and only two, distinct (i. e. linearly independent, which simply means one is not simply "a constant" times the other) eigenfuctions, say fa 1 and fa2 which have the same eigenvalue "a". We then say that "a" is doubly degenerate. If there and N, and only N, linearly independent eigenfunctions which share the same eigenvalue then the degeneracy is N-fold. We can also say that the eigenfuncions belong to an N-fold degenerate set of eigenfuctions. For example the degeneracy of "p-orbitals" is 3-fold, and "d-orbitals" is 5-fold....so these are not ideas irrelavent to CHEMISTRY. If they were we'd happily skip them as mathematical oddities and simply move on. Note also that a set of functions which share an eigenvalue of operator A, most often are non-degenerate with respect to the action of other operators, say B, or even C! An important fact, which we then prove: any linear combination of degenerate functions of operator A is also an eigenfunction of operator A. Proof: Consider A( ⁄i=1N cifai ) = ⁄i=1N ciAfai = ⁄i=1N ciafai = a(⁄i=1N cifai ). QED. Note that linearity of A was used "several times" in this simple demonsteation. Note also that if any one of the funcions fi in the sum didn't have eigenvalue "a", then the a would not factor outsidet he sum, and the result would be false. Said specifically: "linear combinations of non-degenerate eigenfuctions" of an operator are NOT eigenfunctions of that opertor. As the above result has nothing to do with "what linear combination" has been taken, it applies equally to all linear combinations. Can we then generate an infintie number of eigenfuctions of operator A? Of course not: For a double degenerate eigenvalue only two linearly independent eigenfuctions exist, and taking linear combinations does not expand the "space" of degenerate functions beyond two, it just gives different representaions of the same space. An analog in 2D Cartesian space is that any two linearly independent vectors can serve to represent any other vector: but this may be done in an infinite number of way, simply by "rotating" the x,y axies on the page....but nothing new is actually produced by doing that. VIT #2 in the case of degeneracy: "We can always choose to take all of the eigenfunctions of an Hermetian operator to be orthonormal, but we might have to do a bit of work to find the linear combinations of degenerate eigenfunctions that makes this work." Discussion: Where do things not happen "automatically?" as in the case of a non- degenerate eigenvalue of A? Supose "a" is a multiply degenerate eigenvalue, and we examine fa1 and fa2 (note that these superscrips are NOT powers!). Following the Proof of VIT #2 in part A) we find that the key result: (a-a) Ÿt I fa1 HtLM * fa 2HtL ‚ t = 0 does not tell us whether Ÿt I fa1 HtLM * fa 2HtL ‚ t is zero or not, it might be that fa1 and fa2 are orthogonal, but we haven't proven it. It might also be the case that they are NOT orthogonal. BUT, and we state this without detailed proof: if fa1 and fa2 are NOT orthogonal it is always possible to find two linearly independent and orthonormal linear combinations of fa1 and fa2 which ARE orthogonal. We will see examples of both cases as Chem 455A develops. VIT#3 in the case of degeneracy: "We can always choose appropraite linear combinations of degenerate eigenfunctions to obtain simultaneous eigenfunctions of two commuting observables, but we might have to do a bit of work to find those linear combinations." Looking back to part A) the key step was: (B fa L µ fa , which implied that fa was an eigenfunction of B. In the case of degeneracy, we have to replace this by (B fa L µ "ANY linear combination of the degenerate eigenfucntions of A with eigenvalue a". Thus (B fa L is NOT simply a multiple of fa . What do we do? We need to then FIND appropriate linear combinations of the degenerate eigenfucntions of A which ARE also eigenfunctions of B. This can, and we don't give the proof, always be done. We will see examples of this, also, as Chem. 455A proceeds. C) The Measurement Postulate stated in three different ways Statement #1: Given an observable "A" the only results of measurments are the eigenvalues of A. Commentary: this gives us a set (often infinite) of eigenvalues {ai <, which will include anything we measure regarding propert A. However, this doen't tell us how what we might see in a given