Talk:Wave function/Archive 10

Dirac's Principles of quantum mechanics (primarily)

The immediately foregoing section ends with a disorderly posting. It would be out of order for me to reply to it.

On page 19 of the first edition (1930) of his text, Dirac writes:

We now introduce another set of symbols φ₁, φ₂, ... also denoting states. Any state denoted by a ψ symbol ψ_r can be equally well denoted by a φ symbol φ_r having the same suffix.

In the second edition (1935) he writes of his two kinds of vector, the ψ's and the φ's:

... Instead of picturing the ψ's and φ's as vectors in two different vector spaces, we may picture them as two different kinds of vector associated with the same space. The relation between these two kinds of vector is then just the one well known in differential geometry as the relation between covariant and contravariant vectors.

In his 1939 introduction of his bra–ket notation for those same objects, he writes:

... any expression containing an unclosed bracket symbol   ${\displaystyle \langle }$   or   ${\displaystyle \rangle }$   is a vector in Hilbert space, of the nature of a φ or ψ respectively. As names for the new symbols   ${\displaystyle \langle }$   and   ${\displaystyle \rangle }$   to be used in speech, I suggest the words bra and ket respectively.

On page 20 of the first edition (1930), he writes:

... The theory will throughout be symmetrical between the φ's and ψ's. The sum of a φ and a ψ has no meaning and will never appear in the analysis.

       The introduction of a second set of symbols to denote the states may appear to be superfluous, but actually it is necessary when one allows complex coefficients c_r in order to preserve the symmetry between the two roots of −1.

On page 22 of the second edition (1935), he writes:

Each vector ψ_a in the space of ψ's determines uniquely a vector |φ_a⟩ in the space of φ's and vice versa. Thus the space of φ's provides a representation of the states of our dynamical system just as well as the space of ψ's, each state being associated with one direction in the space of φ's. There is, in fact, perfect symmetry between the φ's and ψ's, which symmetry will survive all through the theory.

On page 43 of the fourth edition (1958), he writes:

... Also it is easily seen that the whole theory of functions of an observable is symmetrical between bras and kets and that we could equally well work from the equation

${\displaystyle \langle \xi ^{\prime }|f(\xi )=f(\xi ^{\prime })\langle \xi ^{\prime }|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(38)}$

instead of from (34).

On page 40 of the 4th edition, he wrote:

        The space of bra or ket vectors when the vectors are restricted to be of finite length and to have finite scalar products is called by mathematicians a Hilbert space. The bra and ket vectors that we now use form a more general space than a Hilbert space.

He wasn't over-chatty about Hilbert spaces.

For the peace of mind of some, I may add the following piece of original research. If Dirac had been in the habit of denoting his vector spaces by mathcal symbols (which he wasn't), I guess he might perhaps have written

${\displaystyle \langle \Psi |\in {\mathcal {H}}_{\langle }}$        and        ${\displaystyle \langle \Psi |\notin {\mathcal {H}}_{\rangle }}$        and        ${\displaystyle |\Psi \rangle \in {\mathcal {H}}_{\rangle }}$        and        ${\displaystyle |\Psi \rangle \notin {\mathcal {H}}_{\langle }.}$

He would perhaps have added that ${\displaystyle \langle \Psi |}$ and ${\displaystyle |\Psi \rangle }$ are mutually dual. He said they were a conjugate imaginary pair. Also, I guess he might have said that ${\displaystyle {\mathcal {H}}_{\langle }}$ and ${\displaystyle {\mathcal {H}}_{\rangle }}$ are mutually dual.

There is plenty of literature commentary on this, but I think that should do for now.Chjoaygame (talk) 05:18, 10 March 2016 (UTC)

It really does.not.matter.

"If Dirac had been in the habit of denoting his vector spaces by mathcal symbols (which he wasn't),".

