# I Bracket VS wavefunction notation in QM

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1. Jun 21, 2017

### olgerm

In some sources QM is explained using bracket notation. I quite understand algebra of bracket notation, but I do not understand how is this notation related with physically meaningful things? How is bracket notation related to wavefunction notation?
Could you tell me whether following is true:
to express QM-system in bracket notation you:
1. solve schrödinger equation
$U_{System Potential Energy}(r_1,r_2,r_3,...,r_n,t)-\sum_{n=1}^N((\frac{d^2Ψ(r_1,r_2,r_3,...,r_n,t)}{dx_n^2}+\frac{d^2Ψ(r_1,r_2,r_3,...,r_n,t)}{dy_n^2}+\frac{d^2Ψ(r_1,r_2,r_3,...,r_n,t)}{dz_n^2})*\frac{ħ^2}{m_n})=i*ħ \frac{dΨ(r_1,r_2,r_3,...,r_n,t)}{dt}$
and get wavefunction Ψ.
2. apply fourier transformation to wavefunction.$\Psi(x)=\alpha(1) \cdot sin(1 \cdot x)+\alpha(2) \cdot sin(2 \cdot x)+\alpha(3) \cdot sin(3 \cdot x)...=\sum_{k=0}^\infty(\alpha(k) \cdot sin(k \cdot x))$
or
$\Psi(x)=\alpha(1) \cdot e^{-i \cdot 1 \cdot x}+\alpha(2) \cdot e^{-i \cdot 2 \cdot x}+\alpha(3) \cdot e^{-i \cdot 3 \cdot x}...=\sum_{k=0}^\infty(\alpha(k) \cdot e^{-i \cdot k \cdot x})$
(Which one? How to make fourier transformation to $\Psi(t;x;y;z)$?)
3. make bra of fourier transformation results. $<\psi|=(\alpha(0);\alpha(1);\alpha(2);\alpha(3);...)$
and ket $|\psi>=(\alpha^*(0);\alpha^*(1);\alpha^*(2);\alpha^*(3);...)$.
4. to find momentum of a particle solve equation $\hat{p}|\psi>=p_x \cdot |\psi>$ aka $-i\frac{\partial}{\partial x}|\psi>=p_x \cdot |\psi>$ for $p_x$.
(in which cases the operator must be hermitian?)

Last edited: Jun 21, 2017
2. Jun 21, 2017

One comment on your calculations above is to use complex variables to do the Fourier transform, and also do it in 3 dimensions. Apart from multiple $2 \pi$ factors $\hat{\Psi}(\vec{k})=\iiint \Psi(\vec{r}) e^{-i \vec{k} \cdot \vec{r}} \, d^3 \vec{r}$. (Note: This complex variable form of the F.T. uses Euler's formula $e^{ix}=cos(x)+i sin(x)$). $\\$ I don't have any simple explanation for the comparison of the bracket vs. wavefunction formalisms, but some of the quantum mechanics textbooks describe it fairly well.

3. Jun 21, 2017

### olgerm

So in 3 dimensional space I get one bra and one ket from F.T. per every dimension? or elements of bras and kets are 3-dimensional vectors?

4. Jun 21, 2017

The bra's and ket's are generally 3 dimensional, if they represent a state that has 3 dimensions. To get the wavefunction out of them, you do the following: $\Psi(\vec{x})=<\vec{x}|\Psi>$. The operator $|\vec{x}><\vec{x}|$ integrated over all $d^3 \vec{x}$ is the identity operator. $\\$ I can show you a couple of bits and pieces, but there are others on the Physics Forums website who have more Quantum Mechanics expertise. $\\$ Just one additional item that might be helpful: When you have $<\Psi|\Psi>$, you can throw in the identity operator in between and that gives $<\Psi|\Psi>= \iiint <\Psi|\vec{x}><\vec{x}|\Psi> \, d^3 \vec{x}=\iiint \Psi^{*}(\vec{x}) \Psi(\vec{x}) \, d^3 \vec{x}$. And you can similarly evaluate a form such as $<\Phi|\Psi>$ with the use of the identity operator to get you from the braket notation to the wavefunction notation.

Last edited: Jun 21, 2017
5. Jun 21, 2017

### ivy

Can an electron position probability produce a negative value?

6. Jun 21, 2017

### bhobba

Of course not - the Kolmogorov axioms forbid it. If you are unfamiliar with them look them up.

