Elitzur-Vaidman bombs
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- Опубликовано: 4 июл 2017
- MIT 8.04 Quantum Physics I, Spring 2016
View the complete course: ocw.mit.edu/8-04S16
Instructor: Barton Zwiebach
License: Creative Commons BY-NC-SA
More information at ocw.mit.edu/terms
More courses at ocw.mit.edu
"Bomb is good"
- MIT Professor
lmao
XDDDDD. Still though Dr. Zwiebach is an excellent teacher.
At 3:38, he was referring to the Mach-Zehnder Interferometer, for those who didn't get it; not 'mass center'. The Mach-Zehnder is basically the same set-up with the interferometer discussed in this video but without the bomb.
What he says does t make sense and can't be accurate...according to the set up the photon could could D1 even when the bomb is a dud so whybdoes he say otherwise??
@@leif1075 AFAIK, not, because of destructive interference in the direction of D1. I'm not sure, maybe some who knows this better will correct me.
@@IvanToman wow AFAIK stands for as far as I know right? I've never seen that acronym before! How exciting!
Is it t common like lol or brb or you just came up with it like a trend setting boss?? Off topic lol thanks for answering.
@@leif1075 Yes, correct :)
@@IvanToman oh did you just coin that yourself or its commonly used as far as YOU know? Because AFAIK, AFAIK is not a common acronym..wow did you see what i just did there?!? I just used it twice like a boss..I'm so lame! I'm sorry you had to read this..
This particular idea by elitzur and vaidman blew my mind. Thank you professor Barton zweibach and MIT for sharing this with us, really appreciate it!.
Totally weird and fun. His enthusiasm is infectious. :-)
Rightly said
His enthusiasm questioned me to pay attention to this silly and pointless discussion.
Seven wonders of Quantum world, thank MIT for elaboration!
The Elizur and Weidman: how to have your bomb and eat it too.
Really fascinating illustration!
Hats off to Prof. Barton
This is an alternative to russian roulette.
Hats off to Prof. Barton. Thanks a lot prof.
Dope lecture. Loved it!
I have a “tasty” feeling when people can write nicely (or even draw a straight line) on the blackboard.
I'm thrilled
Thanks a lot prof😍🙌
Mind blowing !
Bombing lecture !
This is insane. This contradicts the world around me.
An oldie but goodie.
Just Amazing..
Oh, this is so fun!
incredible
3:36 Mach-Zehnder interferometer
Also bomb instead of bum lmao
if you close your eyes, it sounds like Borat really made it for himself.
there is nothing like watching a German laughing about a bomb :) such fun!
He is not german. He is peruvian
I would like prof Zwieach to help peruvian students to learn from his life experience and try to follow his professional path
@@mariocortes1203 He is the type of Peruvian whose parents came to Peru in 1945 and 1946
Mikhail Mikhailov
Ouch
@@mariocortes1203 That's right (according to Wikipedia). But there is nothing on his personal life in the article? By accident? He is born in 1954. You know when the Germans (without their bombs and gas-chambers) left for South-America, don't you? But it's great he is giving something back to mankind! en.wikipedia.org/wiki/Barton_Zwiebach
maybe we could see this as if photon particles were riding a wave we can't see, like a buoy on a sea wave. the invisible (to us) wave would be affected by the environment and behave like a wave (interferences etc...). With this perspective, the wave would exist without the photon and would appear to us only when a photon starts riding it. So all the interactions of the wave with the environment (e.g. interferences, probability distribution...) are set before throwing a photon in, and do not depend on the photon in itself.
Robert Spekkens has a paper on this that gives the most intuitive explanation without waves at all. There is an inconsistency in how this is presented, the first beam splitter is treated like a single input logic gate (a single photon) with two outputs (either the upper or lower path) and when they're recombined at the second, it is treated as a two-input logic gate with two possible outputs. This inconsistency goes away if you recognize that the first logic gate is also a two-input logic gate just where one of the inputs is always 1 (a photon) and the other is always 0 (no photon). All you then have to do at that point is to presume that the 0 state can be carried by a beable and even interact with a measuring device despite showing up on that device as a 0, and then you can explain the whole thing in terms of local beables only without even positing any waves at all.
smells like luminiferous ether, which was disproven a long time ago
Laughed hard at the subtitles saying "mass center interferometer"
Classical distribution of energy known as equipartition is, suppose there are 6 modes of dice and energy to be distributed is 21 units. Then each mode have equal amount 3.5 unit of energy. Divide it by mode or frquency, energy per frequency is 7/2, 7/4, 7/6, 7/8, 7/10, 7/12. Thus energy per frequency is decreasing as shown in black body radiation curve.
