# Understanding the copenhagen interpretation

#### spaghetti3451

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,$\varphi$) = $\int d^{3}r ψ^{*}(r)\varphi(r)$

Evolution: TDSE

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

Probability: $\left|ψ\right|^{2}d^{3}r$ Born rule

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

Composite systems: $ψ_{AB} (x_{1},x_{2}) = ψ_{A} (x_{1}) ψ_{B} (x_{2})$ if uncorrelated

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

Any help would be greatly appreciated.

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#### spaghetti3451

Okay, let me give you a headstart. The integral represents an inner product in a Hilbert space. Question is what exactly does it represent.

#### spaghetti3451

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?

#### spaghetti3451

I am beginning to wonder if no one on Physicsforums has the necessary qualifications to answer my question.

#### martinbn

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.

#### sciboudy

can any one explain ? copenhagen interpretation ? and others simply :)

#### genericusrnme

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

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 any one explain ? copenhagen interpretation ? and others simply :) [/QUOTE
http://en.wikipedia.org/wiki/Interpretations_of_quantum_mechanics

#### Fredrik

Staff Emeritus
Gold Member
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).

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#### Khashishi

$\int d^{3}r ψ^{*}(r)\varphi(r)$
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 $abs(\int d^{3}r ψ^{*}(r)\varphi(r))^2$ 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.