experimental situation depends on Y(t), the wavefunction. What we can know about a system is contained the wavefunction (Postulate #1) so there should be a relationship between Y and "which of the possible eigenvalues {ai} we might actually expect to find," if the system is in state Y. So a somewhat more useful statement is: Statement #2: The average value of a large number of independent measurements of the property A, when the system is always prepared with wave function Y, is <A> ≡ Ÿt Y* Aop Y ‚ t, where Y is the normalized wavefunction. The uncertainty DA is then defined as < A2 > - H < A >L2+ , and gives a measure of the "spread" in expected results. Statement #3: However, we can go further, and actaully predict which eigenvalues will actually be observed, and with what probability. This is a lot more infomation that just the average + standard deviation (aka quantum uncertainty.) This is done by analogy with finding the "components of a vector in 3D Euclidean Space by projecting the a vector of interest onto the unit vectors chosen to be the "basis." Suppose we have an ortho-normal set of unit vectors i,j,k: The components of a vector V are given by (v1, v2, v3), and the vector itself may be writen as V = v1 i” + v2 j+ v3k where v1= i”.V ,v2= j .V , and v3 = k .V . Note that we have found the component by "projecting the unit vector onto the vector V". In QM the analog is: The components of a wavefunction Y(t), along the directions defined by the orthonormal eigenfunctions, fi HtL, of an obserable are: ci = Ÿtfi*HtL Y HtL ‚ t, and then Y(t) = ⁄i=1N ci fi HtL. If Y(t) is normalized, and the fi HtL are eigenfunctions of observable "A", with eigenvalue ai , then the probability of observing any one of the eigenvalues, say the j th, is given by Pj= I … cj …M2. If an expansion coefficient, say c14 ,is zero, for Y, we will never observe eigenvalue #14, as there is no "component of Y" in that "direction" in the function space. 2 Chem455A_Lecture12,13.nb Notes on VIT #1,2,3 for the non- degenerate and degenerate cases; and a summary of Quantum Measurment tools. AL The Three Very Imporant Theorems, the case of non- degeneracy. #1 Hermitian operators have real eigenvalues. To prove: A is Hermetian, and has an eigenfucntion fa with eigenvalue "a", then "a" is a real number. Proof: By assumption: Ÿtfa*HtL A faHtL ‚ t = Ÿt HA fa HtLL* faHtL ‚ t, and AfaHtL = afaHtL. The integrations co-ordinates and range of integration, denoted by t, are appropriate to the problem at hand. Then Ÿtfa*HtL a faHtL ‚ t = Ÿt Ha faHtLL* faHtL ‚ t, or, a Ÿtfa*HtL faHtL ‚ t = a* Ÿtfa*HtL faHtL ‚ t, from which a = a*, from which VIT#1 follows. QED #2 Hermitian linear operators have orthogonal eigenfunctions. Let us assume that Hermitian operator A has an eigenfunction fa with non- degenate eigenvalue "a". We need make no assumptions about the possible degeneracy of any of the other eigenfunctions of A. To prove: Ÿtfa*HtL fbHtL ‚ t = Ÿtfb* HtL faHtL ‚ t = 0, where the pair "b, and fbHtL" are any other eigenvalue, eigenfunction of A. By assumption of the non-degeneracy of "a" we have a ≠ b. Proof: As A is Hermitian: Ÿtfb* HtL A faHtL ‚ t = Ÿt HA fb HtLL* faHtL ‚ t, and thus a Ÿtfa*HtL faHtL ‚ t = b Ÿtfa*HtL faHtL ‚ t, as both eigenvalues are real from VIT #1. Subtracting the right side from the left gives: (a - b ) Ÿtfa*HtL faHtL ‚ t = 0. As a ≠b by assumption, it follows that Ÿtfa*HtL faHtL ‚ t = 0. QED #3 Commuting Hermitian linear operaors have simultaneous eigenfunctions. Consider the two distinct operators A,B, such that [A,B] = AB - BA =0. Now assume that one of the operators, say A has an engenfunction fa with non-degenerate eigenvalue a. Note that we need assume nothing about the degeneracy of eigenvalues of B. To prove: fa is automatically an eigenfunction of operator B. Proof: as [A,B] is the "zero operator" it give zero when applied to "any function," specifically: [A,B] fa = 0. Thus, AB fa - BA fa = 0. Using the fact that fa is an eigenfunction of A, this may be re-written as: A (B fa N = a (B fa N , this is equation (*). The non-degeneracy of "a" now implies that these is only a single linearly independent function which satisfies Eqn. (*), and that is fa itself, or a constant multiple of itself. Note that this is the 2nd time in the proof we have used linearity. Thus (B fa N µ fa , where "µ " denotes proportionality. As these