Can Quantum Mechanics be studied without the Copenhagen interpretation

Hello everyone,

I'm a doctoral student of particle physics in ETH Zurich, and have a question in fundamental QM.

I remember a friend telling me that the professor that taught him Quantum Mechanics in ETH Zurich did not believe in the Copenhagen interpretation, and thus taught the class Quantum Mechanics in a way that my friend didn't understand.

Trying to roughly recall what my friend told me, he said that the professor relied a lot on groups theory and complex mathematics, that's what I recall from his description.

So my question is: is that even possible? Through the course of my study of physics I have not seen any interpretation that worked other than the Copenhagen interpretation. Are there other kinds of interpretations that work but people ignore?

Thank you for any efforts.
 

dextercioby

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Having a course based on Leslie Ballentine's book I find illuminating. He's an advocate of the 'Ensemble Interpretation' which I consider to be a viable alternative to the theory built 80 years ago.

Complex mathematics and group theory ? I can't imagine quantum theories without them.
 

UltrafastPED

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I think the professor was a fan of the "shut up and compute" school of thought.

All of the "interpretations" are essentially philosophical discussions. They don't change how you do the calculations - or the results.
 

vanhees71

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The first question what one must ask is what do mean when you say "Copenhagen interpretation", because there are several flavors of it around. For me the Copenhagen interpretation is the minimal statistical interpretation a la Ballentine + the collapse (projection) postulate, which adds the following assumption to the other postulates, describing kinematics and dynamics of quantum theory (including the Born rule assigning the usual probability content to the state kets):

Suppose a quantum system is prepared in a pure state, described by the state vector [itex]|\psi \rangle[/itex], and an observable [itex]A[/itex], described by a self-adjoint operator [itex]\hat{A}[/itex], is precisely measured with the finding of a value [itex]a[/itex]. This value necessarily is an eigenvalue of [itex]\hat{A}[/itex]. Let the eigenspace to that eigenvalue be spanned by an orthonormal basis [itex]|a,\alpha [/itex]. Then after the measurement the system is in the pure state, described by the vector
[tex]|\psi' \rangle=\sum_{\alpha} |a,\alpha \rangle \langle a,\alpha|\psi\rangle.[/tex]

First of all one should note that there is not much necessity for this postulate to apply quantum theory to real systems. So I cannot see that a lecture, leaving it out, could be any more difficult than a lecture using it for some reason as a foundation. In fact, usually it's introduced and than rarely discussed any further, although it's a very problematic postulate. That's why I, personally, prefer the minimal statistical interpretation. Let's see, what's problematic with it.

Supposed this postulate holds. Then there are two cases

(a) The state vector [itex]\psi'[/itex] "after the measurement" is either the same as before, which occurs if and only if [itex]|\psi \rangle[/itex] is an eigenvector of [itex]\hat{A}[/itex] with the measured eigenvalue [itex]a[/itex], i.e., if the system was already prepared such that the observable [itex]A[/itex] has the determined value [itex]a[/itex], and then with 100% probability an (ideal) measurement of [itex]A[/itex] must find this value, and nothing happens to the system by the measurement procedure. This is unproblematic and not very exciting.

(b) The system's state was not prepared such that it is described by any eigenstate of [itex]\hat{A}[/itex]. Then, if [itex]|\psi \rangle[/itex] was chosen to be normalize, [itex]|\psi' \rangle[/itex] consists only of the projection of [itex]|\psi \rangle[/itex] onto the eigenspace of [itex]\hat{A}[/itex] for the measured value [itex]a[/itex]. Thus, [itex]\langle \psi'|\psi' \rangle<1[/itex].

This is problematic, because that means the measurement process is not described by quantum-theoretical time evolution, because implicitly in the collapse postulate you must assume that before as well as after the measurement process you must have the measured quantum system sufficiently separated from the measurement apparatus, i.e., the enlarged state, containing both the object and the measurement device must be a product state. For a unitary time evolution the part of the obejct in this tensor product must be normalized, if it was normalized before it interacted with the measurement apparatus. This means that the interaction between the quantum object and the measurement device is described outside of quantum theory, although there is no reason to believe that the measurement device itself is not also describable by quantum theory.