The fact that you write

"I guess he might perhaps have written"

in nonstandard and obscure notation the Hilbert space and its dual, followed by

"He would perhaps have added that ${\displaystyle \langle \Psi |}$ and ${\displaystyle |\Psi \rangle }$ are mutually dual. He said they were a conjugate imaginary pair. Also, I guess he might have said that ${\displaystyle {\mathcal {H}}_{\langle }}$ and ${\displaystyle {\mathcal {H}}_{\rangle }}$ are mutually dual.

suggests you do not actually know what you are defining, and resort to what others have said. ${\displaystyle {\mathcal {H}}_{\rangle }}$ is the Hilbert space of all allowable states |Ψ⟩ for the system in question, and ${\displaystyle {\mathcal {H}}_{\langle }}$ (in other symbols ${\displaystyle {\mathcal {H}}_{\rangle }^{*}}$) the dual space with elements ⟨Ψ| right? Then you could have said so. On top of your reply last section, that

"According to Dirac they refer to different aspects of the same quantum state, as it is prepared, and as it is observed, two different physical things"

suggests you also misunderstand what the bra-ket notation is for, because p.21 in the 4th edn of Dirac's Principles of QM:

"On account of the correspondence between bra vectors and ket vectors, any state of our dynamical system at a particular time may be specified by the direction of a bra vector just as well as a ket vector. In fact the whole theory will be symmetrical in its essentials between bras and kets."

A ket, and its corresponding bra, describe the same state as you correctly say, but one does not correspond to the quantum state "as prepared", and the other to the quantum state "as observed". Where does Dirac say this? I can't find it in his book, please point to other papers or books. Given a bra-ket equation, you can take the Hermitian conjugate of it and the meaning of the equation is the same. Mathematically: the bras become kets and vice versa, and the operators are replaced by their Hermitian conjugates, for example

${\displaystyle |\psi \rangle \rightarrow \langle \psi |\,,\quad \langle \phi |{\hat {\Omega }}|\psi \rangle \rightarrow \langle \psi |{\hat {\Omega }}^{\dagger }|\phi \rangle }$

and each expression has the same meaning (retains the same information about the physical system) before and after Hermitian conjugation. Hermitian conjugation is a mathematical operation, not physical.

Suppose that bras really did correspond to "observed" and the kets to "prepared". Start from the SE of any quantum state

${\displaystyle i\hbar {\frac {d}{dt}}|\psi \rangle ={\hat {H}}|\psi \rangle }$

Now take the Hermitian conjugate.

${\displaystyle -i\hbar \langle \psi |{\frac {d}{dt}}=\langle \psi |{\hat {H}}^{\dagger }}$

where in the conjugate equation, the derivative acts to the left (because if this bra equation premultiplies a new ket |A⟩, then the derivative must act on ⟨ψ| not |A⟩). What happens? Is the original SE the equation for the prepared state, and the second the observed state?

At least... you (seem to) know correctly that bras and kets are mutually dual. M∧Ŝc²ħεИ_τlk 11:06, 13 March 2016 (UTC)

My non-standard and obscure notation is, as I observed, the dreaded OR. I need to invent it because notation of this kind seems necessary here. The point is that the notation that Editor Yohan7 above tried to make me use does not express my meaning. The non-standard notation does so. The notation ${\displaystyle {\mathcal {H}}}$ and ${\displaystyle {\mathcal {H}}^{*}}$ that Editor Yohan7 tried to make me use is misleading because it wrongly gives priority to ${\displaystyle {\mathcal {H}}}$ for kets as the Hilbert space, with ${\displaystyle {\mathcal {H}}^{*}}$ for bras secondary to it. This hides Dirac's intended symmetry. As you observe, when you copy my many times repeated quotes of Dirac, the theory is symmetrical between bras and kets. My non-standard notation, ${\displaystyle {\mathcal {H}}_{\langle }}$ for bras and ${\displaystyle {\mathcal {H}}_{\rangle }}$ for kets, is adapted to express that symmetry.