Now same amount of energy distributed to same modes as per quantum energy distribution given by Planck's law. According to quantum energy distribution, energy is not distributed equally to modes but in linear multiple of fundamental or minimum energy unit to modes. Thus higher modes have more energy and for above 6 dice modes they are like, 1, 2, 3, 4, 5, 6 where fundamental unit is 1. Dividing these energy for modes by mode number or frequency they are all same, 1 unit. But this is not evident in black body curve as flat line.
I think this experiment need little work to look better. That is, first beam splitter is polished such that it has transmission coefficient of 1/3 and reflection coefficient of 2/3. So probability of photon taking upper path is double of lower path.
Now slightly change in intensity of source also, it has such intensity that emit two photons at time and triggered by switch which let one fire only on triggering.
Now if photons take both paths they interfere and detected by A only. And if photons take upper path then they may not interfere and could be detected at B also.
Now calculating probability, chances of photon detected by B is 5/18 and by A is 13/18. And chances of both photons take upper path is 4/9 and chances of that detector B detects live bomb without exploding it is 1/9 ~ 11%, accuracy is about 4/9 ~ 44%.
And from his model, probability of detecting live bomb is 12.5%, not 25% and accuracy is 33%. In numbers, if total 200 times test is conducted, only 25 times it correctly predict live bomb without exploding it and 50 times it predict falsely, 25 of which are dead.
If someone needs 10 fire-crackers and they are constrained by a budget of buying only 40, you're modification is reducing their chance of success from 25% to 12.5%.
@@vishalmishra3046 no, modification bring chance of success 11% from 12.5% but improve accuracy to 44% from 33%.
Why he doesnt say that what if photon takes up part even though bomb is not functioning? Like why are they assuming it always go both durrectiins if bomb not working but can go only up if working? What am i missing?
It does not matter what is the accuracy or chances of success. The fact this can happen at all is what’s interesting.
I'm not sure if I can handle this knowledge
I can't sleep after learning this
I'm amazed it looks like we would be in a simulation
what is the superposition interaction is enough to trigger the bomb?
As written in every quantum mechanics video
A glitch in the matrix
If the photon gets split into two beams ( no matter a probability construct but still it is a split and it also takes all possible continuous values alpha and beta ), then it not "quantized" in the first place.... So why quantum mechanics ?? I mean what is really quantized here
The photon doesn't split. The wave function is "split" so that if you send one photon at a time through the splitter, the photon will take the upper path half the time and the lower path half of the time.
just because you say 'no matter a probability construct..' doesn't mean it stops mattering. There's a difference between a photon splitting and it's probability function splitting, and you have to try to understand that.
In case the bomb is good and does not explode, what happens inside the photon detector of the bomb when the deflected photon arrives?
A photon detector is simply an absorber. It removes one photon's worth of energy from the free field.
@@schmetterling4477 if the photon always detector makes the same thing, why sometimes the bomb explodes and sometimes not?
@@tommasoboccellari270 The bomb always explodes when its detector removes a photon's worth of energy from the field. That is the definition of "live bomb" used in this nonsense.
@@schmetterling4477 then my first question remains open.
@@tommasoboccellari270 You are probably working with the wrong idea of what a photon is. A photon is the absorbed energy. Energy that has not been absorbed and that remains in the field is not a photon.
Why the probability of getting a photon at D0 or D1 is 1/4 (for the case bomb is good and does not explode ). I think it should be 1/2. Isn't?
(Maybe too late to answer but:)
It’s 1/4 because there’s 1/2 probability the photon goes up and 1/2 probability that it goes to the bomb and detonates it. If it goes up, it goes towards a second beam splitter which further halves the previous 1/2 probability into 1/4 for D0 and 1/4 for D1
Emperor's new Mind by Penrose explains how Quantum Mechanics explains how you can know the effect before the cause.
Sun is emitting light and hot bodies glow light. But spectrum of sun is continuous which can be observed with crystal or cd. Different colors are in thick bands and not thin line. So how energy is quantized in hot solid bodies. By quantization means different colors are separated by dark regions which is not observed. Temperature of body can be calculated by light emitting from body of different colors and they do integration which is for continuous function.
