The h Bar

ayc

02:35

@JohnRennie I got stuck in Normalization topic in Griffiths intro to Q.Mech .The math is new to me and it also has a lot of technical details.What should I do now?....Would you recommend skipping the topic or learning the math first?

vzn

02:48

@Secret very big news, cited the quanta article on it earlier. what do you mean "agrees with standard formulation"? the authors dont phrase it exactly in this way, but its clearly an experimental proof QM is incomplete (but also in line with other recent experiments). its a subquantum effect/ hidden variable imagined/ foreseen by Bohm. quantamagazine.org/…

Secret

> Our findings, which agree with theoretical predictions essentially without adjustable parameters, support the modern quantum trajectory theory

Unless modern quantum trajectory theory is the new thing in town

vzn

@Secret the "modern quantum trajectory theory" is not classic QM afaik...

Secret

Ah I see...

vzn

[5][6][7][8][9] seem to be from quantum optics, not familiar with them...

Secret

neither do I, but I think after reading the article, I will get some idea. But first I need to load this pile of PhD data onto the cluster...

vzn

02:53

curiously those refs are 1992-1999.

Secret

03:06

In other news:

in Mathematics, 4 mins ago, by Secret

Adam's message is still plain English even for those who don't understand them, but for me, 3 years of cataloguing conversations I had with many people seemed to suggest they actually read a strange language when they do not somehow "click" with my thoughts process. In other words, they literally cannot understand my language, not just not making sense

I literally speak alien unless you somehow have something in common with me

In fact, for people who are not even anti aligned with me, they will not comprehend the above message and probably found it written in alien gibberish

in Mathematics, Jun 3 at 10:34, by Secret

I am inaccessibly insane

Put it simply, you cannot understand me unless you are also an inaccessible cardinal to begin with

knowyourmeme.com/memes/on-the-internet-nobody-knows-youre-a-dog

May 28 at 10:17, by Ajay Mishra

are you one of the bots?

The answer is: I am not a bot, but I am an inaccessible cardinal

vzn

03:32

Carmichael [5] An Open Systems Approach to Quantum Optics link.springer.com/book/10.1007/978-3-540-47620-7

Secret

03:58

@ vzn: Quantum joke: I cannot communicate with anyone who does not form interference patterns with me

There are 10 ways to form an interference pattern

John Rennie

04:19

@ayc which chapter/section/page?

I can have a look in my copy of Griffiths. I'd be surprised if it was using any very complicated maths.

Abhas Kumar Sinha

@Secret ugh?

John Rennie

04:48

@ayc do you mean section 1.4 (that's in my second edition of Griffiths)? If so it's quite short so we can go through it here if you want.

John Rennie

06:23

That's an odd way to answer a question.

2

ayc

06:42

@JohnRennie Yes the section 1.4....shall we discuss it now?

John Rennie

@ayc yes, I'm free now. Where is the first point you get stuck?

ayc

@JohnRennie Hes says that a glance at equation 1.1(schordinger's equation) reveals that if psi is a solution ,so too is A(psi) where A is any complex constant.What does that mean?

John Rennie

@ayc It's simpler than you think. Suppose you start with some solution to the Schrodinger equation $H\psi = i\hbar d\psi/dt$. Now define a new function $\phi = A\psi$ where $A$ is a constant - $A$ can be a complex number. OK so far?

ayc

@JohnRennie ok

John Rennie

Now we put our new function $\phi$ into Schrodinger's equation to see if it is a solution. We get $H\phi = i\hbar d\phi/dt$ which is $H(A\psi) = i\hbar d(A\psi)/dt$.

But because $A$ is a constant this can be rewritten as:

$$ A H\psi = A i\hbar \frac{d\psi}{dt} $$

Are you happy with this step? I can go into more detail if necessary.

ayc

06:50

@JohnRennie ok for now,but whats the purpose of doing this?

John Rennie

Because now we can divide both sides by $A$ to get:

$$ H\psi = i\hbar \frac{d\psi}{dt} $$

And we know this is true because we started with the assumption that $\psi$ is a solution of the SE.

So that means our new function $\phi = A\psi$ is also a solution of the SE.

What we find is that if you have a solution of the SE, $\psi$, then you can define a new function by multiplying $\psi$ by any constant and the new function is also a solution of the SE.

@ayc OK so far?

ayc

@JohnRennie ok for now

John Rennie

But if this is true how can we ever find a solution of the SE that makes physical sense, because we have an infinite number of possible solutions i.e. $\psi$ multiplied by an infinite number of possible constants $A$. We need some method to find out what value of $A$ corresponds to real life.

That is what Griffiths is doing in this section. He is saying that we need some extra condition to find out what $A$ should be, and that extra condition is normalisation.

ayc

@JohnRennie I hope you have the book with you right now.The equation 1.20:could the integral on L.H.S give values greater than 1?