"He would perhaps have added that ${\displaystyle \langle \Psi |}$ and ${\displaystyle |\Psi \rangle }$ are mutually dual. He said they were a conjugate imaginary pair. Also, I guess he might have said that ${\displaystyle {\mathcal {H}}_{\langle }}$ and ${\displaystyle {\mathcal {H}}_{\rangle }}$ are mutually dual. Dirac wasn't over-chatty about Hilbert spaces, so I can only guess what he might have said if he had used the mathcal symbols.

suggests you do not actually know what you are defining, and resort to what others have said. Argumentum ad verecundiam, no reply. ${\displaystyle {\mathcal {H}}_{\rangle }}$ is the Hilbert space of all allowable states |Ψ⟩ for the system in question, and ${\displaystyle {\mathcal {H}}_{\langle }}$ (in other symbols ${\displaystyle {\mathcal {H}}_{\rangle }^{*}}$) the dual space with elements ⟨Ψ| right? No, wrong, as I have indicated just above, in that they are mutually dual, and your prejudice in favor of the kets as the states is misleading. Dirac writes in the 1930 edition on page 62: "We can without inconsistency suppose that each fundamental φ and the conjugate imaginary fundamental ψ of an orthogonal representation are to be both pictured by the same real vector." Then you could have said so. Why would I hide my meaning by using the misleading notation that you advocate and that expresses your prejudice? But yes, you have it right that ${\displaystyle {\mathcal {H}}_{\langle }^{*}={\mathcal {H}}_{\rangle }}$. On top of your reply last section, that

"According to Dirac they refer to different aspects of the same quantum state, as it is prepared, and as it is observed, two different physical things"

suggests you also misunderstand what the bra-ket notation is for, because p.21 in the 4th edn of Dirac's Principles of QM:

"On account of the correspondence between bra vectors and ket vectors, any state of our dynamical system at a particular time may be specified by the direction of a bra vector just as well as a ket vector. In fact the whole theory will be symmetrical in its essentials between bras and kets."}} At last you are registering what I have repeatedly quoted from Dirac.

A ket, and its corresponding bra, describe the same state as you correctly say, but one does not correspond to the quantum state "as prepared", and the other to the quantum state "as observed". Where does Dirac say this? I can't find it in his book, please point to other papers or books. My partial mistake. Though he says something close to it in the first edition, Dirac doesn't exactly say this in later editions. It is many years since I learnt from Feynman's text, that I will shortly quote, that kets refer to the starting state or the state as prepared, and bras to the final state or state as observed. In reading Dirac I read it into the text from habitual thinking following Feynman. For example, it is a natural reading of the following sentences, from pages 23–24 of the first edition of Dirac: "Our introduction of products of φ's with ψ's has so far been entirely a mathematical question, with no physical implications. A physical meaning will now be given to the product φ_rψ_s. Consider that maximum observation of the state φ_r for which there is a certainty of a particular result being obtained. We have seen that such a maximum observation always exists. Suppose now this maximum observation to be made on the system in the state ψ_s. There will be a certain probability of the same result being obtained, which we call the probability of agreement of ψ_s with φ_r. It is a number that depends only on the two states ψ_s and φ_r." In Dirac's bra–ket notation, it is an arbitrary convention to choose to denote (a), states as prepared, by kets, and then (b), states as observed, by bras. Feynman says: "In other words, the two brackets 〈 〉 are a sign equivalent to “the amplitude that”; the expression at the right of the vertical line always gives the starting condition, and the one at the left, the final condition." ^[1] Auletta, Fortunato, Parisi say: "In this context, we see that kets may be thought of as input states, whereas bras as output states of a certain physical evolution or process."^[2] Landau and Lifshitz write:

${\displaystyle \langle n|f|m\rangle \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(11.17)}$

This symbol is written so that it may be regarded as "consisting" of the quantity f and the symbols |m⟩ and ⟨n| which respectively stand for the initial and final states as such (independently of the representation of the wave functions of the states).^[3]

