Quantum Two 1 2 Time Independent Approximation Methods

Quantum Two 1 2 Time Independent Approximation Methods

Quantum Two 1 2 Time Independent Approximation Methods 3 4

A Stronger Variational Theorem 5 The strong form of the variational theorem can be stated as follows: The mean value of is stationary at each of its eigenstates| So what does this mean? Maybe this will help: this same theorem is often also expressed in the language of the calculus of variations as follows: If | is a non-zero vector then the linear variation

of about the state | vanishes if and only if | is an eigenstate of . Hmmm. So what does that mean? 6 The strong form of the variational theorem can be stated as follows: The mean value of is stationary at each of its eigenstates| So what does this mean? Maybe this will help: this same theorem is often also expressed in the language of the calculus of variations as follows: If | is a non-zero vector then the linear variation

of about the state | vanishes if and only if | is an eigenstate of . Hmmm. So what does that mean? 7 The strong form of the variational theorem can be stated as follows: The mean value of is stationary at each of its eigenstates| So what does this mean? Maybe this will help: this same theorem is often also expressed in the language of the calculus of variations as follows: If | is a non-zero vector then the linear variation

of about the state | vanishes if and only if | is an eigenstate of . Hmmm. So what does that mean? 8 The strong form of the variational theorem can be stated as follows: The mean value of is stationary at each of its eigenstates| So what does this mean? Maybe this will help: this same theorem is often also expressed in the language of the calculus of variations as follows: If | is a non-zero vector then the linear variation

of about the state | vanishes if and only if | is an eigenstate of . Hmmm. So what does that mean? 9 The strong form of the variational theorem can be stated as follows: The mean value of is stationary at each of its eigenstates| So what does this mean? Maybe this will help: this same theorem is often also expressed in the language of the calculus of variations as follows: If | is a non-zero vector then the linear variation

of about the state | vanishes if and only if | is an eigenstate of . Hmmm. So what does that mean? 10 The strong form of the variational theorem can be stated as follows: The mean value of is stationary at each of its eigenstates| So what does this mean? Maybe this will help: this same theorem is often also expressed in the language of the calculus of variations as follows: If | is a non-zero vector then the linear variation

of about the state | vanishes if and only if | is an eigenstate of . Hmmm. So what does that mean? 11 Recall that a function of a single variable is locally stationary at each point where the derivative vanishes, usually a maximum, a minimum, or a point of inflection with zero slope. Thus a Taylor series expansion of about one of these points has the property that the linear variation about such a point vanishes.

In other words, right near the function is flat, it is not increasing and not decreasing. Thats what stationary means. 12 Recall that a function of a single variable is locally stationary at each point where the derivative vanishes, usually a maximum, a minimum, or a point of inflection with zero slope. Thus a Taylor series expansion of about one of these points has the property that the linear variation about such a point vanishes. In other words, right near the function is flat, it is not increasing and not

decreasing. Thats what stationary means. 13 Recall that a function of a single variable is locally stationary at each point where the derivative vanishes, usually a maximum, a minimum, or a point of inflection with zero slope. Thus a Taylor series expansion of about one of these points has the property that the linear variation about such a point vanishes. In other words, right near the function is flat, it is not increasing and not decreasing. Thats what stationary means.

14 Recall that a function of a single variable is locally stationary at each point where the derivative vanishes, usually a maximum, a minimum, or a point of inflection with zero slope. Thus a Taylor series expansion of about one of these points has the property that the linear variation about such a point vanishes. In other words, right near the function is flat, it is not increasing and not decreasing. Thats what stationary means. 15

So the strong variational theorem basically says that if you evaluate the mean value along any path | in Hilbert space that passes through one of the eigenstates | of , then the rate of change of will be zero at the moment it actually passes through that eigenstate. Thus, will not be increasing and it will not be decreasing at that point. Its linear variation about that point in Hilbert space will vanish. 16 So the strong variational theorem basically says that if you evaluate the mean

value along any path | in Hilbert space that passes through one of the eigenstates | of , then the rate of change of will be zero at the moment it actually passes through that eigenstate. Thus, will not be increasing and it will not be decreasing at that point in Hilbert space. Its linear variation about that point in Hilbert space will vanish. 17 So the strong variational theorem basically says that if you evaluate the mean value along any path | in Hilbert space that passes through one of the

eigenstates | of , then the rate of change of will be zero at the moment it actually passes through that eigenstate. Thus, will not be increasing and it will not be decreasing at that point in Hilbert space. Its linear variation about the point | vanishes if and only if | is an eigenstate of . 18 To understand and prove this, let | be an arbitrary normalizable state of the system, and consider a family of kets |() = | + |

that differ from the original state | = |(0) by a small amount, | |() | | in which | is a fixed but arbitrary normalizable state and is a real parameter whose variation describes a trajectory through state space passing through the state | when .