funcions (B fa M and fa are proportional there is some constant, b, such that: Bfa = bfa. Thus fa is an eigenfunction of operator B. QED. REMARK: note that the eigenvalues a and b are completely unrelated! They usually have different units, and so can't even be compared. For example one might be an "energy," the other an "angular momentum." B) The Three Very Imporant Theorems, the case of degeneracy. #1 Eigenvalues of Hermitian operators are real. This result has nothing to do with degeneracy, so the proof is the same! PREAMBLE to VIT #2 and #3 in the case of degeneracy. Before discussing VIT #2 and #3, which are actually "changed" by the presence of degeneracy, a few useful remarks about "degenerate" eigenvalues and "degenerate" eigenfuctions may be of use. Suppose operator A has two, and only two, distinct (i. e. linearly independent, which simply means one is not simply "a constant" times the other) eigenfuctions, say fa 1 and fa2 which have the same eigenvalue "a". We then say that "a" is doubly degenerate. If there and N, and only N, linearly independent eigenfunctions which share the same eigenvalue then the degeneracy is N-fold. We can also say that the eigenfuncions belong to an N-fold degenerate set of eigenfuctions. For example the degeneracy of "p-orbitals" is 3-fold, and "d-orbitals" is 5-fold....so these are not ideas irrelavent to CHEMISTRY. If they were we'd happily skip them as mathematical oddities and simply move on. Note also that a set of functions which share an eigenvalue of operator A, most often are non-degenerate with respect to the action of other operators, say B, or even C! An important fact, which we then prove: any linear combination of degenerate functions of operator A is also an eigenfunction of operator A. Proof: Consider A( ⁄i=1N cifai ) = ⁄i=1N ciAfai = ⁄i=1N ciafai = a(⁄i=1N cifai ). QED. Note that linearity of A was used "several times" in this simple demonsteation. Note also that if any one of the funcions fi in the sum didn't have eigenvalue "a", then the a would not factor outsidet he sum, and the result would be false. Said specifically: "linear combinations of non-degenerate eigenfuctions" of an operator are NOT eigenfunctions of that opertor. As the above result has nothing to do with "what linear combination" has been taken, it applies equally to all linear combinations. Can we then generate an infintie number of eigenfuctions of operator A? Of course not: For a double degenerate eigenvalue only two linearly independent eigenfuctions exist, and taking linear combinations does not expand the "space" of degenerate functions beyond two, it just gives different representaions of the same space. An analog in 2D Cartesian space is that any two linearly independent vectors can serve to represent any other vector: but this may be done in an infinite number of way, simply by "rotating" the x,y axies on the page....but nothing new is actually produced by doing that. VIT #2 in the case of degeneracy: "We can always choose to take all of the eigenfunctions of an Hermetian operator to be orthonormal, but we might have to do a bit of work to find the linear combinations of degenerate eigenfunctions that makes this work." Discussion: Where do things not happen "automatically?" as in the case of a non- degenerate eigenvalue of A? Supose "a" is a multiply degenerate eigenvalue, and we examine fa1 and fa2 (note that these superscrips are NOT powers!). Following the Proof of VIT #2 in part A) we find that the key result: (a-a) Ÿt I fa1 HtLM * fa 2HtL ‚ t = 0 does not tell us whether Ÿt I fa1 HtLM * fa 2HtL ‚ t is zero or not, it might be that fa1 and fa2 are orthogonal, but we haven't proven it. It might also be the case that they are NOT orthogonal. BUT, and we state this without detailed proof: if fa1 and fa2 are NOT orthogonal it is always possible to find two linearly independent and orthonormal linear combinations of fa1 and fa2 which ARE orthogonal. We will see examples of both cases as Chem 455A develops. VIT#3 in the case of degeneracy: "We can always choose appropraite linear combinations of degenerate eigenfunctions to obtain simultaneous eigenfunctions of two commuting observables, but we might have to do a bit of work to find those linear combinations." Looking back to part A) the key step was: (B fa L µ fa , which implied that fa was an eigenfunction of B. In