To me that doesn't make much sense, and that's why (among the other reason that there's also trouble in the context of relativity, i.e., the EPR paradoxon) I prefer the minimal interpretation. The above problem simply doesn't occur in it since it simply doesn't make the collapse assumption. It doesn't matter what happens to the system and/or the measurement apparatus after the measurement. Quantum theory simply tells you the probabilities to find a certain outcome of the measurement of an observable, given the state of the measured system. That's all you need to make the connection between the quantum theoretical formalism and the real world in the lab.

I think not that your friend's lecture was (subjectively) "bad", because the professor didn't adhere to the one or the other flavor of the Copenhagen interpretation. In fact, I once taught Quantum Mechanics II, and did not refer to the Copenhagen interpretation but used the ensemble interpretation throughout. According to my evaluation the students were pretty satisfied with my treatment, and that although I also gave them a lot of group and Lie algebra representation theory to swallow, mostly focusing on space-time symmetries, i.e., on the Galileo (non-relativstic spacetime) and the Poincare (special relativistic spacetime) groups.

However, I wouldn't do group theory in Quantum Mechanics I, but I'd also use the ensemble interpretation. Group and Lie algebra representation theory a pretty advanced topic, and in QM I it is much more important to give a good idea about both the formal structure of quantum theory, introduce the mathematics of the Schrödinger Equation, eigenvalue problems, orthonormal function systems in [itex]\mathrm{L}^2[/itex], the abstract Hilbert space formulation (bras and kets) and to give a lot of the standard examples in simple applications (one-dimensional problems like potential pots (infinitely and finitely high), potential steps, harmonic oscillator), motion in central potentials (including the non-relativistic hydrogen atom), scattering theory (Lippmann-Schwinger equation). Of course, you have to discuss angular momentum (at least orbital angular momentum), which is already representation theory of the Lie algebra so(3)=su(2).

For QM II, however, I consider group and Lie algebra representation theory mandatory. A lot of misunderstandings, particularly in relativistic QT, comes from a lack of group-theoretical foundations. Last but not least Noether's theorem gives a convincing argument for why the observables are represented the way they are (e.g., why is [itex]\hat{\vec{p}}=-\mathrm{i} \vec{\nabla}[/itex] in the position representation? Answer: Momentum is the generator for spatial translations, which holds as well in quantum mechanics as in classical mechanics). In other words, the Lie algebra of the symmetry groups provide the commutator relations for the algebra of observables, which is on the very foundation of the theory.
 
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Thank you dextercioby and UltrafastPED; your comments were helpful.

vanhees71: I'm not sure how to answer your argument, but you left the main topic and went far away; maybe I'm wrong, but you probably have some problems understanding Quantum Mechanics. I'm not willing to go into a discussion on the basics of quantum theory, that's not the purpose of this post; I hope it was clear, but... I hope we won't just waste our time on a chit-chat on something that's clearly explained on Wikipedia's page "QM".

My first comment to your answer is that a state <psi'|psi'> cannot be less than unity unless the space, over which you do the expansion, is incomplete. This is satisfied by its simplest form in the probability current conservation. You cannot go from one state to another without having the probability transfer from one state to another. You're destroying very fundamental laws in physics like Noether's theorem with what you're saying. A trivial example to this, is when you say that you have a 3D space described by the xy-plane (yes, it's a stupid example but sufficient to the point), and then you do a cross product for two vectors on this plane, and then complain saying "OMG... where did the result go? why is it not described by my magical 3D xy-plane?"... well, of course you can't find your result, because your 3D space is incomplete and isn't compatible with the operator you've used, where you need a third axis, the z-axis, to find your result proportional to its unit vector. It's exactly the same argument in your example! If your |psi> subspace components are complete and describe every possible state in your system, then <psi'|psi'> must be unity all the time (of course everything should be normalized).

My second comment, I'm not saying that the course was "bad", I'm nothing but a doctoral student and shouldn't pass judgements on ETH professors... the guy has a gazillion publications... but I just wanted to know about more models/interpretations of QM. But apparently, like UltrafastPED's said, the interpretations are philosophical issues; but I still wonder why the course should be that hard for my friend (he's smart, BTW) when they all should be the same, but different in philosophical issues.