In the early days of QM they got negative probabilities so they knew something was wrong. It turned out to be positive probabilities of antiparticles. If you are interested in that story start a new thread.

Thanks
Bill

7. Jun 21, 2017

### bhobba

There is no simple explanation.

Its this. Suppose you have a general Ket |u> then we can expand it in eigenvalues of position so |u> = ∫ f(x) |x> dx. By definition f(x) is called the wave function. From the so called Born Rule its easy to show |f(x)|^2 is the probability of when observed you get position x.

But the math behind it is very advanced and deep. I gave an overview in this thread - but its probably gibberish unless you have studied linear algebra and preferably Hilbert spaces:

I do recommend you study linear algebra and distribution theory. They are a must for any applied mathematician including physics. For linear algebra there is tons of stuff on the internet eg:
http://linear.ups.edu/
http://quantum.phys.cmu.edu/CQT/chaps/cqt03.pdf

Read them in the above order.

For distribution theory get a copy of:
https://www.amazon.com/Theory-Distributions-Nontechnical-Introduction/dp/0521558905

Its worth it for the section on Fourier transforms alone. Without it it becomes bogged down in issues of convergence. If I was a teacher of Fourier Transforms, which I am not, but if I was I would do by teaching Distribution Theory.

Thanks
Bill

Last edited: Jun 22, 2017
8. Jun 22, 2017

### olgerm

Where are the multipliers, which relate every frequency and amplitude? These are noted as $\alpha$ in my post. I also put these multipliers into bra.
Is it just standard practice to use exponent form ( $\Psi(x)=\sum_{k=0}^\infty(\alpha(k) \cdot e^{-i \cdot k \cdot x})$) in fourier transformation, or by using other form ($\Psi(x)=\sum_{k=0}^\infty(\alpha(k) \cdot sin(k \cdot x))$) I got bra with which some QM formulas do not hold up?

Last edited: Jun 22, 2017
9. Jun 22, 2017

### bhobba

The Fourier transform is defined in many sources - even good old Wikipedia:
https://en.wikipedia.org/wiki/Fourier_transform

If F is the Fourier transform of a function f(x) denoted by F(f(x)) is a transform from one function to another. Their is another similar transform called in inverse Fourier transformation F' and interestingly, provided some assumptions are made about f(x) then F'F = 1. Let F(f(x)) = f'(x) so you have f(x) = F'(f'(x)). Intuitively when you look at the equation it means, physically, you can decompose a function into the sum of a large number of wavelike functions. That is its intuitive meaning.

In QM I mentioned any state can be decomposed into eigenkets of position |u> = ∫f(x) |x> dx. It can be also be decomposed into eigenkets of momentum p, |u> = ∫f'(p) |p> dp. Interestingly f(x) and f'(p) are related by the Fourier transform. Their is an intuitive reason for it, but I will leave you, using the references I gave, figure that one out. You understand better what you nut out for yourself. Hint the wave-function of a state of definite momentum is e^ipx - now look at the Fourier transform and remember the intuitive interpretation I gave.

Thanks
Bill

Last edited: Jun 22, 2017
10. Jun 22, 2017

### ivy

Schrödinger transforms the electron matter wave into a probability wave (Greene, p. 105) then uses a spherical coordinate system (r,Θ,φ,) representation in a wave equation,

-(h2/2u)∇"Ψ(r,Θ,φ) + V(r,Θ,φ) + V(r,Θ,φ)Ψ(r,Θ,φ) = EΨ(r,Θ,φ)

but the original atomic electron matter wave could not be represented in a spherical coordinate system. Using a spherical coordinate system to represent an atomic electron of the particle-in-box transformation is a deception. Furthermore, Schrodinger's wave equation is used to derive the equations of the atomic orbitals that is based on Schrodinger's interfering probability waves but an electron position probability can only represent a positive value or zero and cannot depict a negative value that is required in representing destructive wave interference used to derive the equations of the atomic orbitals which proves the derivation of the atomic orbital equations using Schrodinger's wave equation is physically invalid. Does this not diametrically prove that quantum mechanism is invalid?

11. Jun 22, 2017

@ivy The Q.M. commutator algebra with $[\vec{x},\vec{p_x}]=i \hbar$ is kind of a strange bird, and I believe Max Born was awarded the Nobel Prize for this and other contributions. In any case, as much as some of the results and some of the Q.M. fundamentals are certainly not intuitive, the Q.M. has seen tremendous success at many levels and is not likely to be unseated.