the case of degeneracy, we have to replace this by (B fa L µ "ANY linear combination of the degenerate eigenfucntions of A with eigenvalue a". Thus (B fa L is NOT simply a multiple of fa . What do we do? We need to then FIND appropriate linear combinations of the degenerate eigenfucntions of A which ARE also eigenfunctions of B. This can, and we don't give the proof, always be done. We will see examples of this, also, as Chem. 455A proceeds. C) The Measurement Postulate stated in three different ways Statement #1: Given an observable "A" the only results of measurments are the eigenvalues of A. Commentary: this gives us a set (often infinite) of eigenvalues {ai <, which will include anything we measure regarding propert A. However, this doen't tell us how what we might see in a given experimental situation depends on Y(t), the wavefunction. What we can know about a system is contained the wavefunction (Postulate #1) so there should be a relationship between Y and "which of the possible eigenvalues {ai} we might actually expect to find," if the system is in state Y. So a somewhat more useful statement is: Statement #2: The average value of a large number of independent measurements of the property A, when the system is always prepared with wave function Y, is <A> ≡ Ÿt Y* Aop Y ‚ t, where Y is the normalized wavefunction. The uncertainty DA is then defined as < A2 > - H < A >L2+ , and gives a measure of the "spread" in expected results. Statement #3: However, we can go further, and actaully predict which eigenvalues will actually be observed, and with what probability. This is a lot more infomation that just the average + standard deviation (aka quantum uncertainty.) This is done by analogy with finding the "components of a vector in 3D Euclidean Space by projecting the a vector of interest onto the unit vectors chosen to be the "basis." Suppose we have an ortho-normal set of unit vectors i,j,k: The components of a vector V are given by (v1, v2, v3), and the vector itself may be writen as V = v1 i” + v2 j+ v3k where v1= i”.V ,v2= j .V , and v3 = k .V . Note that we have found the component by "projecting the unit vector onto the vector V". In QM the analog is: The components of a wavefunction Y(t), along the directions defined by the orthonormal eigenfunctions, fi HtL, of an obserable are: ci = Ÿtfi*HtL Y HtL ‚ t, and then Y(t) = ⁄i=1N ci fi HtL. If Y(t) is normalized, and the fi HtL are eigenfunctions of observable "A", with eigenvalue ai , then the probability of observing any one of the eigenvalues, say the j th, is given by Pj= I … cj …M2. If an expansion coefficient, say c14 ,is zero, for Y, we will never observe eigenvalue #14, as there is no "component of Y" in that "direction" in the function space. Chem455A_Lecture12,13.nb 5 Notes on VIT #1,2,3 for the non- degenerate and degenerate cases; and a summary of Quantum Measurment tools. AL The Three Very Imporant Theorems, the case of non- degeneracy. #1 Hermitian operators have real eigenvalues. To prove: A is Hermetian, and has an eigenfucntion fa with eigenvalue "a", then "a" is a real number. Proof: By assumption: Ÿtfa*HtL A faHtL ‚ t = Ÿt HA fa HtLL* faHtL ‚ t, and AfaHtL = afaHtL. The integrations co-ordinates and range of integration, denoted by t, are appropriate to the problem at hand. Then Ÿtfa*HtL a faHtL ‚ t = Ÿt Ha faHtLL* faHtL ‚ t, or, a Ÿtfa*HtL faHtL ‚ t = a* Ÿtfa*HtL faHtL ‚ t, from which a = a*, from which VIT#1 follows. QED #2 Hermitian linear operators have orthogonal eigenfunctions. Let us assume that Hermitian operator A has an eigenfunction fa with non- degenate eigenvalue "a". We need make no assumptions about the possible degeneracy of any of the other eigenfunctions of A. To prove: Ÿtfa*HtL fbHtL ‚ t = Ÿtfb* HtL faHtL ‚ t = 0, where the pair "b, and fbHtL" are any other eigenvalue, eigenfunction of A. By assumption of the non-degeneracy of "a" we have a ≠ b. Proof: As A is Hermitian: Ÿtfb* HtL A faHtL ‚ t = Ÿt HA fb HtLL* faHtL ‚ t, and thus a Ÿtfa*HtL faHtL ‚ t = b Ÿtfa*HtL faHtL ‚ t, as both eigenvalues are real from VIT #1. Subtracting the right side from the left gives: (a - b ) Ÿtfa*HtL faHtL ‚ t = 0. As a ≠b by assumption, it follows that Ÿtfa*HtL faHtL ‚ t = 0. QED #3 Commuting Hermitian linear operaors have simultaneous eigenfunctions. Consider the two distinct operators A,B, such that [A,B] = AB - BA =0. Now