Please don't misunderstand my impatience, I'm just bad in formulating "nice" sentences. If I were in front of you, you'll know from the smile on my face that this is nothing but a friendly chat with a beer in my hand :)

Thank you all :) ; and I still would love to hear more about this!
 
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UltrafastPED

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Sometimes the presentation is unclear because you were expecting something else ... there are lots of text books because each author feels that there is a better way to say it, or to learn it, or a better choice of topics.

I remember giving a lecture on the fundamentals of analytical mechanics which clearly missed the mark - so next time I backed up and started with a few mechanical examples to illustrate generalized coordinates and degrees of freedom - my students had no difficulty once we they were clear on the jargon and its meaning.
 

cgk

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TheDestroyer, you asked about two unrelated issues (Copenhagen interpretation and a group-theoretical approach to QM) and van Hees gave two (good) answers to both of them. In particular, he explained why (a) the Copenhagen interpretation is just a small part of QM and in practice not really needed to do QM (something you explicitly asked about), (b) why it can be viewed as problematic and why many researchers choose to not believe in it, (c) why Lie groups should not be avoided in QM even though they can be confusing for new students.

You might want to consider trying to understand his answer to *your* question. Your style of posting is not appropriate, even more so if you are wrong (and you are). And, just to make this clear:
Please don't understand my impatience, I'm just bad in formulating "nice" sentences. If I were in front of you, you'll know from the smile on my face that this is nothing but a friendly chat with a beer in my hand :)
is not an excuse. Formulating answers like above in a professional setting (e.g., to your colleagues or to researchers in conferences) is a certain path to leaving your institution without a degree.
 
I think the professor was a fan of the "shut up and compute" school of thought.

All of the "interpretations" are essentially philosophical discussions. They don't change how you do the calculations - or the results.
shut up is just another interpretation, physical theory without interpretation is just math.

in any case physics is grounded on philosophy.

.
 

UltrafastPED

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Physics is the study of how things work - applied at many levels, from astrophysics to mechanics to atoms and light. In this study you will learn how to think about problems and how to solve them.

Once upon a time the philosophers of ancient Greece were the "keepers of all knowledge" - but I don't use Plato, Aristotle, Kant, or Nietzsche when I'm setting up an optical or electron beam experiment.
 
Physics is the study of how things work - applied at many levels, from astrophysics to mechanics to atoms and light. In this study you will learn how to think about problems and how to solve them.

Once upon a time the philosophers of ancient Greece were the "keepers of all knowledge" - but I don't use Plato, Aristotle, Kant, or Nietzsche when I'm setting up an optical or electron beam experiment.
HOW = Reason or Purpose.
 

UltrafastPED

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I disagree; WHY = reason or purpose. You use philosophy (or religion) to answer why. How is just building blocks with rules - which is what logic and mathematics give you, along with the empirical knowledge which we have acquired over the years.

I suspect that we won't reach agreement in a forum like this. It would take a tavern and a few beers!
 

vanhees71

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Thank you dextercioby and UltrafastPED; your comments were helpful.

vanhees71: I'm not sure how to answer your argument, but you left the main topic and went far away; maybe I'm wrong, but you probably have some problems understanding Quantum Mechanics. I'm not willing to go into a discussion on the basics of quantum theory, that's not the purpose of this post; I hope it was clear, but... I hope we won't just waste our time on a chit-chat on something that's clearly explained on Wikipedia's page "QM".

My first comment to your answer is that a state <psi'|psi'> cannot be less than unity unless the space, over which you do the expansion, is incomplete. This is satisfied by its simplest form in the probability current conservation. You cannot go from one state to another without having the probability transfer from one state to another. You're destroying very fundamental laws in physics like Noether's theorem with what you're saying. A trivial example to this, is when you say that you have a 3D space described by the xy-plane (yes, it's a stupid example but sufficient to the point), and then you do a cross product for two vectors on this plane, and then complain saying "OMG... where did the result go? why is it not described by my magical 3D xy-plane?"... well, of course you can't find your result, because your 3D space is incomplete and isn't compatible with the operator you've used, where you need a third axis, the z-axis, to find your result proportional to its unit vector. It's exactly the same argument in your example! If your |psi> subspace components are complete and describe every possible state in your system, then <psi'|psi'> must be unity all the time (of course everything should be normalized).

