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Basic question on quantum mechanics and choice

  1. May 25, 2013 #1
    Suppose we are given two boxes and we are told that there is a ball in either one of them.
    We are free to pickup any one of the boxes - but only one.

    Then we are allowed to measure or check what is in the picked box and the other one.
    Normally we would have 50% chance of finding the ball in the box we picked.

    Is it possible to build the system in such a way that the ball always turns up in the box that was not picked up.

    i.e if we pickup a box we will never find the ball inside it but if we check the other box the ball would be in it.

    Even if we repeat the experiment, the ball should not turn up in the box that was picked but always turn up in the other one.

    I know this would be impossible using large objects like a ball, but if we replcae the ball with an electron which can be in a qauntum state, is this possible?
  2. jcsd
  3. May 25, 2013 #2


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    No - that's not possible - classically or in QM. It would require clairvoyance. There is a problem with predicting the future - its in the future which, despite the zillions that could be made from the stock market etc if you could do it, no one has quite figured how to determine. Basically your conscious decisions are independent of the behavior of either classical or quantum objects.

    Don't be fooled by this consciousness creates realty new age bunkum you see in some popular accounts of QM - its just that - bunkum - or rather popularization's designed to bedazzle the general reader. QM is actually not like that - its actually even weirder - or much more sensible - depending on how you look at it.

    Last edited: May 25, 2013
  4. May 25, 2013 #3
    The actual probability of finding the ball in the picked box would be 0.5.
    Is it atleast possible to reduce this?
    So if we carry out this experiment many number of times we would see that most of the time the ball turns up in the other box and very few times in the box we picked.
  5. May 25, 2013 #4


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    Of course not - since your choice is freely taken there is no way the particle can know what you will do then do something different or the same.

    Last edited: May 25, 2013
  6. May 25, 2013 #5
    The issue is that the scenario you're describing is one of classic probability. This is true regardless of whether or not the object inside in classical or quantum. Quantum effects require superposition, which your scenario doesn't have.

    Let's assign state ##|0\rangle## for "the electron is in box 1, box 2 is empty" and ##|1\rangle## for "the electron is in box 2, box 1 is empty". Now I present you with two boxes and say there's a 50% chance I've given you ##|0\rangle## and a 50% chance I've given you ##|1\rangle##. (Actually, you didn't say 50% in your original problem, you just said "the ball is in either one of them", but it was implied later. Note that just because I tell you the ball is in either box doesn't mean the probability of it being in one or the other is 50%. You're starting with the assumption of equal probability). Then, quantum mechanically, you would say I've given you a mixed state described by the density matrix ##\rho = \frac{1}{2}|0\rangle \langle0| + \frac{1}{2}|1\rangle \langle1|##. This is just a multiple of the identity matrix, so it will be the same regardless of what unitary operations you apply to the state. There aren't any terms that can interfere with each other; probability is completely classical.

    Suppose, instead, I gave you two boxes (experimentally, it would be easier if the "boxes" can be taken more abstractly; say, as two different energy levels of an atom) and say they're in an equal superposition of "electron in box 1" and "electron in box 2". This is not the same thing as saying "the electron is in either box". For instance, if I tell you that I've prepared the boxes in the state ##\frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)## then, with the application of some unitary operators to the system, you can ensure the electron is in whichever box you prefer when you do your measurement. However, if you don't do anything to the boxes before opening one, you will indeed find that it's in box 1 or box 2 with equal probability of 50%. This state is described by the density matrix ##\rho = \frac{1}{2}|0\rangle \langle0| + \frac{1}{2}|0\rangle \langle1| + \frac{1}{2}|1\rangle \langle0| + \frac{1}{2}|1\rangle \langle1|##. The matrix now has off-diagonal terms, which represent the possibility for the interference that makes quantum mechanical probabilities so distinctive.
    Last edited: May 25, 2013
  7. May 25, 2013 #6
    I want to note that one should be careful when devising these thought experiments. Quantum mechanically, the particle isn't 100% in either box until you check. It's not that it's in one or the other, and then you check to find out which, is really in both and neither at the same time until you check.

    The only way to collapse the ball into a specific state is pick a box, look inside, put it back. Now you can always pick the box with nothing in it, given there is no time evolution.

    Always picking the wrong box in a system of two boxes is the same as always picking the correct box, you just have to know where it is. The problem is there is no wrong or right box until you check. Even the ball doesn't know where it is until you check.
  8. May 25, 2013 #7
    Thanks for the detailed answer. Can you also Please explain how these unitary operators can be
    physically implemented.
  9. May 25, 2013 #8
    That depends entirely on the physical implementation of the system you're considering (i.e. what your quantum states refer to). Since you're describing a scenario involving two quantum states (a "qubit") it falls under the auspices of quantum computing. To convert ##\frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)## to a ##|0\rangle## state you would use a Hadamard gate. To convert it to a ##|1\rangle## state you would use a 180° phase shift gate followed by a Hadamard.

    Actually doing these operations requires finding a suitable (and physically realizable) Hamiltonian that yields one of these unitary gates as its corresponding time evolution operator. The nature of such a Hamiltonian depends on what your quantum state vectors are referring to (electron spin states, energy states of some particle, photon polarizations, etc.) Here is a short overview of physical implementations of quantum computers.
  10. May 26, 2013 #9
    Maybe I am not reading your post right ....however

    While all the above answers are correct, there is one minor thing that got overlooked.

    I say minor because it's not related to the "fundamentals/core" of QM

    It's not possible quantum mechanically, however it's possible classically.

    the probabilities between the two boxes can be manipulated, classically. for example - you can have a coin that shows heads 30% of the time and tails 70% of the time; same with the boxes

    Classically you can do whatever you want - whatever probabilities you want between 0 and 1 for any of the boxes/scenario.

    you can even change the probabilities in QM experiment (you can change the probabilities of photon/electron landing on various co-ordinates on the screen) however then you would be using the classical part...for example you can have various kinds of interference patterns in single particle double slit experiment.

    However you cannot transfer information FTL that way. Nor can you predict the future....because, for example, you don't know (and can never predict) the sequence of heads and tails, even though you know 30/70 probability ratio.

    The "fundamental randomness" in QM cannot be predicted or controlled in any way.

    Did I overlook some requirement/condition in your post?
    Last edited: May 26, 2013
  11. May 26, 2013 #10


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    Unless I'm misunderstanding you (quite possible) LastOneStanding addressed this in the second paragraph of post #5 above.

    There's no reason why we cannot prepare the system in the (suitably normalized) state [itex]\sqrt{.7}\left|0\right\rangle + \sqrt{.3}\left|1\right\rangle[/itex]
  12. May 26, 2013 #11
    Thanks Nugatory, agreed. I saw posts 2 and 4 and wanted to know if they were correct in saying that it was not possible to change the probabilities, since no one commented on them
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