assume that one of the operators, say A has an engenfunction fa with non-degenerate eigenvalue a. Note that we need assume nothing about the degeneracy of eigenvalues of B. To prove: fa is automatically an eigenfunction of operator B. Proof: as [A,B] is the "zero operator" it give zero when applied to "any function," specifically: [A,B] fa = 0. Thus, AB fa - BA fa = 0. Using the fact that fa is an eigenfunction of A, this may be re-written as: A (B fa N = a (B fa N , this is equation (*). The non-degeneracy of "a" now implies that these is only a single linearly independent function which satisfies Eqn. (*), and that is fa itself, or a constant multiple of itself. Note that this is the 2nd time in the proof we have used linearity. Thus (B fa N µ fa , where "µ " denotes proportionality. As these funcions (B fa M and fa are proportional there is some constant, b, such that: Bfa = bfa. Thus fa is an eigenfunction of operator B. QED. REMARK: note that the eigenvalues a and b are completely unrelated! They usually have different units, and so can't even be compared. For example one might be an "energy," the other an "angular momentum." B) The Three Very Imporant Theorems, the case of degeneracy. #1 Eigenvalues of Hermitian operators are real. This result has nothing to do with degeneracy, so the proof is the same! PREAMBLE to VIT #2 and #3 in the case of degeneracy. Before discussing VIT #2 and #3, which are actually "changed" by the presence of degeneracy, a few useful remarks about "degenerate" eigenvalues and "degenerate" eigenfuctions may be of use. Suppose operator A has two, and only two, distinct (i. e. linearly independent, which simply means one is not simply "a constant" times the other) eigenfuctions, say fa 1 and fa2 which have the same eigenvalue "a". We then say that "a" is doubly degenerate. If there and N, and only N, linearly independent eigenfunctions which share the same eigenvalue then the degeneracy is N-fold. We can also say that the eigenfuncions belong to an N-fold degenerate set of eigenfuctions. For example the degeneracy of "p-orbitals" is 3-fold, and "d-orbitals" is 5-fold....so these are not ideas irrelavent to CHEMISTRY. If they were we'd happily skip them as mathematical oddities and simply move on. Note also that a set of functions which share an eigenvalue of operator A, most often are non-degenerate with respect to the action of other operators, say B, or even C! An important fact, which we then prove: any linear combination of degenerate functions of operator A is also an eigenfunction of operator A. Proof: Consider A( ⁄i=1N cifai ) = ⁄i=1N ciAfai = ⁄i=1N ciafai = a(⁄i=1N cifai ). QED. Note that linearity of A was used "several times" in this simple demonsteation. Note also that if any one of the funcions fi in the sum didn't have eigenvalue "a", then the a would not factor outsidet he sum, and the result would be false. Said specifically: "linear combinations of non-degenerate eigenfuctions" of an operator are NOT eigenfunctions of that opertor. As the above result has nothing to do with "what linear combination" has been taken, it applies equally to all linear combinations. Can we then generate an infintie number of eigenfuctions of operator A? Of course not: For a double degenerate eigenvalue only two linearly independent eigenfuctions exist, and taking linear combinations does not expand the "space" of degenerate functions beyond two, it just gives different representaions of the same space. An analog in 2D Cartesian space is that any two linearly independent vectors can serve to represent any other vector: but this may be done in an infinite number of way, simply by "rotating" the x,y axies on the page....but nothing new is actually produced by doing that. VIT #2 in the case of degeneracy: "We can always choose to take all of the eigenfunctions of an Hermetian operator to be orthonormal, but we might have to do a bit of work to find the linear combinations of degenerate eigenfunctions that makes this work." Discussion: Where do things not happen "automatically?" as in the case of a non- degenerate eigenvalue of A? Supose "a" is a multiply degenerate eigenvalue, and we examine fa1 and fa2 (note that these superscrips are NOT powers!). Following the Proof of VIT #2 in part A) we find that the key result: (a-a) Ÿt I fa1 HtLM * fa 2HtL ‚ t = 0 does not tell us whether Ÿt I