19 To understand and prove this, let | be an arbitrary normalizable state of the system, and consider a family of kets |() = | + | that differ from the original state | = |(0) by a small amount, | |() | | in which | is a fixed but arbitrary normalizable state

and is a real parameter whose variation describes a trajectory through state space passing through the state | when . 20 To understand and prove this, let | be an arbitrary normalizable state of the system, and consider a family of kets

|() = | + | that differ from the original state | = |(0) by a small amount, | |() | | in which | is a fixed but arbitrary normalizable state and is a real parameter whose variation describes a trajectory through state space passing through the state |

when . 21 To understand and prove this, let | be an arbitrary normalizable state of the system, and consider a family of kets |() = | + | that differ from the original state | = |(0) by a small amount, | |() | | in which

| is a fixed but arbitrary normalizable state and is a real parameter whose variation describes a trajectory through state space passing through the state | when . 22 To understand and prove this, let | be an arbitrary normalizable state of the

system, and consider a family of kets |() = | + | Denote the deviation of |() from the original state | = |(0) by | |() | | in which | is a fixed but arbitrary normalizable state and is a real parameter whose variation describes a trajectory

through state space passing through the state | when . 23 Denote the mean value of with respect to the varied state |(), by in which we have included the normalization so that we do not have to worry about constraining the variation to normalized states, and in terms of which we introduce the value

With these definitions, we can then prove the following statement: The state | is an eigenstate of if and only if for all arbitrary local variations | =| about the state | . 24 Denote the mean value of with respect to the varied state |(), by in which we have included the normalization so that we do not have to worry about constraining the variation to normalized states, and in terms of which we

introduce the value With these definitions, we can then prove the following statement: The state | is an eigenstate of if and only if for all arbitrary local variations | =| about the state | . 25 Denote the mean value of with respect to the varied state |(), by

in which we have included the normalization so that we do not have to worry about constraining the variation to normalized states, and in terms of which we introduce the value With these definitions, we can then prove the following statement: The state | is an eigenstate of if and only if for all arbitrary local variations | =| about the state | . 26

Denote the mean value of with respect to the varied state |(), by in which we have included the normalization so that we do not have to worry about constraining the variation to normalized states, and in terms of which we introduce the value With these definitions, we can then prove the following statement: The state | is an eigenstate of if and only if for all arbitrary local variations | =| about the state | .

27 Denote the mean value of with respect to the varied state |(), by in which we have included the normalization so that we do not have to worry about constraining the variation to normalized states, and in terms of which we introduce the value With these definitions, we can then prove the following statement: The state | is an eigenstate of if and only if

for all arbitrary local variations | = | about the state | . 28 To prove the statement, we first compute the derivative of using the product rule. Introducing the obvious notation we have (since is independent of ) Multiply through by | | and use the two identities above to obtain

29 To prove the statement, we first compute the derivative of using the product rule. Introducing the obvious notation we have (since is independent of ) Multiply through by | | and use the two identities above to obtain 30

To prove the statement, we first compute the derivative of using the product rule. Introducing the obvious notation we have (since is independent of ) Multiply through by | | and use the two identities above to obtain 31 To prove the statement, we first compute the derivative of using the product rule. Introducing the obvious notation

we have (since is independent of ) Multiply through by | | and use the two identities above to obtain 32 To prove the statement, we first compute the derivative of using the product rule. Introducing the obvious notation we have (since is independent of )

Now multiply through by | | and use the two identities above to obtain 33 To prove the statement, we first compute the derivative of using the product rule. Introducing the obvious notation we have (since is independent of ) Now multiply through by | | and use the two identities above to obtain

34 Using our definition of , the last term in this expression becomes: thus giving the key relation Now note that if | is an actual eigenstate of its eigenvalue must be equal to so that the right hand side of the key relation above vanishes for arbitrary |.