My second comment, I'm not saying that the course was "bad", I'm nothing but a doctoral student and shouldn't pass judgements on ETH professors... the guy has a gazillion publications... but I just wanted to know about more models/interpretations of QM. But apparently, like UltrafastPED's said, the interpretations are philosophical issues; but I still wonder why the course should be that hard for my friend (he's smart, BTW) when they all should be the same, but different in philosophical issues.

Please don't misunderstand my impatience, I'm just bad in formulating "nice" sentences. If I were in front of you, you'll know from the smile on my face that this is nothing but a friendly chat with a beer in my hand :)

Thank you all :) ; and I still would love to hear more about this!
I think cgk has already given the appropriate comments to this posting. I just want to add that you should formulate your questions more clearly if you want the very specific answers you expect. I understood your question in the way I answered it. Sorry if I didn't meet your expectations.

I just want to correct an obvious misunderstanding from your part and to clarify the issue also for the other readers of this thread.

The (generalized) eigenvectors [itex]|a,\alpha \rangle[/itex] of a self-adjoint operator [itex]\hat{A}[/itex], where [itex]a[/itex] runs over the spectrum of [itex]\hat{A}[/itex] and [itex]\alpha[/itex] summarizes one or more additional parameters labeling the "degeneracy". These parameters can be thought of as denoting the eigenvalues of other self-adjoint independent pairwise commuting operators that, together with [itex]\hat{A}[/itex] (with which they must commute too) form a complete set of compatible observables.

The subspace [itex]\text{Eig}(\hat{A},a)[/itex] spanned by all eigenvectors to a single eigenvalue [itex]a[/itex] of the self-adjoint operator [itex]\hat{A}[/itex] usually is a proper subspace of the Hilbert space. Thus, if you project the vector [itex]|\psi \rangle[/itex] onto this subspace,
[tex]|\psi' \rangle=\sum_{\alpha} |a,\alpha \rangle \langle a,\alpha|\psi \rangle,[/tex]
in general you have
[tex]\|\psi' \| \leq \|\psi \|,[/tex]
and the equality sign holds if and only if [itex]|\psi' \rangle \in \text{Eig}(\hat{A},a)[/itex].

Further, the subspace [itex]\text{Eig}(\hat{A},a)[/itex] is the full Hilbert space [itex]\mathcal{H}[/itex] if and only if [itex]\hat{A}=a \text{id}[/itex].

If you want to learn about the physics part of the debate about the interpretations of quantum theory, you should consult the already mentioned textbook by Ballentine. Another very good book on this topic is

A. Peres, Quantum Theory: Concepts and Methods, Kluwer (2002)

Another very good book is Weinberg's newest textbook, written in his typical "no-nonsense approach" to quantum theory, giving a very clear exposition of various interpretations of quantum theory early in the book.

S. Weinberg, Lectures on Quantum Mechanics, Cam. University Press (2013)

I have no recommendation concerning philosophy of quantum mechanics except that you first should be very familiar with good physical textbooks on the subject, because philosophers tend to obscure the issues more than clarifying them ;-)).
 
I think cgk has already given the appropriate comments to this posting. I just want to add that you should formulate your questions more clearly if you want the very specific answers you expect. I understood your question in the way I answered it. Sorry if I didn't meet your expectations.

I just want to correct an obvious misunderstanding from your part and to clarify the issue also for the other readers of this thread.

The (generalized) eigenvectors [itex]|a,\alpha \rangle[/itex] of a self-adjoint operator [itex]\hat{A}[/itex], where [itex]a[/itex] runs over the spectrum of [itex]\hat{A}[/itex] and [itex]\alpha[/itex] summarizes one or more additional parameters labeling the "degeneracy". These parameters can be thought of as denoting the eigenvalues of other self-adjoint independent pairwise commuting operators that, together with [itex]\hat{A}[/itex] (with which they must commute too) form a complete set of compatible observables.

