Understanding the copenhagen interpretation

In summary, (ψ,\varphi) = \int d^{3}r ψ^{*}(r)\varphi(r) represents the inner product between two wavefunctions in a Hilbert space, and it is used to calculate probabilities and perform transformations between different bases. This is shown under 'state' because it is a crucial concept in understanding the quantum mechanical state of a system.
  • #1
spaghetti3451
1,344
33
This is an extract from the lecture notes I took for the 'Foundations of QM' third year course.



Copenhagen QM - classical-quantum division

State: wavefunction ψ(x); (ψ,[itex]\varphi[/itex]) = [itex]\int d^{3}r ψ^{*}(r)\varphi(r)[/itex]

Evolution: TDSE

Observables ([itex]\hat{x},\hat{p},\hat{H}[/itex]): A = A-dagger; {x,p} = 1 [itex]\rightarrow[/itex] [itex][\hat{x},\hat{p}][/itex] = i[itex]\hbar[/itex]

Probability: [itex]\left|ψ\right|^{2}d^{3}r[/itex] Born rule

Measurement: collapse of ψ - can't assume system possesses properties if not measured

Composite systems: [itex]ψ_{AB} (x_{1},x_{2}) = ψ_{A} (x_{1}) ψ_{B} (x_{2})[/itex] if uncorrelated



I am wondering what (ψ,[itex]\varphi[/itex]) = [itex]\int d^{3}r ψ^{*}(r)\varphi(r)[/itex] means and why it is shown under 'state'.

Any help would be greatly appreciated.
 
Last edited:
Physics news on Phys.org
  • #2
Okay, let me give you a headstart. The integral represents an inner product in a Hilbert space. Question is what exactly does it represent.
 
  • #3
I am wondering if you could be more specific.

The closer the inner product to one, the greater the two states are related to each other, that is, operators acting on them will return the same eigenvalue with higher and higher probabilities. Is that it?
 
  • #4
I am beginning to wonder if no one on Physicsforums has the necessary qualifications to answer my question.
 
  • #5
The answer is simple. It doesn't show up under 'state', only $\psi(x)$ shows up under 'state'. Your confusion comes from the way you take notes.
 
  • #6
can any one explain ? copenhagen interpretation ? and others simply :)
 
  • #7
I'm not sure what your question is, the copenhagen interpretation is nothing more than an interpretation

A state is not a wavefunction, a wavefunction is a state in some basis, usually the position basis.

State: wavefunction ψ(x); (ψ,φ) = ∫d3rψ∗(r)φ(r)

A wavefunction is indeed a [itex]\psi (x)[/itex], this is not a state however
A state is [itex]|\alpha \rangle[/itex], it is related to a position wavefunction by [itex]\psi(x) = \langle x | \alpha \rangle [/itex]

The '(ψ,φ) = ∫d3rψ∗(r)φ(r)' part is just the definition of the inner product. The inner product in this picture is the probability of a system being prepared in state φ, after being measured being found in state ψ. (well, it's the mod squared of this that's the probability really)

If you take a simple example with two orthogonal states a,b then we have (A,B) = 0. We can prepare our system in a state S=c(A+B) where c is a normalisation constant, then we'd have (A,c(A+B)) = c(A,A)+c(A,B) = c, so the probability of finding our system S in a state A after measurment is |c|^2.

I am beginning to wonder if no one on Physicsforums has the necessary qualifications to answer my question.
You shouldn't be throwing around stuff like that, you're question has already been answered by two other people.


can anyone explain ? copenhagen interpretation ? and others simply :) [/QUOTE
http://en.wikipedia.org/wiki/Interpretations_of_quantum_mechanics
 
  • #8
failexam said:
I am beginning to wonder if no one on Physicsforums has the necessary qualifications to answer my question.
Obviously, a lot of people here are more than qualified.

The set of wavefunctions is a vector space, let's call it V, and the map from ##V\times V\to\mathbb C## that you defined is a semi-inner product on that space. The pair (V,the map you defined) is a semi-inner product space.

The semi-inner product is as important as the wavefunctions, since you use it to calculate probabilities. For example, when you perform a measurement of A on each member of a large collection of systems that have all been prepared in a way that's consistent with the wavefunction ψ, QM predicts that the average result will be (ψ,Aψ).
 
Last edited:
  • #9
[itex]\int d^{3}r ψ^{*}(r)\varphi(r)[/itex]
This represents how much overlap there is between these two wave states. Typically, you would have an expression like this if you are converting from one basis to another and you want to calculate elements of the transformation matrix. If your initial state is given by |ψ>, then [itex]abs(\int d^{3}r ψ^{*}(r)\varphi(r))^2[/itex] gives you the probability of measuring it in the other state. Maybe this isn't what you wanted to know, but then you need to learn to ask better questions.
 

Related to Understanding the copenhagen interpretation

1. What is the Copenhagen interpretation?

The Copenhagen interpretation is a widely accepted interpretation of quantum mechanics, developed by Niels Bohr and Werner Heisenberg in the 1920s. It proposes that particles do not have definite properties until they are observed, and that the act of measurement itself can influence the outcome of an experiment.

2. How does the Copenhagen interpretation differ from other interpretations of quantum mechanics?

The Copenhagen interpretation differs from other interpretations, such as the Many-Worlds interpretation, in its treatment of the role of the observer in determining the outcome of an experiment. It also emphasizes the importance of probabilistic outcomes in quantum mechanics.

3. What is the role of the observer in the Copenhagen interpretation?

In the Copenhagen interpretation, the observer plays a crucial role in determining the outcome of an experiment. This is because the act of measurement collapses the wave function, causing the particle to have a definite state. Without the observer, the particle would exist in a state of superposition, with all possible outcomes simultaneously.

4. What are the implications of the Copenhagen interpretation?

The Copenhagen interpretation has several implications, including the idea that reality is fundamentally indeterminate and that the observer plays a crucial role in shaping reality. It also highlights the limitations of our understanding of the quantum world and the need for more research in this area.

5. What is the current status of the Copenhagen interpretation?

The Copenhagen interpretation is still widely accepted by many physicists and remains a dominant perspective on quantum mechanics. However, it has also been subject to criticism and alternative interpretations continue to be proposed and studied. Ultimately, the debate over the interpretation of quantum mechanics remains an ongoing and active area of research in physics.

Similar threads

  • Quantum Interpretations and Foundations
Replies
33
Views
3K
  • Quantum Interpretations and Foundations
2
Replies
47
Views
2K
  • Quantum Interpretations and Foundations
4
Replies
109
Views
7K
  • Quantum Interpretations and Foundations
Replies
2
Views
984
Replies
11
Views
1K
  • Quantum Interpretations and Foundations
Replies
21
Views
2K
  • Quantum Interpretations and Foundations
3
Replies
76
Views
4K
  • Quantum Interpretations and Foundations
Replies
12
Views
2K
  • Quantum Interpretations and Foundations
Replies
34
Views
4K
  • Calculus and Beyond Homework Help
Replies
3
Views
219
Back
Top