35 Using our definition of , the last term in this expression becomes: thus giving the key relation Now note that if | is an actual eigenstate of its eigenvalue must be equal to so that the right hand side of the key relation above vanishes for arbitrary |. 36

Using our definition of , the last term in this expression becomes: thus giving the key relation Now note that if | is an actual eigenstate of its eigenvalue must be equal to so that the right hand side of the key relation above vanishes for arbitrary |. 37

Using our definition of , the last term in this expression becomes: thus giving the key relation Now note that if | is an actual eigenstate of its eigenvalue must be equal to so that the right hand side of the key relation above vanishes for arbitrary |. 38 Using our definition of , the last term in this expression becomes:

thus giving the key relation Now note that if | is an actual eigenstate of its eigenvalue must be equal to so that the right hand side of the key relation above vanishes for arbitrary |. 39 Using our definition of , the last term in this expression becomes:

thus giving the key relation Now note that if | is an actual eigenstate of its eigenvalue must be equal to so that the right hand side of the key relation above vanishes for arbitrary |. 40 So if | is an eigenstate of , then Since, in this case, the eigenstate | is nonzero, the only way the left hand side

can vanish is if for arbitrary variations | | about the eigenstate | of . This proves the "if" part of the statement. If | is an eigenstate of , the mean value of is stationary at that point in Hilbert space. We now need to show that this can only happen if | is an eigenstate of . 41 So if | is an eigenstate of , then

Since, in this case, the eigenstate | is nonzero, the only way the left hand side can vanish is if for arbitrary variations | | about the eigenstate | . This proves the "if" part of the statement. If | is an eigenstate of , the mean value of is stationary at that point in Hilbert space. We now need to show that this can only happen if | is an eigenstate of . 42 So if | is an eigenstate of , then

Since, in this case, the eigenstate | is nonzero, the only way the left hand side can vanish is if for arbitrary variations | | about the eigenstate | . This proves the "if" part of the statement. If | is an eigenstate of , the mean value of is stationary at that point in Hilbert space. We now need to show that this can only happen if | is an eigenstate of . 43

So if | is an eigenstate of , then Since, in this case, the eigenstate | is nonzero, the only way the left hand side can vanish is if for arbitrary variations | | about the eigenstate | . This proves the "if" part of the statement. If | is an eigenstate of , the mean value of is stationary at that point in Hilbert space. We now need to show that this can only happen if | is an eigenstate of . 44

To prove the "only if" part of the theorem, assume that the derivative of with respect to does indeed vanish for arbitrary variations about the (now assumed arbitrary) state | . In this case our key expression reduces to the statement that for arbitrary . But this must then be true for any particular we might choose. Making the inspired choice so

we find that 45 To prove the "only if" part of the theorem, assume that the derivative of with respect to does indeed vanish for arbitrary variations about the (now assumed arbitrary) state | . In this case our key expression reduces to the statement that for arbitrary .

But this must then be true for any particular we might choose. Making the inspired choice so we find that 46 To prove the "only if" part of the theorem, assume that the derivative of with respect to does indeed vanish for arbitrary variations about the (now assumed

arbitrary) state | . In this case our key expression reduces to the statement that for arbitrary . But this must then be true for any particular we might choose. Making the inspired choice so we find that

47 To prove the "only if" part of the theorem, assume that the derivative of with respect to does indeed vanish for arbitrary variations about the (now assumed arbitrary) state | . In this case our key expression reduces to the statement that for arbitrary . But this must then be true for any particular we might choose. Making the inspired choice

so we find that 48 To prove the "only if" part of the theorem, assume that the derivative of with respect to does indeed vanish for arbitrary variations about the (now assumed arbitrary) state | . In this case our key expression reduces to the statement that

for arbitrary . But this must then be true for any particular we might choose. Making the inspired choice so we find that 49 To prove the "only if" part of the theorem, assume that the derivative of with

respect to does indeed vanish for arbitrary variations about the (now assumed arbitrary) state | . In this case our key expression reduces to the statement that for arbitrary . But this must then be true for any particular we might choose. Making the inspired choice so we find that

50 This last relation is equivalent to the statement that which means that the vector must itself vanish, so | is then necessarily an eigenstate of with eigenvalue . Thus, | is an eigenstate of with eigenvalue whenever the derivative vanishes for arbitrary variations | |, completing the proof of the strong variational

theorem. 51 This last relation is equivalent to the statement that which means that the vector must itself vanish, so | is then necessarily an eigenstate of with eigenvalue . Thus, | is an eigenstate of with eigenvalue whenever the derivative vanishes for arbitrary variations | |, completing the proof of the strong variational

theorem. 52 This last relation is equivalent to the statement that which means that the vector must itself vanish, so | is then necessarily an eigenstate of with eigenvalue . Thus, | is an eigenstate of with eigenvalue whenever the derivative vanishes for arbitrary variations | |, completing the proof of the strong variational

theorem. 53 This last relation is equivalent to the statement that which means that the vector must itself vanish, so | is then necessarily an eigenstate of with eigenvalue . Thus, | is an eigenstate of with eigenvalue whenever the derivative vanishes for arbitrary variations | |, completing the proof of this stronger variational

theorem. This variational theorem forms the basis for the variational method, outlined in the next segment. 54 55

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