Last edited: Jun 22, 2017
12. Jun 22, 2017

### bhobba

Heisenberg came up with the commutator stuff but later it was extended by Dirac - it was part of his not so well known q number approach that was later incorporated into his transformation theory that united all the various versions under the one formalism. Max Born came up with the appropriately named Born Rule that enshrined the probability part of QM. Mathematically the transformation theory using Dirac delta functions and the like was on shaky ground according to the math known at the time, and Von-Neumann came up with an approach that was mathematically rigorous. However through the work of a number of great 20th century mathematicians Dirac's approach was put on firm ground using what is called Rigged Hilbert Spaces rather than the Hilbert space approach of Von-Neumann. Here is a summary of it history about that time:

Thanks
Bill

13. Jun 22, 2017

@bhobba Thank you. The paper is extremely good reading. Some of the mathematical details are beyond my present level, but it is interesting to read how the top physicists and mathematicians of that period resolved some very difficult mathematical puzzles in order to put Q.M. on solid footing.

14. Jun 22, 2017

### bhobba

That's easily fixed.

It will be a long road, but to start with get a hold of the following:
https://www.amazon.com/Theory-Distributions-Nontechnical-Introduction/dp/0521558905

What it contains should be known to all applied mathematicians and physicists - its that important. It will enrich your reading of texts that use this sort of stuff freely without saying exactly what it is. Griffiths for example doesn't say exactly what is going on with the Dirac Delta function - simply he gives his personal guarantee that coming to grips with it (again without saying exactly what is going on) will enrich their physics a lot.

Thanks
Bill

Last edited: Jun 22, 2017
15. Jun 22, 2017

My calculus and mathematics background is quite good, but when they start referring to things like isomorphisms, it is hard to follow all of the details. I expect that these were quite difficult mathematical puzzles though, or it wouldn't have taken these extremely brilliant people a couple of years to resolve them.

16. Jun 22, 2017

### bhobba

Isomorphism is easy - I am surprised you dont know it.

Its from set theory - it simply means the elements of one set can be put into 1-1 corrsponence with another.

Its used a lot in that fascinating mathematical area of infinity which also ties in with your calculus:
http://www.math.helsinki.fi/logic/sellc-2010/ws/guangzhou-boban.pdf

Put simply a set is infinite if it can be put into 1-1 correspondence with a proper subset. Integers, reals and rationals are all infinite. But now for something really interesting - the rationals are a different type of infinity to the reals - this is Cantors famous diagonal argument. There are all sorts of infinities.

Enough said - its over to you now - as a homework helper I am sure you can nut it all out for yourself. Here is a good textbook - and cheap to:
https://www.amazon.com/Schaums-Outline-Theory-Related-Topics/dp/0070381593

Thanks
Bill

17. Jun 22, 2017

### Staff: Mentor

It does not. You may be missing the distinction between probability amplitudes which are complex numbers so are neither positive nor negative, and probabilities which of course are restricted to zero and positive values.

Please start a new thread if you want to further explore your question - we're coming close to an attempted hijack of this thread.

18. Jun 23, 2017

### olgerm

Fourier transformation of wavefunction is a function(that has uncountably possible values), but bra is vector(not function. and has countable elements). Isn't it Contradiction?
On Wikipedia the Forier transformation is defined only for one argument functions. How to get bra and ket from for 4 argument wavefunction $\Psi(t;x;y;z)$? Should I use Fourier-Stieltjes transformation?
which convention should be used to get bra from wavefunction?

Is that position ket and momentum ket are related by the Fourier transform because De Broigle formula $\lambda=\frac{h}{p}$?
$F(f_x)(\lambda)$ is amplitude of wavelike function with wavelenght $\lambda$
and $\frac{h}{F(f_x)(\lambda)}$is probability amplitude of momentum p.

19. Jun 23, 2017

### bhobba

No - because in Rigged Hilbert Spaces you can have basis that are indexed by real numbers - not just countable indices.

Thanks
Bill

20. Jun 23, 2017

### bhobba

You are on the right track. Further hint - what are wave-packets in De-Broglies (outdated) ideas made of?

Like I said with this one I am not going to spell it out.

Thanks
Bill