# How to understand braket two state?

1. Feb 24, 2010

### luxiaolei

Hi all, can anyone help me understand how to understand braket two different state?

<x|x'> what is the physics meaning behind?
and also <x|V|x'>

2. Feb 24, 2010

### Frame Dragger

3. Feb 24, 2010

### Fredrik

Staff Emeritus
Your questions don't seem to have anything to do with bra-ket notation. The notation is explained here. The most important part is to note that $\langle\phi|\psi\rangle$ is just the inner product of $|\phi\rangle$, and $|\psi\rangle$ (which are both vectors).

The physical intepretation of an inner product is given by the rule that says that if the system is in state $|\psi\rangle$ when you measure an observable represented by the operator A, and $|a\rangle$ is the eigenvector of A with eigenvalue a, then the probability that the result of the measurement will be a is $|\langle a|\psi\rangle|^2$.

$\langle\phi|A|\psi\rangle$ doesn't have any physical significance that I can think of when the two states are different, but $\langle\psi|A|\psi\rangle$ is the average value of a large number of measurements of A on systems that are all prepared in the state $|\psi\rangle$. This follows from the probability rule I mentioned above and

$$\langle\psi|A|\psi\rangle=\sum_a\langle\psi|A|a\rangle\langle a|\psi\rangle=\sum_a a \langle\psi|a\rangle\langle a|\psi\rangle=\sum_a a|\langle a|\psi\rangle|^2$$

4. Feb 24, 2010

### luxiaolei

@Fredrik, Thanks for replay, I am actually learning cross-section in particle physics, and I met this <K|V|K'> where K is the incoming wave vector, K' is the outgoing wave vector, and V is the target potential, and I am hardly understand it...

5. Feb 24, 2010

### SpectraCat

The physical significance is that the square modulus of this bra-operator-ket expression represents the probability that the state resulting from the application of operator A to state $$\psi$$ will be found to be in state $$\phi$$. If $$\psi$$ and $$\phi$$ are non-degenerate eigenstates of A, then that probability will be uniformly zero, otherwise it may be non-zero in the general case.

For the OP's problem, |<K|V|K'>|2 is the probability that the incoming state with wave-vector K', interacting with the potential V, will be found in the scattered state with wave-vector K.

6. Feb 24, 2010

### luxiaolei

@all. Thanks all! great explanation! I got it!

7. Feb 24, 2010

### peteratcam

That interpretation is only valid if A is unitary, which in general it is not.

8. Feb 24, 2010

### Fredrik

Staff Emeritus
...and if it isn't, divide with $\|A|\psi\rangle\|^2$ to get the correct probability.

9. Feb 25, 2010

### peteratcam

That procedure is inconsistent. For example if A is hermitian, there is no guarantee that $\|A|\psi\rangle\|^2=\|A|\phi\rangle\|^2$ whereas by the properties of the inner product, we would require $$|\langle \psi|A|\phi\rangle|^2=|\langle \phi|A|\psi\rangle|^2$$

I still maintain my objection that 'the state resulting from the application of operator A to state $\psi$' doesn't make any sense except for a unitary A. 'State' has a physical meaning which should be respected. Meaning can be given to the matrix element, for example in transition rates between states for perturbations, but not the meaning that SpectraCat gave.

10. Feb 25, 2010

### SpectraCat

Yeah ... to call it a probability is claiming too much in the general case. It is more correct to call it "the unnormalized projection of the state resulting from the application of A to $$\psi$$ onto the state $$\phi$$" ... that is ugly to write or say, and the physical meaning is less clear, but I believe it is correct in the general case.

$$|\langle \phi|A|\psi\rangle|^{2}$$ is only a strictly a probability in the case where $$\psi$$ and $$\phi$$ are normalized, and A is unitary, as you say. However if, as in the OP's case, $$\psi$$ and $$\phi$$ are vectors from the same basis, then $$|\langle \phi|A|\psi\rangle|^{2}$$ will be proportional to the probability, with the normalization condition Frederik mentioned. This will also be true if $$\phi$$ is an eigenstate of A, which is actually what I was thinking of when I wrote my original reply.