The subspace [itex]\text{Eig}(\hat{A},a)[/itex] spanned by all eigenvectors to a single eigenvalue [itex]a[/itex] of the self-adjoint operator [itex]\hat{A}[/itex] usually is a proper subspace of the Hilbert space. Thus, if you project the vector [itex]|\psi \rangle[/itex] onto this subspace,
[tex]|\psi' \rangle=\sum_{\alpha} |a,\alpha \rangle \langle a,\alpha|\psi \rangle,[/tex]
in general you have
[tex]\|\psi' \| \leq \|\psi \|,[/tex]
and the equality sign holds if and only if [itex]|\psi' \rangle \in \text{Eig}(\hat{A},a)[/itex].

Further, the subspace [itex]\text{Eig}(\hat{A},a)[/itex] is the full Hilbert space [itex]\mathcal{H}[/itex] if and only if [itex]\hat{A}=a \text{id}[/itex].

If you want to learn about the physics part of the debate about the interpretations of quantum theory, you should consult the already mentioned textbook by Ballentine. Another very good book on this topic is

A. Peres, Quantum Theory: Concepts and Methods, Kluwer (2002)

Another very good book is Weinberg's newest textbook, written in his typical "no-nonsense approach" to quantum theory, giving a very clear exposition of various interpretations of quantum theory early in the book.

S. Weinberg, Lectures on Quantum Mechanics, Cam. University Press (2013)

I have no recommendation concerning philosophy of quantum mechanics except that you first should be very familiar with good physical textbooks on the subject, because philosophers tend to obscure the issues more than clarifying them ;-)).
Thank you for your answer. I'm confused a little bit, and apparently I'm not as good as I thought in Quantum Mechanics, even after reading 3 Quantum Field Theory books and having a QFT course.

Can you please clarify, why should ∥ψ′∥≤∥ψ∥? Do the books you mentioned explain this paradigm of the problem? I'd be thankful if you could give an example on this issue.

And what does [itex]\hat{A}=a\mbox{id}[/itex] mean?

Thank you.
 

dextercioby

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It's a projection of a vector onto another vector. It can't be bigger than the projected vector. Think about 6th grade plane geometry and its theorem of Pitagora.
 
It's a projection of a vector onto another vector. It can't be bigger than the projected vector. Think about 6th grade plane geometry and its theorem of Pitagora.
Ah, you're right. I'm sorry. That identity is obvious, I don't know why I got confused.

But doesn't this mean already that if ∥ψ′∥<∥ψ∥, then ∑|a,α><a,α| < 1, which means that the vector space isn't complete and still requires more bases, bringing us back to my cross-product example?
 
How is just building blocks with rules - which is what logic and mathematics give you, along with the empirical knowledge which we have acquired over the years.

I suspect that we won't reach agreement in a forum like this. It would take a tavern and a few beers!
well said !

http://www.philosophy-index.com/philosophy/branches/

Branches of Philosophy:
Metaphysics, which deals with the fundamental questions of reality.
Epistemology, which deals with our concept of knowledge, how we learn and what we can know.
Logic, which studies the rules of valid reasoning and argumentation
Ethics, or moral philosophy, which is concerned with human values and how individuals should act.
Aesthetics or esthetics, which deals with the notion of beauty and the philosophy of art.

-----
http://en.wikipedia.org/wiki/Philosophy

.
 
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Nugatory

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... and to vanhees71's point, that the Copenhagen interpretation is problematic.
Yes, especially if you take into account the information in chapter 8 of David Albert's book "Quantum Mechanics and Experience"/
 
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shut up is just another interpretation, physical theory without interpretation is just math.
Errrrr. Its based on two axioms as found in for example Ballentine's book. These are clear and unambiguous statements about nature:

1. Observable's are Hermitian operators whose eigenvalues are the possible outcomes of an observation.

2. A positive operator P of unit trace exists, called the state, such that the expected value of the observable O is Tr(OP).

in any case physics is grounded on philosophy
Hmmmm. Philosophers might like to think so but actual physicists probably don't - its more likely they think its grounded in math and correspondence with experiment. Of course that in itself is a philosophy so really its a tautology - but of course that's not quite the sense its meant, rather its meant its not founded on philosophical dialectic.