@JohnRennie Griffiths says yes in the next paragraph

John Rennie

@ayc We have already said that if $\psi$ is a solution then so is $\phi = A\psi$. Yes?

ayc

07:01

@JohnRennie yes

John Rennie

So let's evaluate that integral for $\phi$ ...

$$ \int |\phi^2| dx = \int |(A\psi)^2|dx = \int A^2|\psi^2|dx = A^2\int |\psi^2| dx $$

And since $\int |\psi^2|dx = 1$ that means our integral for $\phi$ can give any answer we want just by choosing a suitable value for $A$. It could be greater than 1 or less than 1.

ayc

@JohnRennie What would each choice of A represent?..I mean ,whats the difference between choosing A as 1/2 or 1/4?

John Rennie

This is the key point.

From a mathematical point of view $A$ can have any value. But we want our equation to describe something real e.g. an electron. And for this to be the case we have to have a specific value for $A$.

Section 1.4 is explaining how we find out what the value of $A$ has to be.

ayc

@JohnRennie and this process of finding A is called normalisation?

John Rennie

@ayc exactly, yes.

The point is that $|\psi^2|$ is a probability density.

That means $|\psi^2|dx$ gives the probability of find the particle in the small distance $dx$ i.e. at a position between $x$ and $x+dx$.

Does this make sense so far?

ayc

07:10

@JohnRennie yes

John Rennie

So if we integrate $|\psi^2|dx$, i.e. sum up all the probabilities, the result must be one because the probability of finding the particle somewhere between $-\infty$ and $+\infty$ has to be one.

ayc

@JohnRennie yes!

John Rennie

So suppose we find some solution $\psi$ to the SE by whatever method - maybe we just guess it. Then we know that all fuctions $\phi = A\psi$ are solutions.

But we must have $\int |\phi^2|dx = 1$, and that means $\int |(A\psi^2)|dx = 1$.

So we solve this equation to calculate $A$ and that is the value that our real physical wavefunction has to have. There is only one possible value for $|A|$.

ayc

@JohnRennie Could you explain the same thing in another situation.We are now trying to locate an electron around nucleus.

John Rennie

That's a little more complicated because it's three dimensional, but yes I can go through that if you want.

ayc

07:18

@JohnRennie I would love to.But, are there any prerequisites?

John Rennie

@ayc we'll have to use polar coordinates so you need to be happy using these instead of the Cartesian coordinates x,y,z. Apart from that there aren't any prerequisites.

ayc

@JohnRennie Yes,I think I have basic knowledge in polar coordinates.Let's continue

John Rennie

OK. For the 1s orbital in a hydrogen atom it turns out that the solution is $\psi_{1s} = A e^{-r/a_0}$ where $A$ is a constant and $a_0$ is the Bohr radius.

I won't go into how we show this. Basically just substitute it into the SE for a hydrogen atom and you can show it is a solution.

OK so far?

ayc

@JohnRennie ok

John Rennie

Now in 3D the probability of finding the electron in an infinitesimal volume $dV$ is $P = |\psi^2|dV$

So it's like the 1D example we started with, but in 3D the particle can be anywhere in a tiny volume $dV$ rather than a tiny 1D distance $dx$.

So our normalisation equation becomes:

$$ \int |\psi^2|dV = 1 $$

And for our $1s$ orbital $\psi = Ae^{-r/a_0}$ this gives:

$$ A^2 \int e^{-2r/a_0} dV = 1 $$

OK so far?

ayc

07:27

@JohnRennie ok

John Rennie

The only tricky bit is figuring out what the volume element $dV$ has to be. Can you guess? Or shall I go through it?

ayc

@JohnRennie idk...(4pir^2)(dr)?

John Rennie

BOOM! :-)

Exactly! :-)

It's the volume of the spherical shell from $r$ to $r+dr$.

So we just substitute $dV = 4 \pi r^2 dr$ into our equation to get:

$$ 4\pi A^2 \int_0^\infty e^{-2r/a_0} r^2 dr = 1 $$

And we solve that to get the value of $A$.

It's not a particularly simple integral, though not especially hard either. I guess you'd use integration by parts though personally I'd just feed it into Mathematica.

ayc

@JohnRennie If I were to locate an electron in an atom what are the general steps that I have to follow?

John Rennie

@ayc can you clarify what you mean by locate an electron?

The electrons in atoms do not have a location. They are spread out over the whole atom like a fuzzy cloud.

ayc

07:36

@JohnRennie To determine the quantum numbers(n,l,m,s).If we have these numbers we can know how the electron is distributed right!..Yes: "locate" was not the right word .My bad!

John Rennie

When we solve the Schrodinger equation for a hydrogen atom we get a series of solutions of increasing energy i.e. the 1s, 2s, 2p, etc, etc

The actual form of these equations is somewhat tedious to calculate. If you're interested the first few solutions are given on this web site.

ayc

@JohnRennie Do we find out which orbital the electron is in using SE?

John Rennie

The SE tells you what possible orbitals the electron could have, but which orbital the electron is actually in has to be determined by experiment.

That is, if you have a collection of hydrogen atoms then different atoms may have their electrons in different orbitals.