fa1 HtLM * fa 2HtL ‚ t is zero or not, it might be that fa1 and fa2 are orthogonal, but we haven't proven it. It might also be the case that they are NOT orthogonal. BUT, and we state this without detailed proof: if fa1 and fa2 are NOT orthogonal it is always possible to find two linearly independent and orthonormal linear combinations of fa1 and fa2 which ARE orthogonal. We will see examples of both cases as Chem 455A develops. VIT#3 in the case of degeneracy: "We can always choose appropraite linear combinations of degenerate eigenfunctions to obtain simultaneous eigenfunctions of two commuting observables, but we might have to do a bit of work to find those linear combinations." Looking back to part A) the key step was: (B fa L µ fa , which implied that fa was an eigenfunction of B. In the case of degeneracy, we have to replace this by (B fa L µ "ANY linear combination of the degenerate eigenfucntions of A with eigenvalue a". Thus (B fa L is NOT simply a multiple of fa . What do we do? We need to then FIND appropriate linear combinations of the degenerate eigenfucntions of A which ARE also eigenfunctions of B. This can, and we don't give the proof, always be done. We will see examples of this, also, as Chem. 455A proceeds. C) The Measurement Postulate stated in three different ways Statement #1: Given an observable "A" the only results of measurments are the eigenvalues of A. Commentary: this gives us a set (often infinite) of eigenvalues {ai <, which will include anything we measure regarding propert A. However, this doen't tell us how what we might see in a given experimental situation depends on Y(t), the wavefunction. What we can know about a system is contained the wavefunction (Postulate #1) so there should be a relationship between Y and "which of the possible eigenvalues {ai} we might actually expect to find," if the system is in state Y. So a somewhat more useful statement is: Statement #2: The average value of a large number of independent measurements of the property A, when the system is always prepared with wave function Y, is <A> ≡ Ÿt Y* Aop Y ‚ t, where Y is the normalized wavefunction. The uncertainty DA is then defined as < A2 > - H < A >L2+ , and gives a measure of the "spread" in expected results. Statement #3: However, we can go further, and actaully predict which eigenvalues will actually be observed, and with what probability. This is a lot more infomation that just the average + standard deviation (aka quantum uncertainty.) This is done by analogy with finding the "components of a vector in 3D Euclidean Space by projecting the a vector of interest onto the unit vectors chosen to be the "basis." Suppose we have an ortho-normal set of unit vectors i,j,k: The components of a vector V are given by (v1, v2, v3), and the vector itself may be writen as V = v1 i” + v2 j+ v3k where v1= i”.V ,v2= j .V , and v3 = k .V . Note that we have found the component by "projecting the unit vector onto the vector V". In QM the analog is: The components of a wavefunction Y(t), along the directions defined by the orthonormal eigenfunctions, fi HtL, of an obserable are: ci = Ÿtfi*HtL Y HtL ‚ t, and then Y(t) = ⁄i=1N ci fi HtL. If Y(t) is normalized, and the fi HtL are eigenfunctions of observable "A", with eigenvalue ai , then the probability of observing any one of the eigenvalues, say the j th, is given by Pj= I … cj …M2. If an expansion coefficient, say c14 ,is zero, for Y, we will never observe eigenvalue #14, as there is no "component of Y" in that "direction" in the function space. 6 Chem455A_Lecture12,13.nb Notes on VIT #1,2,3 for the non- degenerate and degenerate cases; and a summary of Quantum Measurment tools. AL The Three Very Imporant Theorems, the case of non- degeneracy. #1 Hermitian operators have real eigenvalues. To prove: A is Hermetian, and has an eigenfucntion fa with eigenvalue "a", then "a" is a real number. Proof: By assumption: Ÿtfa*HtL A faHtL ‚ t = Ÿt HA fa HtLL* faHtL ‚ t, and AfaHtL = afaHtL. The integrations co-ordinates and range of integration, denoted by t, are appropriate to the problem at hand. Then Ÿtfa*HtL a faHtL ‚ t = Ÿt Ha faHtLL* faHtL ‚ t, or, a Ÿtfa*HtL faHtL ‚ t = a* Ÿtfa*HtL faHtL ‚ t, from which a = a*, from which VIT#1 follows. QED #2 Hermitian linear operators have orthogonal eigenfunctions. Let us assume that Hermitian operator A has an eigenfunction