By Dirac's symmetry, it would be valid, though it is not customary, to make the alternative choice, with kets as output states, and bras as input states mutatis mutandis. Perhaps I need to say initial state = input state = starting condition = state as prepared, and final state = output state = final condition = state as observed. Dirac does distinguish preparation and observation, as is customary. For example, for observation: "... for any state there must exist one maximum observation which will for a certainty lead to one particular result, and conversely, if we consider any possible result of a maximum observation, there must exist a state of the system for which this result for the observation will be obtained with certainty."^[4] For preparation: "In practice the conditions could be imposed by a suitable preparation of the system, consisting perhaps in passing it through various kinds of sorting apparatus, such as slits and polarimeters, the system being left undisturbed after the preparation."^[5] The important physics here is that, as I tried recently to convey, the preparation-observation set-up is symmetrical. One can interchange the oven and the anti-oven, and get the same results if the the filters and so forth commute. To be a good quantum analyzer, something such as a prism or crystal must have a sort of Helmholtz reciprocity. That means it can be turned 180 degrees and work the same. That's why bras and kets are symmetrical and why compatible degrees of freedom are encoded by commuting operators. They say "simultaneously measured" but they mean 'observed through contiguous analyzers'. "Simultaneously measured" is an abuse of language almost characteristic of quantum mechanics.

• Auletta, G., Fortunato, M., Parisi, G. (2009). Quantum Mechanics, Cambridge University Press, Cambridge UK, ISBN 9781107665897.
• Feynman, R.P., Leighton, R.B., Sands, M. (1963). The Feynman Lectures on Physics, Volume 3, Addison-Wesley, Reading MA, available here.
• Landau, L., Lifshitz, E. (1974/1977). Quantum Mechanics: Non-Relativistic Theory, 3rd edition, translated from Russian into English by J.B. Sykes, J.S. Bell, Pergamon, Oxford UK, ISBN 0-08-020940-8.

Given a bra-ket equation, you can take the Hermitian conjugate of it and the meaning of the equation is the same. Mathematically: the bras become kets and vice versa, and the operators are replaced by their Hermitian conjugates, for example

${\displaystyle |\psi \rangle \rightarrow \langle \psi |\,,\quad \langle \phi |{\hat {\Omega }}|\psi \rangle \rightarrow \langle \psi |{\hat {\Omega }}^{\dagger }|\phi \rangle }$

and each expression has the same meaning (retains the same information about the physical system) before and after Hermitian conjugation. Hermitian conjugation is a mathematical operation, not physical.

Suppose that bras really did correspond to "observed" and the kets to "prepared". Start from the SE of any quantum state

${\displaystyle i\hbar {\frac {d}{dt}}|\psi \rangle ={\hat {H}}|\psi \rangle }$

Now take the Hermitian conjugate.

${\displaystyle -i\hbar \langle \psi |{\frac {d}{dt}}=\langle \psi |{\hat {H}}^{\dagger }}$

where in the conjugate equation, the derivative acts to the left (because if this bra equation premultiplies a new ket |A⟩, then the derivative must act on ⟨ψ| not |A⟩). What happens? Is the original SE the equation for the prepared state, and the second the observed state?}} The Schrödinger equation describes the evolution of the isolated system, not referring to the contact with the outside world through the oven and anti-oven, i.e. producer and destroyer. These two opposite contacts are two different physical processes, both necessary for physical experiments, not described by the Schrödinger equation. The evolution of the isolated system is not real evolution unless the Hamiltonian is explicitly time-dependent. This is because the oven and anti-oven can be interchanged without affecting the results. There are two signs for the square root of minus one. Time can run backwards or forwards. That's why people make muddles on causality in quantum mechanics.

Perhaps it may be useful to add the Dirac's first edition (1930) was considered too abstract for many readers, and he responded by making later edition less abstract, though he commented that readers who like abstraction for its own sake might still prefer the 1930 edition.