A circle is locus of points which are equidistant from a point called as focus or centre. Joining of locus is curve, where points are discrete but curve or function is continuous. So, do circles are continuous or discrete, continuous means curve is smooth. Similarly calculus is done with discretization of interval but on continuous function.
Yes, quantum mechanics shows that wave seems as continuous but made of discrete small particles, operated as smooth. Similarly interaction of energy and matter is participated by discrete particles known as atoms but there is no restriction on amount of matter and energy, continuous.
what is the meaning of- sign in matrix of beam splitter
Should be a phase change by 180 degrees. I forgot where it comes from, but it exists in the classical electromagnetic wave treatment as well, if I remember correctly.
Can someone explain to me why do we square (multiply by the complex conjugate) the probability amplitudes in order to find the probability of the outcome?
That has been explained in the earlier lectures. Basically, the state is represented as a vector with cos and sin components along the possible outcomes which are represented as x and y axis respectively and sin^2 + cos^2 = 1, so that square is considered as probabilities of possible outcomes.
This is called the Born rule, it is one of the Axioms of QM. But to motivate: if you want probabilities the number have to be 0 or larger than zero (which complex numbers do not fulfill), which the absolut value fulfills. They also should add up to 1 which they do, as Compilations Mania explained below. (generally this is the generalized pythagorean theorem in hilbert space)
Gleason's theorem is the mathematical proof as to why you have to square it. Squaring it was originally assumed as just a law, its own axiom, alongside the Schrodinger equation, but Gleason's theorem shows that there is no other way it could be if you accept the Schrodinger equation as an axiom. The process of squaring the wave function to get the probability distribution is called the Born rule.
Does this have any relation with the entanglement?
No, this is a phenomena related to superposition.
Here's a thought...
Adding the concept of superposition, maybe there's a difference between:
1. A single photon in a single path then the detector works 100% of the time.
2. Bomb detector interaction with a "split photon", i.e. unlike case (1) the particle that hits the detector is not a "full photon" but a part of the superposition.
Maybe what is missing here is the understanding of: - the probability of an interaction of "half a photon" with the detector.
We know that the effect of a "full photon" on the detector is 100% deterministic. Boom.
But let's assume that the probability of the detector (observer) is either:
50% detonate.
25% not detonate, and do nothing to the half photons, they reach D0
25% not detonate, and phase shift the 2 half photons so they reach D1.
This means that we still know nothing about the bomb, is it alive or dead.
The only "hole" in this hypothesis is the question: in the dud, is the detector behaving differently when the bomb is a dud, or is it the very same detector?
If the former then hypothesis still stands.
Sorry to say so but your concepts is missing the collapse of the wavefunction. The particle that hits the detector is always what you call "full photon". How we now that well in 3.1 of these courses the photoelectric effect will be explained that should make it clear (That is what Einstein received the nobel price not for SRT or ART) why the concept "half photons" is not going to work just by choosing the detector right.
The only way i see to keep your concept alive would be to say these "half photons" are still entangled with each other and one partner get energy to step back to a "full photon" and the other "half photon" is giving its energy (then energy is conserved) and we still measure a "full photon" at the detector as to be done. But that would mean to argue with an effect from quantum mechanic against the quantum mechanic original explanation of the seen effect. (still you could use phase shifters and polarized filters to rule the concept of "half photons" out, but it would hold within the described experiment).
But do not worry that strangeness/weirdness is exactly the reason for the so often used famous words: "If you think you understand quantum mechanics then you do not understand quantum mechanics." (which by the way is half true and half not)
@@Techmagus76 There are no "half photons" in this universe, my friend. :-)
prof. adams did something like this on his first lecture for 8.04 2013
Waves undergo interference when they meet.Why photon needs 2nd splitter to interfere.
if nothing blocks both paths of the photon , it go through both paths and interfere with it self in the second beam spliter witch is made in a way to only allow constructive interfernce in the direction of D0 , however if somthing block one of the photons paths , the wavefunction collapses and the photons travels as a particle , when it hits the second beam spliter , it has 50/50 of going to D1/D0
Single photons don't interfere. Think about it for a little bit.
@@lepidoptera9337 You are falsely assuming there is a single photon here and not two beables. Beam splitters have two inputs and two outputs. How can you convert 1 bit of information into 2? Quantum logic gates are unitary, inputs and outputs always match in terms of number. If you recognize that there are two beables upon the input, then there are always two beables midway, and two beables upon the output, and there is no mystery.