fa with non- degenate eigenvalue "a". We need make no assumptions about the possible degeneracy of any of the other eigenfunctions of A. To prove: Ÿtfa*HtL fbHtL ‚ t = Ÿtfb* HtL faHtL ‚ t = 0, where the pair "b, and fbHtL" are any other eigenvalue, eigenfunction of A. By assumption of the non-degeneracy of "a" we have a ≠ b. Proof: As A is Hermitian: Ÿtfb* HtL A faHtL ‚ t = Ÿt HA fb HtLL* faHtL ‚ t, and thus a Ÿtfa*HtL faHtL ‚ t = b Ÿtfa*HtL faHtL ‚ t, as both eigenvalues are real from VIT #1. Subtracting the right side from the left gives: (a - b ) Ÿtfa*HtL faHtL ‚ t = 0. As a ≠b by assumption, it follows that Ÿtfa*HtL faHtL ‚ t = 0. QED #3 Commuting Hermitian linear operaors have simultaneous eigenfunctions. Consider the two distinct operators A,B, such that [A,B] = AB - BA =0. Now assume that one of the operators, say A has an engenfunction fa with non-degenerate eigenvalue a. Note that we need assume nothing about the degeneracy of eigenvalues of B. To prove: fa is automatically an eigenfunction of operator B. Proof: as [A,B] is the "zero operator" it give zero when applied to "any function," specifically: [A,B] fa = 0. Thus, AB fa - BA fa = 0. Using the fact that fa is an eigenfunction of A, this may be re-written as: A (B fa N = a (B fa N , this is equation (*). The non-degeneracy of "a" now implies that these is only a single linearly independent function which satisfies Eqn. (*), and that is fa itself, or a constant multiple of itself. Note that this is the 2nd time in the proof we have used linearity. Thus (B fa N µ fa , where "µ " denotes proportionality. As these funcions (B fa M and fa are proportional there is some constant, b, such that: Bfa = bfa. Thus fa is an eigenfunction of operator B. QED. REMARK: note that the eigenvalues a and b are completely unrelated! They usually have different units, and so can't even be compared. For example one might be an "energy," the other an "angular momentum." B) The Three Very Imporant Theorems, the case of degeneracy. #1 Eigenvalues of Hermitian operators are real. This result has nothing to do with degeneracy, so the proof is the same! PREAMBLE to VIT #2 and #3 in the case of degeneracy. Before discussing VIT #2 and #3, which are actually "changed" by the presence of degeneracy, a few useful remarks about "degenerate" eigenvalues and "degenerate" eigenfuctions may be of use. Suppose operator A has two, and only two, distinct (i. e. linearly independent, which simply means one is not simply "a constant" times the other) eigenfuctions, say fa 1 and fa2 which have the same eigenvalue "a". We then say that "a" is doubly degenerate. If there and N, and only N, linearly independent eigenfunctions which share the same eigenvalue then the degeneracy is N-fold. We can also say that the eigenfuncions belong to an N-fold degenerate set of eigenfuctions. For example the degeneracy of "p-orbitals" is 3-fold, and "d-orbitals" is 5-fold....so these are not ideas irrelavent to CHEMISTRY. If they were we'd happily skip them as mathematical oddities and simply move on. Note also that a set of functions which share an eigenvalue of operator A, most often are non-degenerate with respect to the action of other operators, say B, or even C! An important fact, which we then prove: any linear combination of degenerate functions of operator A is also an eigenfunction of operator A. Proof: Consider A( ⁄i=1N cifai ) = ⁄i=1N ciAfai = ⁄i=1N ciafai = a(⁄i=1N cifai ). QED. Note that linearity of A was used "several times" in this simple demonsteation. Note also that if any one of the funcions fi in the sum didn't have eigenvalue "a", then the a would not factor outsidet he sum, and the result would be false. Said specifically: "linear combinations of non-degenerate eigenfuctions" of an operator are NOT eigenfunctions of that opertor. As the above result has nothing to do with "what linear combination" has been taken, it applies equally to all linear combinations. Can we then generate an infintie number of eigenfuctions of operator A? Of course not: For a double degenerate eigenvalue only two linearly independent eigenfuctions exist, and taking linear combinations does not expand the "space" of degenerate functions beyond two, it just gives different representaions of the same space. An analog in 2D Cartesian space is that any two linearly independent vectors can serve to represent