At least... you (seem to) know correctly that bras and kets are mutually dual. Argumentum ad verecundiam. No reply.Chjoaygame (talk) 17:57, 13 March 2016 (UTC)

Please clarify/define what "with ${\displaystyle {\mathcal {H}}^{*}}$ for bras secondary to it" followed by "No, wrong, as I have indicated just above, in that they [ ${\displaystyle {\mathcal {H}}_{\rangle }}$ and ${\displaystyle {\mathcal {H}}_{\langle }}$ ] are mutually dual" mean.

It isn't that I have "only just" acknowledged what Dirac has written, because I don't "dislike"/disregard what he has written, never have or will. That symmetry quote refers to the Hermitian conjugate as the correspondence between kets and bras, something you seem to keep avoid discussing.

Please understand this. A quantum state as a bra or ket is not "as measured/observed". It just is what it is - and corresponds to the same quantum state, the same physical entity. The correspondence between the two is mathematical, not physical.

For someone who relies so heavily on Dirac's quotes, don't you realize that the Dirac quote on the symmetry between bras and kets, AND your insistence that bras and kets are physically different, is a contradiction?

This may be useful (certainly more useful than your unintelligible "ovens" and "anti-ovens"): In the expression ⟨n|f|i⟩ as you mention, the observed state happens to be a bra ⟨n|, the initial state is a ket |i⟩. But this is a mathematical expression for an integral/sum. If you have ⟨n| on its own, then this does not automatically make it an observed state.

The causality issue has nothing to do with the present topic. Nevertheless... Time reversal is just negating the time parameter. Complex/Hermitian conjugation only approximately corresponds to time reversal, because time often just happens to be multiplied by i, e.g. the phase factor e^−iEt/ħ for stationary states.

Finally - "Argumentum ad verecundiam" is exactly what you are doing. Deferring to the pioneers. M∧Ŝc²ħεИ_τlk 12:39, 15 March 2016 (UTC)

Please clarify/define what "with ${\displaystyle {\mathcal {H}}^{*}}$ for bras secondary to it" followed by "No, wrong, as I have indicated just above, in that they [ ${\displaystyle {\mathcal {H}}_{\rangle }}$ and ${\displaystyle {\mathcal {H}}_{\langle }}$ ] are mutually dual" mean. I read '${\displaystyle {\mathcal {H}}}$ is the state space and ${\displaystyle {\mathcal {H}}^{*}}$ its dual' as intending that kets have priority that negates Dirac's symmetry. It seems to rule out that bras might be points in the state space with kets merely their duals. "${\displaystyle {\mathcal {H}}_{\rangle }}$ and ${\displaystyle {\mathcal {H}}_{\langle }}$ are mutually dual" means '${\displaystyle {\mathcal {H}}_{\rangle }^{*}={\mathcal {H}}_{\langle }}$ and ${\displaystyle {\mathcal {H}}_{\langle }^{*}={\mathcal {H}}_{\rangle }}$', as I wrote above. The sentence '${\displaystyle {\mathcal {H}}}$ is the state space and ${\displaystyle {\mathcal {H}}^{*}}$ its dual' gives the impression of denying that bras denote states with kets merely their duals. I don't see you rejecting that denial.

It isn't that I have "only just" acknowledged what Dirac has written, because I don't "dislike"/disregard what he has written, never have or will. That symmetry quote refers to the Hermitian conjugate as the correspondence between kets and bras, something you seem to keep avoid discussing. I wrote above "You have it right that ${\displaystyle {\mathcal {H}}_{\langle }^{*}={\mathcal {H}}_{\rangle }}$".

Please understand this. A quantum state as a bra or ket is not "as measured/observed". It just is what it is - and corresponds to the same quantum state, the same physical entity. The correspondence between the two is mathematical, not physical. I am persuaded by Feynman, by L&L, and by Auletta, G., Fortunato, M., & Parisi, G..

For someone who relies so heavily on Dirac's quotes, don't you realize that the Dirac quote on the symmetry between bras and kets, AND your insistence that bras and kets are physically different, is a contradiction? Symmetry isn't necessarily identity.