@@amihartz Where do you see quanta during unitary evolution???? That the same photons will come out that you sent in is not even remotely true. Doppler effect will change the energy of the quanta. Reflection will change the momenta. High energy interactions change the total number of quanta. The world IS NOT linear quantum optics. Not that photon number is conserved in quantum optics, either, see thermal states and spontaneous parametric down conversion. You need to start learning real physics and stop messing around with the trivial corner cases.
@@lepidoptera9337 You just spouted off a lot of random things with no relevance to what I stated. It's like if I made a comment about US history and so you start rattling off information about cake recipes. There is zero relevance here other than you just listing unrelated facts to pretend you're knowledgeable and making a point.
Phenomena of interference by light, producing dark fringes is result of how we disturb or measure light.
If two slits are not cut abruptly but in gradual manner like gaussian curve, there will be no interference. Distribution of gaussian function represented in harmonics is again gaussian, so there is no dark fringe.
Same can be said about interference by division of amplitude. As splitter is uniform and not gradual so it happens. Also fringes are not local.
So light has no property of wave and not described as particle, but disturbance of particles. So light cannot travel without matter, leaving side whether it has electric or magnetic field.
"So light cannot travel without matter" well then how does it travel through vacuum
After all matter in vacuum is beyond negligible
@@saifahmad141 it can be proved in many ways that wave cant travel without medium, because it is disturbance of medium and thus speed of wave depends upon two factors, nature of source which defines wavelength and nature of medium which defines frequency.
In case of light, they thought that Faraday's law and Ampere's corrected law proves that light doesnt need medium, while they only states that change in field in time causes another field in space.
@@neillibertine3044 yes but again medium dependent waves are simply mechanical but when it comes to light is an em wave so it changes the story light is just variations in electric and magnetic field perp. To the dirn of actual wave correct me If I'm wrong which I definitely maybe because I didn't understand you sorry but pls could you explain what you are trying to say
@@saifahmad141 okay try to put in simple words. Voltage cause current, and current indicates voltage. But without resistance, relation between voltage and current is not establish, and that resistance is medium dependent and inverse of resistance is conductance, which depends upon speed of charge. If resistance becomes zero then v¹ and v² produce same amount of current, there is no causal relationship between voltage and current. Now replace it with electric and magnetic field.
This is supplement, there is no such thing as EM wave, all are mechanical in nature. Surprised, okay what is voltage or electric field, amount of charge and magnetic field is amount of current. These amount of whatever transported from one point to another is mechanical overall, but the force behind the motion is electromagnetic.
Yep image u send the bomb and it not exploded. How happy your teacher
5:45 why that d0 probability is 1 instead of 1/4?
D0 is the only option when no path is known for the photon and when the bomb is bad, no path can be known. He doesn't explain this well in this video - I assume that is described before this video begins.
@@DaleSackrider he didn't even use 'superposition' word
Hi,
This is because if the detector is not working then the arrangement is similar to the bomb being not there and it becomes the same as described in the just previous lecture L2.4 around 3 minutes.
if nothing blocks both paths of the photon , it go through both paths and interfere with it self in the second beam spliter witch is made in a way to only allow constructive interfernce in the direction of D0 , however if somthing block one of the photons paths , the wavefunction collapses and the photons travels as a particle , when it hits the second beam spliter , it has 50/50 of going to D1/D0
I am sorry to say, that your lab will explode half of the time xD
Where is the FULL LECTURE PLEASE with setup.before this?
Here is the playlist for the course: ruclips.net/p/PLUl4u3cNGP60cspQn3N9dYRPiyVWDd80G. The course materials can be found on MIT OpenCourseWare at: ocw.mit.edu/8-04S16. Best wishes on your studies!
@@mitocw Thank you I was trying to find which part comes right before this one so I fully understand the set up
The playlist matches the order of the course... which means the video before this one is: ruclips.net/video/37-GdFJGSXs/видео.html
@@mitocw WAIT his conclusion Makes NO SENSE..the fact that a photon hits D1 does not necessarily the bomb is working..itnculd just mean the photon took the lower path and passed through the dud bomb..so why does he say this?