The different orbitals have different energies. The 1s orbital has the lowest energy, the 2s and 2p the next lowest, then the 3s/3p/3d, and so on.

ayc

07:52

@JohnRennie SO to know where abouts of an electron I do an experiment and I found out the energy the electron has and then match it with energy of an orbital that I already know and then extract quantum numbers(n,l,m) and then learn how it is distributed around the nucleus?..

John Rennie

Yes, exactly! :-)

ayc

@JohnRennie My entire life was a lie.I had wrong ideas on SE.How stupid was I!

John Rennie

@ayc :-) That's a bit harsh!

ayc

@JohnRennie I have few more questions...I'll be back after lunch..Will you be free?

John Rennie

Quantum mechanics is very unlike classical mechanics and it seems really weird at first. What you'll find is the more you study it the more natural it seems.

2

@ayc yes, I'm around for several hours more today.

ayc

07:57

@JohnRennie Cool!

Slereah

morning

eranreches

08:13

hello

Captain Bohemian

10:40

today is a holiday? I didn't know that until I went out finding the postiffice and a lot of stores are not open.

fortunately that eatery I planned to go to was open, and I heard the TV news there talking about Dragon's Festival. Maybe toady is Dragon's Festival holiday. The weather has become so hot these days, being the characteristic of Dragon's Festival.

Dragon's Festival marks the start of summer.

Slereah

I mean every day is a holiday to some

June 7

June 7 is the 158th day of the year (159th in leap years) in the Gregorian calendar. 207 days remain until the end of the year. == Events == 421 – Emperor Theodosius II marries Aelia Eudocia at Constantinople (Byzantine Empire). 879 – Pope John VIII recognizes the Duchy of Croatia under Duke Branimir as an independent state. 1002 – Henry II, a cousin of Emperor Otto III, is elected and crowned King of Germany. 1099 – First Crusade: The Siege of Jerusalem begins. 1420 – Troops of the Republic of Venice capture Udine, ending the independence of the Patria del Friuli. 1494 – Spain and Portugal sign...

many such holidays

Captain Bohemian

11:02

long ago I have noticed a lot of days on the calendar are noted with some historic events, but most of those days are not marked red as holidays.

Dragon's Festival is May 5 on lunar calendar, so it may escape my attention.

Akash. B

15:40

@JohnRennie are you there?

John Rennie

@Akash.B hi

Akash. B

Hi i have a question

@JohnRennie have you seen a ride in a park called see saw?

It caused me to think about the gallelio 's experiment

Feather drop experiment

John Rennie

@Akash.B in the UK a seesaw is what the rest of the world calls a teeter totter (I think).

Have you go a link to show me what the seesaw ride is like?

Akash. B

Gallelio told that the gravity pulls everything uniformly

Okat 1 min

JMac

@JohnRennie FYI those are synonymous in Canada in my experience, both terms are understood

Akash. B

15:45

m.indiamart.com/proddetail/playground-seesaw-14382922573.html

@JohnRennie

So my question is

John Rennie

@Akash.B ah, yes, OK that is what we call a seesaw in the UK.

Akash. B

If gravity pulls everything uniformly, irrespective of its weight then it should remain in air, right?

@JohnRennie

John Rennie

@Akash.B what Galileo said was that the gravitational acceleration is the same for everything i.e. if you drop two objects they fall at the same rate.

This does not mean the gravitational force is the same for everything.

Akash. B

well i misunderstood

John Rennie

The gravitational force is given by $F = mg$ where $m$ is the object's mass and $g$ is the gravitational acceleration of 9.81 m/s^2.

The reason all objects fall at the same rate is that Newton's second law tells us that the force on an object and its acceleration are linked by $F = ma$.

Akash. B

15:51

Ok thank you

John Rennie

But from the gravitational force we have $F = mg$, and if we set the two forces equal we get $ma = mg$ and dividing both sides by $m$ gives $a = g$.

So the acceleration of all falling objects is equal to $g$ regardless of their mass.

Akash. B

Oh! good

Novalium Company

17:29

I was thinking about creating a neuroscience forum, but I realized that in 2019, with all the social media, reddit and so on, it really isn't worth it.

vzn

@NovaliumCompany cyberspace is "full of eyeballs... spread thin... attn is the new scarcity" :(

vzn

18:11

...

in theory salon, 21 secs ago, by vzn

‘I Am Mother’: Netflix’s Stunning and Ambitious Killer-Robots Movie / dailybeast

John P

18:53

Hi, would someone please help me with solving this problem? physics.stackexchange.com/questions/484784/…

bolbteppa

21:40

@Slereah t'Hooft p. 34 seems to say that given an arbitrary Lagrangian including ghosts the way to know whether it came from a Gauge-invariant Lagrangian is whether it possesses BRST symmetry, i.e. a non-BRST invariant ghost Lagrangian must not come from a gauge-invariant Lagrangian and so is potentially meaningless

and that in the early days they thought they simply had to check the connection between a ghost-form of an action and a gauge-invariant form of an action directly to know whether there was a link until BRST symmetry was discovered as a way to sidestep this direct checking, thus seems like the real reason for caring about BRST

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