This may be useful (certainly more useful than your unintelligible "ovens" and "anti-ovens"): In the expression ⟨n|f|i⟩ as you mention, the observed state happens to be a bra ⟨n|, the initial state is a ket |i⟩. But this is a mathematical expression for an integral/sum. If you have ⟨n| on its own, then this does not automatically make it an observed state. No comment.

The causality issue has nothing to do with the present topic. Nevertheless... Time reversal is just negating the time parameter. Complex/Hermitian conjugation only approximately corresponds to time reversal, because time often just happens to be multiplied by i, e.g. the phase factor e^−iEt/ħ for stationary states. Too complicated to pursue here.

Finally - "Argumentum ad verecundiam" is exactly what you are doing. Deferring to the pioneers. By argumentum ad verecundiam I mean that you try to shame me. Wikipedia editing requires careful comparison of sources to find reliable ones. Dirac is a candidate.Chjoaygame (talk) 04:13, 16 March 2016 (UTC)

Other sources

Shankar (2nd edition, 1994), on page 11, discusses two ways of seeing the inner or scalar product. One is as between two vectors of the same space, with no involvement of a dual space. The other is between two vectors from dual spaces. Of the latter, he writes "Inner products are really defined only between bras and kets and hence from elements of two distinct but related vector spaces." Unlike Dirac, Shankar does not emphasize the symmetry between bras and kets. But he does not explicitly deny it. On page 121, Shankar seems to depart from Dirac. He writes "As far as the state vector |ψ⟩ is concerned, there is just one space, the Hilbert space, in which it resides." He does not explicitly say that the corresponding bra ⟨ψ| is not also a state vector. Nevertheless, one could easily get the impression from Shankar that he does not make the point made by Dirac, L&L, and Feynman, that bras and kets represent states as observed and as prepared. I think there is a case for going against Shankar, with Dirac, L&L, and Feynman. They have a physical reason for their way, while Shankar is silent as to his reason, if he has one. I don't think 'modernity' is a useful reason. Looking at other sources: Merzbacher (2nd edition, 1970) is also silent on the point, so far as I have seen; Sakurai & Napolitano (2nd edition, 2011) are also silent on the point, as far as I have seen, though on page they express deference to "the master" Dirac, and profess to "follow" him. Zettili (2nd edition, 2009) is also silent on the point. Even the unusually careful Messiah (1961) is silent, so far as I have seen on flicking through his text. It seems that to follow Dirac closely here, one would need to rely on quality against quantity. I favor Dirac, L&L (3rd edition, 1977, p. 35), and Feynman, with their physical reason, which I find convincing, against the others with their silence. Perhaps a further survey of sources might be useful, but not by me right now.Chjoaygame (talk) 11:50, 10 March 2016 (UTC)

On page 69, Walter Greiner <(2001), Quantum Mechanics: An Introduction, 4th edition, Springer, Berlin, ISBN 3-540-67458-6> writes: "... the element ⟨ψ₁| is called a "bra" and |ψ₂⟩ is called a "ket" ... Both are vectors (state vectors) in a linear vector space."

Dirac, his notation, and his book

In physics we have a number of heroes. Dirac was a hero, and should be treated as such. But in physics, we do not have prophets. It would be ridiculous to go to Einstein to seek answers to present day questions about general relativity. It is likewise ridiculous to assume that every formulation in Dirac's pioneering book should be taken and the final formulation, word by word, of topics quantum mechanics. It is widely acknowledged that Dirac's theory is the correct one. See for example Weinberg ("The Quantum Theory of Fields" vol 1, and also "Lectures on Quantum Mechanics"), and Weinberg's knowledge of QM is a lifetime deeper than Dirac's, so he should know.