I like puzzles
I did not understand, Why would the probability be zero for detector D1 when the bomb is defective?
if nothing blocks both paths of the photon , it go through both paths and interfere with it self in the second beam spliter witch is made in a way to only allow constructive interfernce in the direction of D0 , however if somthing block one of the photons paths , the wavefunction collapses and the photons travels as a particle , when it hits the second beam spliter , it has 50/50 of going to D1/D0
@@enjoyhealthnow1 How did you come up with this bullshit? Photons don't travel, they are not particles and wave functions never collapse. This isn't even a quantum experiment. :-)
Because if bomb is defective the photon simply passes like before and hence the situation gets reduced to the earlier case where we and professor already solved the matrices and computed probabilities.see basically the thing is we cannot be sure that the bomb explodes without it actually exploding but by using this method we can have a probabilistic view on it which to be precise there is 1/4 chance of bomb being good but not exploding despite . And hence in the end we can only probalistic view whether it could explode but didn't and is a good one but could be used for later . In the end he said there are ways to improve this probability and improve accuracy
@@lepidoptera9337 so am I right to think when there is no block the photon is actually chosen only one path but since we can't be sure we have probabilistic views about both paths or does the photon actually take both paths like a wave ?
@@saifahmad141 Photons are energy. Energy can only be exchanged once, i.e. a photon can only be measured once. Something that has a path requires at least two measurements. Photons can not have a path and nobody has ever measured the path of a photon.
9:15- BUT, knowing the bomb is good has 1/2 of the chance MIT would be blown up. is the game worth it? 😁
Which physics course is this ?
As the video description states, this course is MIT 8.04 Quantum Physics I, Spring 2016. View the complete course: ocw.mit.edu/8-04S16. Best wishes on your studies!
@@mitocw can we get all other courses on Quantum Physics?
Sounds more like Schrodinger's bomb.
I don't understand why did you consider the bomb would work, if it went through detector 1??? Because there was no photons that went through the bomb??? So, my question is why did you conclude that the bomb works, when no photos passed through the photon detector in the bomb, and just went straight towards the detector 1??
Thank you, I hope I get an answer.
There's a 0% chance that the photons reach either detector if the bomb doesn't work, therefore if a photon does reach a either D0 or D1 then it means it does work.
@@jamesleliveld9957 Thank you, but he said that if it goes to D0, then it does not work?
I was wondering the same exact thing, hopefully it is covered in a later lecture
@DR @@funeralhomeengineer7691 What he said was that photons could be detected in D0 regardless, but if a photon is detected in D1 then the bomb must be good because if the bomb is defective no photons could be detected in D1 as he explained in the previous video.
Okay so the real solution here is build new bombs.
👁️👁️👁️
*The probability calculations are inaccurate* due to unreasonable assumptions. e.g. The probability that a random bomb works or is defective cannot be assumed to be equal.
For example - if you have 100 bombs and for the purpose of testing to determine probability, you're open to exploding a random sample of 10 of them (and 7 actually explode), it is likely that P(bomb is defective) = 30%. This has a material impact on the probability outcomes of the experiment.
It does not matter, the reasoning is the same if 0% of the bombs are good, or if 100% of the bombs are good. If you detect a photon at D1 it means that specific bomb you tested is a "good" bomb
@@astrolillo That's a cheating. A detection at D1 means that bomb is NOT BAD. 100% that the detector in this bomb is NOT BAD. All the other conlusions that make "THE BOMB IS GOOD" are just assumtions that somehow considered to be 100% true. They are "this is a bomb" , "this kind of bombs never ever fail", "the only case this bombs fail is the detector" ... and so on. This idea of measurment without interaction have presumtion that there's absolute confidence about the object of measurement. Ok, I have a result: "it's heavy"! WHAT is heavy? Well, I'm 95% sure it's 10kg. Ok, so you are 95% sure it's a kettlebell. But there's 5% chance it's a rock.
@@shoutitallloud the bomb is described as either good and it will explode if photon passes through it, consuming photon, or bad, when it does nothing and photon passes through unchanged. There is nothing else to this.
@@MikeKrasnenkov This is “gedunken experiment”. It’s just your assumptions.
@@shoutitallloud This is not my experiment and thus not my assumptions. If you disagree with them then you are not talking about the same thing.
DO NOT TRY THIS AT HOME
- FROM EARTH 251
In the segment of the video where we supposed bomb is defective, isn't the probability of photons reaching detector D0 is 1/2 and photons reaching detector D1 also 1/2...? This is bcoz if bomb is defective then photon can pass through it easily and give half half chances of appearing at the detectors.. beam splitter will allow them to go to any one of the detectors...