I am not sure what you mean here ... what is the physical meaning of "state" that should be respected? ... to me it just means an arbitrary quantum state, i.e. a vector in a Hilbert space, which corresponds to some wavefunction in the position representation. There is no particular stipulation that such an entity must be normalized in the general case as far as I am aware. Are you happier if I call them kets instead of states?

Furthermore, matrix elements for transition rates have their meaning for precisely the reason I expressed in my first post .. they represent probabilities that the initial state will be found in the final state after the action of the perturbation ... of course that is because a transition matrix is a unitary matrix expressed in an orthonormal basis.

11. Feb 25, 2010

### peteratcam

Suppose $$|\psi\rangle$$ is some normalised ket. I'd happily call it a state. Multiply it by 5, and I'll forgive you calling it a state, but I'd prefer just ket, or perhaps unnormalised state. (I get suspicious...why is it unnormalised to begin with? where has it come from to be unnormalised? I don't like unnormalised states they shouldn't be with us.)

Multiply it by 7 metres, or 5 Joules, or 2i meters per second, and I get very very uneasy calling it a state, because it can't be normalised to the number 1, so the probability interpretation gets completely messed up. The unitary operators we meet in QM are 'dimensionless' in a physics sense. But most of the others aren't (eg, Hamiltonian, angular momentum) and so the quantity you get by sticking the operator in the middle of an inner product has the dimensions of that operator. In the practice of doing QM the only legitimate places such inner products occur are such that the operator is dimensionless, or when the result is an average of the quantity.

You reserve the right to think I'm mad.

12. Feb 25, 2010

### SpectraCat

No, it's probably a very good way to think about it, because it keeps one from making the mistake of over-simplification (as I did in my first post). The issue of states multiplied by dimensioned-quantities is interesting .. I hadn't really thought about it that way before. I suppose this is because we generally pull all of the dimensioned quantities out of the equation using the appropriate constants of proportionality, so that all we deal with in the bra-ket expressions are the dimensionless functional forms of the states and operators. Still, it is worth keeping in mind for "debugging" of future mathematical derivations. Thanks!

13. Feb 25, 2010

### Fredrik

Staff Emeritus
Peteratcam, I don't think your argument is correct. There's no need to bring dimensions into the picture. It's perfectly valid to think of everything as dimensionless*. If $|\psi\rangle$ is a vector in the Hilbert space we're working with, then so is $A|\psi\rangle$, and

$$|\chi\rangle=\frac{A|\psi\rangle}{\|A|\psi\rangle\|}$$

is a vector of unit norm. There's no reason why the things we can say about a state vector that corresponds to a known state preparation procedure wouldn't also be valid for one that doesn't.

*) For example, a length measurement in units of meters is just telling us the dimensionless number of meter sticks that you'd need to fill up the measured distance. Measurements done with different units are handled by using a different self-adjoint operator to represent the measuring device mathematically. For example, if A is the operator that corresponds to a measuring device that presents the answer in m2, then 10000A corresponds to the same device modified to present the answer in cm2.

14. Feb 26, 2010

### peteratcam

There are more equations one can write down than make sense physically, and looking at the dimensions of quantities has been a trick in a physicist's toolbox for a long time in distinguishing between a string of symbols and a piece of physics. If someone asked me a basic mechanics question, and in my answer I had added an energy to a distance, then it would be complete nonsense. I was marking some QM work recently where the question was to calculate a probability, and someone had the answer a/9 (where 'a' was a length). Again, it is obvious that the answer is wrong.

All I'm saying (which I don't think is controversial) is that keeping track of physics dimensions, even with operators in Hilbert space, helps in interpreting the physical meaning, if any, of expressions we write down.

15. Feb 5, 2011

thanks