There is a reason that books have been written the past 80 years. Dirac's presentation can be improved significantly on. In particular, Dirac's text is a graduate level text and requires considerable mathematical background to interpret correctly. This is very clear, if not from anything else, from this talk page. "Dirac wasn't over-chatty about Hilbert spaces." Right. Then don't try to learn about Hilbert spaces from Dirac. You'll get things wrong. You'll get quantum mechanics (the physical interpretation and its mathematical formulation) wrong. The same goes for Feynman (another of the few true heroes) and his (pretty awful) lecture notes.

In particular, you (Chjoaygame) read things (physics not found elsewhere) into the Dirac notation. They aren't there.

Modern educational methods are better than those of the 1930:s. Modern ways of presenting physics are better than those of the 1930:s. Mathematics has become the irreversibly final tool and language in expressing physics. You may not like it, but you can't turn the flow of time around.

Filling the article, and in particular this talk page, with hundreds of Dirac quotes based on misunderstandings makes Dirac look like babbling fool. I don't like that at all. YohanN7 (talk) 13:30, 15 March 2016 (UTC)

For instance,

... Instead of picturing the ψ's and φ's as vectors in two different vector spaces, we may picture them as two different kinds of vector associated with the same space. The relation between these two kinds of vector is then just the one well known in differential geometry as the relation between covariant and contravariant vectors.

This is interpretable by a mathematician or a mathematically inclined physicist (and actually also by me). Do you really think this is the way to learn about Hilbert spaces and quantum mechanics? That "same space" to which bras and kets are associated is not what you'd think. It is not at all bra's and kets lumped together in the same Hilbert space or anything of the sort. YohanN7 (talk) 13:47, 15 March 2016 (UTC)

I'm not suggesting putting bras and kets in the same Hilbert space. I have explicitly written the contrary above, and Dirac, as I cited him above, explicitly says they cannot be added together. Bras and kets are heads and tails of coins that are the mathematical signifiers of the state space.Chjoaygame (talk) 22:25, 15 March 2016 (UTC)Chjoaygame (talk) 03:21, 16 March 2016 (UTC)Chjoaygame (talk) 04:24, 16 March 2016 (UTC)

The time when the exact nature of your interpretations could be worth discussing is long gone. No doubt you will go on and on and on about this and somehow have it look like that you were perfectly right all the time. This is not the issue. YohanN7 (talk) 12:43, 16 March 2016 (UTC)

Perhaps it may help when I say that for the sake of good will I have just now looked over Chapters 1–4 of Shankar. I have also recently looked at that text. Also I have in the not-too-distant past several times read Weinberg's view on interpretation in his Lectures.Chjoaygame (talk) 06:38, 16 March 2016 (UTC)

Does "for the sake of good will" mean that you step down from the Dirac pedestal temporarily and try to see things from the inferior perspective? It sure sounds that way. But good for you. We might even be able in the far future to agree on a complete sentence if we share some common understanding of the subject. YohanN7 (talk) 12:43, 16 March 2016 (UTC)

Pure states and von Neumann's projection postulate

A state of a system may be defined as an undisturbed motion that is restricted by as many conditions or data as are theoretically possible without mutual interference or contradiction. In practice, the conditions could be imposed by a suitable preparation of the system, consisting perhaps of passing it through various kinds of sorting apparatus, such as slits and polarimeters, the system being undisturbed after preparation.^[1]

When we measure a real dynamical variable ξ, the disturbance involved in the act of measurement causes a jump in the state of the dynamical system. From physical continuity, if we make a second measurement of the same dynamical variable ξ immediately after the first, the result of the second measurement must be the same as that of the first. Thus after the first measurement has been made, there is no indeterminacy in the result of the second.^[2]

von Neumann (1932) makes his famous projection postulate on pp. 200–201:

( P.) The probability that in the state φ the quantities with the operators R₁, ... ,R_l take on values from the respective intervals I₁ ... I_l is

|| E₁(I₁) ... E_l(I_l) φ ||²

where E₁(λ), ..., E_l(λ) are the resolutions of the identity belonging to R₁, ... ,R_l respectively.