any other vector: but this may be done in an infinite number of way, simply by "rotating" the x,y axies on the page....but nothing new is actually produced by doing that. VIT #2 in the case of degeneracy: "We can always choose to take all of the eigenfunctions of an Hermetian operator to be orthonormal, but we might have to do a bit of work to find the linear combinations of degenerate eigenfunctions that makes this work." Discussion: Where do things not happen "automatically?" as in the case of a non- degenerate eigenvalue of A? Supose "a" is a multiply degenerate eigenvalue, and we examine fa1 and fa2 (note that these superscrips are NOT powers!). Following the Proof of VIT #2 in part A) we find that the key result: (a-a) Ÿt I fa1 HtLM * fa 2HtL ‚ t = 0 does not tell us whether Ÿt I fa1 HtLM * fa 2HtL ‚ t is zero or not, it might be that fa1 and fa2 are orthogonal, but we haven't proven it. It might also be the case that they are NOT orthogonal. BUT, and we state this without detailed proof: if fa1 and fa2 are NOT orthogonal it is always possible to find two linearly independent and orthonormal linear combinations of fa1 and fa2 which ARE orthogonal. We will see examples of both cases as Chem 455A develops. VIT#3 in the case of degeneracy: "We can always choose appropraite linear combinations of degenerate eigenfunctions to obtain simultaneous eigenfunctions of two commuting observables, but we might have to do a bit of work to find those linear combinations." Looking back to part A) the key step was: (B fa L µ fa , which implied that fa was an eigenfunction of B. In the case of degeneracy, we have to replace this by (B fa L µ "ANY linear combination of the degenerate eigenfucntions of A with eigenvalue a". Thus (B fa L is NOT simply a multiple of fa . What do we do? We need to then FIND appropriate linear combinations of the degenerate eigenfucntions of A which ARE also eigenfunctions of B. This can, and we don't give the proof, always be done. We will see examples of this, also, as Chem. 455A proceeds. C) The Measurement Postulate stated in three different ways Statement #1: Given an observable "A" the only results of measurments are the eigenvalues of A. Commentary: this gives us a set (often infinite) of eigenvalues {ai <, which will include anything we measure regarding propert A. However, this doen't tell us how what we might see in a given experimental situation depends on Y(t), the wavefunction. What we can know about a system is contained the wavefunction (Postulate #1) so there should be a relationship between Y and "which of the possible eigenvalues {ai} we might actually expect to find," if the system is in state Y. So a somewhat more useful statement is: Statement #2: The average value of a large number of independent measurements of the property A, when the system is always prepared with wave function Y, is <A> ≡ Ÿt Y* Aop Y ‚ t, where Y is the normalized wavefunction. The uncertainty DA is then defined as < A2 > - H < A >L2+ , and gives a measure of the "spread" in expected results. Statement #3: However, we can go further, and actaully predict which eigenvalues will actually be observed, and with what probability. This is a lot more infomation that just the average + standard deviation (aka quantum uncertainty.) This is done by analogy with finding the "components of a vector in 3D Euclidean Space by projecting the a vector of interest onto the unit vectors chosen to be the "basis." Suppose we have an ortho-normal set of unit vectors i,j,k: The components of a vector V are given by (v1, v2, v3), and the vector itself may be writen as V = v1 i” + v2 j+ v3k where v1= i”.V ,v2= j .V , and v3 = k .V . Note that we have found the component by "projecting the unit vector onto the vector V". In QM the analog is: The components of a wavefunction Y(t), along the directions defined by the orthonormal eigenfunctions, fi HtL, of an obserable are: ci = Ÿtfi*HtL Y HtL ‚ t, and then Y(t) = ⁄i=1N ci fi HtL. If Y(t) is normalized, and the fi HtL are eigenfunctions of observable "A", with eigenvalue ai , then the probability of observing any one of the eigenvalues, say the j th, is given by Pj= I … cj …M2. If an expansion coefficient, say c14 ,is zero, for Y, we will never observe eigenvalue #14, as there is no "component of Y" in that "direction" in the function space. Chem455A_Lecture12,13.nb 7