To the OP of course it can be studied without reference to Copenhagen or indeed any interpretation for that matter. And Ballentine is a good book for that view, although he does compare Copenhagen to the Ensemble which he champions - as do I - but I add in decoherence for a twist.

Actually a SUPERB book to come to grips with interpretations is not actually about them - it's Decoherence and the Quantum-to-Classical Transition by Schlosshauer

Thanks
Bill
 
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9,076
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Once upon a time the philosophers of ancient Greece were the "keepers of all knowledge" - but I don't use Plato, Aristotle, Kant, or Nietzsche when I'm setting up an optical or electron beam experiment.
Aren't that the truth.

And we have the case of Kant who believed Euclidean Geometry was a-priori and so intimidated was Gauss that he held off publishing his papers on Non Euclidean geometry which is in fact just as valid as Euclidean geometry - it was mathematicians and physicists that discovered this and its importance to describing the world while philosophers were still arguing.

Seriously physics, mathematics and science in general long ago divorced themselves from philosophy.

Thanks
Bill
 
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Yes, especially if you take into account the information in chapter 8 of David Albert's book "Quantum Mechanics and Experience"/
Ballentine's book examines it quite well. I don't entirely agree with his analysis that purports to show Copenhagen has issues with taking the state as fundamental, but its a good place to get to grip with the issues.

Thanks
Bill
 
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Physics is the study of how things work - applied at many levels, from astrophysics to mechanics to atoms and light. In this study you will learn how to think about problems and how to solve them.

Once upon a time the philosophers of ancient Greece were the "keepers of all knowledge" - but I don't use Plato, Aristotle, Kant, or Nietzsche when I'm setting up an optical or electron beam experiment.
I felt I had to jump in and say I fully agree with this quote. Very nicely put, IMHO.
 
... and to vanhees71's point, that the Copenhagen interpretation is problematic.
I'm sorry, but I still don't see the problem. Whenever you see a state you haven't accounted for, can't you simply add it and renormalize again? Why can't this be done?
 

vanhees71

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Ah, you're right. I'm sorry. That identity is obvious, I don't know why I got confused.

But doesn't this mean already that if ∥ψ′∥<∥ψ∥, then ∑|a,α><a,α| < 1, which means that the vector space isn't complete and still requires more bases, bringing us back to my cross-product example?
You got confused, because you didn't read my formula carefully enough. Of course you have
[tex]\sum_{a,\alpha} |a,\alpha \rangle \langle a,\alpha|=\text{id},[/tex]
but I summed only over [itex]\alpha[/itex]! BTW [itex]\text{id}[/itex] is the identity operator, i.e.,
[itex] \text{id} |\psi \rangle = |\psi \rangle[/itex]. You could also write [itex]\hat{1}[/itex].
 
You got confused, because you didn't read my formula carefully enough. Of course you have
[tex]\sum_{a,\alpha} |a,\alpha \rangle \langle a,\alpha|=\text{id},[/tex]
but I summed only over [itex]\alpha[/itex]! BTW [itex]\text{id}[/itex] is the identity operator, i.e.,
[itex] \text{id} |\psi \rangle = |\psi \rangle[/itex]. You could also write [itex]\hat{1}[/itex].
Thanks for the answer. But why would you think that an incomplete basis should give you a unity probability, thus claiming that the Copenhagen interpretation is problematic? I mean you're simply giving your state/probability current a path to escape to a state that you haven't taken into account, and then complain that the state "has disappeared"! It's more or less like the cross product example! you do a cross product on a plane and then you ignore the third axis where your resulting vector will be... well, of course the result won't make sense... because you're not looking where the state is located. Right?

In other words: I'm not surprised that ∥ψ′∥<∥ψ∥ when you use an incomplete basis, but I'm surprised that you call it a problem when you use an incomplete basis and yet expect that ∥ψ′∥==∥ψ∥ should be true.

Sorry for being persistent, but I have to understand this. Thank you for your time and efforts.
 

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