Kushagra Uniyal If both the pathways are clear ( case of defective bomb) all the photons will end up only in DO. Review the previous video. Photon goes to DO and D1 only when there is a block I the lower pathway :)
if nothing blocks both paths of the photon , it go through both paths and interfere with it self in the second beam spliter witch is made in a way to only allow constructive interfernce in the direction of D0 , however if somthing block one of the photons paths , the wavefunction collapses and the photons travels as a particle , when it hits the second beam spliter , it has 50/50 of going to D1/D0
Voltage cause current, and current indicates voltage. But without limited resistance or conductance, relation between voltage and current is not establish, and that resistance is medium dependent and inverse of resistance is conductance, which depends upon speed of charge. If resistance becomes zero then v¹ and v² produce same amount of current, there is no causal relationship between voltage and current. Now replace it with electric and magnetic field.
This is supplement, there is no such thing as EM wave, all are mechanical in nature. Surprised, okay what is voltage or electric field, amount of charge and magnetic field is instantaneous amount of charge transported. These amount of whatever transported from one point to another is mechanical overall, but the force behind the motion is electromagnetic. Thus it is untrue that EM wave is independent of medium, it depends upon amount of some quality of medium like charge to be disturbed. Mechanical wave like sound is local transportation of matter or density and EM wave is local transportation of charge or charge density.
Hi,
Thank you for the content firstly. secondly, in the first minute, you mention that the concept of a bomb was created in Tal Aviv, honestly, this example needs to be reviewed and changed since in this university those concepts of bombs were created to kill and wipe out Palestinians in Gaza and cleanse them from the land, as a scientist, this example is not suitable and not appropriate.
Thank you.
Do test but disconnect bomb from detonator.
If you can do that you don't need this test
Sus
Thank you, German Bill Gates!
Of course, if you are an even halfway intelligent student, then you will have noticed that the bombs are perfectly classical objects. They explode when they remove energy from the field and they don't explode when they don't. They are physically identical to a piece of black cardboard that is either there or that isn't. Now you can ask yourself, how does a piece of black cardboard introduce Planck's constant into this experiment? And the answer is that it doesn't, which means that there are zero quantum effects in here.
OK, admittedly, there was probably not one student in that lecture hall who could think that well. :-)
Sorry but I didn't understand what you are trying to say if I'm right by your thought process its impossible to determine whether the bomb will explode or not but professor is showing how we can have a probabilistic view of now much chance does the bomb have of exploding without exploding!
In other words we are trying to formulate probabilities of something happening without it actually happening whereas classical mechanics can only guarantee the bomb exploding only after it explodes! I got confused 😕
@@saifahmad141 The professor is repeating a stupid analysis by a stupid paper. The bombs behave exactly like a piece of cardboard. So this experiment is nothing else than you holding a black paper into a two way interferometer. The only quantum effect that is happening here is at the detector and the detector basically just shows you the classical expectation value for the intensities that you get from Maxwell's equations. That's it. There is no quantum mechanics anywhere here.
@@saifahmad141 Look, the smart physicist solves the bomb problem this way:
1) Tape black tape over the bomb's detector.
2) Turn the light on.
3) Carry the bomb to the ordinance disposal site.
4) Rip the tape off using a long string.
5) See if the bomb explodes.
The result is a 100% test without any damage.
Only a stupid physicist would build an imperfect interferometer around bombs. OK? :-)
@@lepidoptera9337 after point 5, youre still not aware if the bomb is good without it not exploding right? Can you elaborate on what you mean when you say there arent quantum effects on this? Yes the bomb is a classical object, but the interferometer setup gives us a small probability of detecting the good bomb without it exploding right?
With such experiments, Quantum mechanics teaches us how to lead a happy and fulfilling life by inspiring us with magic (un-intuitive) ways in which reality works.
Here is the *Mantra* - Just work on improving the probability of getting what you want in life without expecting to get anything significant.
Life will give you a lot more than what you ever expected in your life - happier surprises. (Regarding expectation, I'm assuming that you're not a quantum physicist else life may not be able to beat your higher and more accurate expectations at least not by a significant margin, esp. when averaged over your entire life !! )
Classic Israeli 😂
This is seriously a complete waste of time… in essence I didn’t learn anything from this fantasy bomb 💣.
This is seriously a complete waste of time… in essence I didn’t learn anything from this fantasy bomb 💣.