...we postulate P. for all commuting R₁, ... ,R_l . Then the E₁(I₁), ..., E_l(I_l) commute, and therefore E₁(I₁) ... E_l(I_l) is a projection (THEOREM 14. in II. 4.), and the probability in question becomes

P  =  || E₁(I₁) ... E_l(I_l) φ ||²  =  (E₁(I₁) ... E_l(I_l) φ , φ)

According to Weinberg on page 26:

...Thus if a system is in a state represented by a wave function ψ, and we make a measurement that puts the system in any one of a set of states represented by orthonormal wave functions ψ_n (which may or may not be energy eigenfunctions) then the probability that the system will be found to be in a particular state represented by the wave function ψ_m is

${\displaystyle P(\psi \rightarrow \psi _{m})=\left|\int \psi _{m}^{*}\psi \right|^{2}.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1.5.18)}$

This can be taken as the fundamental interpretive postulate of quantum mechanics.^[1]

Weinberg in 2013 is using the traditional term "measurement" as Dirac used it in 1958, to express the von Neumann projection postulate.

Greiner, W. (2001), Quantum Mechanics: An Introduction, 4th edition, Springer, Berlin, ISBN 3-540-67458-6

P. 438:

Axiom 1: As a result of the measurement of an observable, only one of the eigenvalues of the corresponding operator can be found. After the measurement, the system occupies that state which corresponds to the measured eigenvalue.

P. 469:

Now we ask how measurement influences the system. Let the measurement have the result q_l. Immediately after the measurement, we assume it to be repeated. If the measurements are to make physical sense, we have to claim that the experimental result does not change: the second measurement also has to result in the value q_l.

Greiner in 2001 is using the traditional term "measurement" as Dirac used it in 1958, to express the von Neumann projection postulate.

Page 5 of the 3rd edition of L&L (Pergamon 1977):

We shall now formulate the meaning of a complete description of a state in quantum mechanics. Completely described states occur as a result of the simultaneous measurement of a complete set of physical quanti­ties.^[1]

L&L in 1977 used the traditional term "measurement" as Dirac used it in 1958, to express the meaning of a complete description of a state.

On page 235, Cohen-Tannoudji, Diu & Laloë (2nd edition, 1973/1977) write:

Similarly, we can construct devices, intended to prepare a quantum system, in such a way that they only allow the passage of one state, corresponding to a particular eigenvalue of each of the observables of the complete set chosen.^[1]

These authors do not here use the traditional term "measurement" for their preparation procedure, though they rely on devices which prepare by selecting or filtering.

The traditional term "measurement", as used by Dirac, Greiner, Landau & Lifshitz, and Weinberg, is, regrettably, an oft-encountered abuse of language. A better term would be 'purification by selection or filtration', or Dirac's "sorting apparatus".

This proposed better term is not original research. For example, Park & Band in 1992 wrote:

P. 658:

It is not unusual to find excellent didactical presentations of the quantal algorithm embedded in a kind of structureless philosophical void which offers the student no clue as to the empirical significance of the formalism. ... We believe that the student of quantum theory can successfully weather the above-mentioned and other common affronts to his intellect if at some point in the educational process, preferably early, he is introduced to the preparation-measurement format of experimental science and then immediately taught the particular manner in which quantum theory copes with physical problems by employing that framework.

P. 659:

Nevertheless, to this day the preparation process is still often misnamed "measurement" ...

P. 662:

A further selection may be imposed ... We call this an almost "pure" state because the result of any subsequent momentum measurement has an almost foregone conclusion. ... When we have completed this selective preparation of a pure state, we have, according to conventional wisdom, already performed the measurement. ... Conventional discussions based on the von Neumann mathematical scheme stopped at the pure state preparation and asserted that the measurement act had left the particle in the pure state corresponding to the result of the measurement. This is really nothing worse than a semantic error.^[1]

Perhaps this may be useful.Chjoaygame (talk) 13:25, 30 March 2016 (UTC)