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[Explore Stack Internal](https://stackoverflow.co/internal/?utm_medium=referral&utm_source=physics-community&utm_campaign=side-bar&utm_content=explore-teams-compact-popover) # [What actually is superposition?](https://physics.stackexchange.com/questions/757650/what-actually-is-superposition) [Ask Question](https://physics.stackexchange.com/questions/ask) Asked 2 years, 11 months ago Modified [2 years, 11 months ago](https://physics.stackexchange.com/questions/757650/what-actually-is-superposition?lastactivity "2023-04-01 14:38:01Z") Viewed 1k times 1 \$\\begingroup\$ What does [superposition](https://en.wikipedia.org/wiki/Quantum_superposition) actually mean? Can something like an atom actually be in two different states at once or do we just not know which state it is in? Also, how can our act of observing something force it to "decide" (whatever that means) which state it is in? What controls this randomness? - [quantum-mechanics](https://physics.stackexchange.com/questions/tagged/quantum-mechanics "show questions tagged 'quantum-mechanics'") - [hilbert-space](https://physics.stackexchange.com/questions/tagged/hilbert-space "show questions tagged 'hilbert-space'") - [definition](https://physics.stackexchange.com/questions/tagged/definition "show questions tagged 'definition'") - [superposition](https://physics.stackexchange.com/questions/tagged/superposition "show questions tagged 'superposition'") - [quantum-states](https://physics.stackexchange.com/questions/tagged/quantum-states "show questions tagged 'quantum-states'") [Share](https://physics.stackexchange.com/q/757650 "Short permalink to this question") Cite [Improve this question](https://physics.stackexchange.com/posts/757650/edit) Follow [edited Apr 1, 2023 at 7:34](https://physics.stackexchange.com/posts/757650/revisions "show all edits to this post") [](https://physics.stackexchange.com/users/2451/qmechanic) [Qmechanic](https://physics.stackexchange.com/users/2451/qmechanic)♦ 226k5252 gold badges646646 silver badges2\.6k2\.6k bronze badges asked Mar 31, 2023 at 17:22 [](https://physics.stackexchange.com/users/363199/ladan) [Ladan](https://physics.stackexchange.com/users/363199/ladan) 5744 bronze badges \$\\endgroup\$ 4 - 4 \$\\begingroup\$ "Superposition" is a synonym for "sum". If you are asking whether an atom can be in two states zat once, you haven't understood what "state" means. \$\\endgroup\$ WillO – [WillO](https://physics.stackexchange.com/users/4993/willo "19,588 reputation") 2023-03-31 18:14:42 +00:00 Commented Mar 31, 2023 at 18:14 - 1 \$\\begingroup\$ I don't know why this question is being downvoted. IMHO it does reflect the beginner's confusion about quantum mechanics fairly well. It does pose a serious challenge, of course. There is no simple and accurate answer to any of the questions. One basically already has to know quantum mechanics to understand the correct answers. \$\\endgroup\$ FlatterMann – [FlatterMann](https://physics.stackexchange.com/users/346466/flattermann "3,243 reputation") 2023-03-31 19:35:27 +00:00 Commented Mar 31, 2023 at 19:35 - 1 \$\\begingroup\$ Your answer is here [physics.stackexchange.com/a/211528/226902](https://physics.stackexchange.com/a/211528/226902) and here [physics.stackexchange.com/a/211476/226902](https://physics.stackexchange.com/a/211476/226902) \$\\endgroup\$ Quillo – [Quillo](https://physics.stackexchange.com/users/226902/quillo "6,469 reputation") 2023-04-01 10:55:54 +00:00 Commented Apr 1, 2023 at 10:55 - \$\\begingroup\$ If you hit the "c" key on the piano really hard you get the note c .... but you also get some c of the next octave up and some c of the next next octave up etc etc. The vibrating string is vibrating at multiple frequencies ... it's in superposition. \$\\endgroup\$ PhysicsDave – [PhysicsDave](https://physics.stackexchange.com/users/144747/physicsdave "2,978 reputation") 2023-04-01 20:16:04 +00:00 Commented Apr 1, 2023 at 20:16 [Add a comment](https://physics.stackexchange.com/questions/757650/what-actually-is-superposition "Use comments to ask for more information or suggest improvements. Avoid answering questions in comments.") \| ## 4 Answers 4 Sorted by: [Reset to default](https://physics.stackexchange.com/questions/757650/what-actually-is-superposition?answertab=scoredesc#tab-top) 1 \$\\begingroup\$ Superposition is a mathematical property, nothing more. It is a property of linear operators. Simple addition(\$+\$) and multiplication(\$\\cdot\$) are examples of linear operators. A linear operator \$f\$ has the following property: for any two values \$X\$ and \$Y\$, and any two scalar constants \$a\$ and \$b\$, \$f(aX + bY) = a f(x) + b f(y)\$. Any linear operator shares this property. As for quantum mechanics, there's nothing physically "special" about superposition. A particle is in the state it is in. Nothing more, nothing less. And the operators that describe its evolution over time are simply the operators. Nothing more, nothing less. However, we observe that many of the operators we care about are indeed linear. This means they obey the above property. This is a convenience for calculation. It means that whenever we see a \$f(aX, bY)\$, we can replace it with \$a f(x) + b f(y)\$ in the same way we can replace \$2+3\$ with \$5\$ to simplify an algebraic equation. By very cleverly choosing our \$X\$ and \$Y\$, we can make the computation simpler. For instance, for a complicated linear operator, computing \$f(\[5, 7\])\$ might be difficult, but if we break it apart into \$f(5 \[1, 0\] + 7 \[0, 1\])\$, we can leverage the superposition principle to say that that is equal to \$5f(\[1, 0\]) + 7f(\[0, 1\])\$ Often the behavior of these operators is much easier to understand for simple orthogonal inputs, like \$\[1, 0\]\$ and \[0, 1\]\$. Indeed, in many cases the equations start to look like the classical ones that we are used to. And therein lies the rub with superposition. It almost makes things *too* easy to understand. It turns complicated quantum realities into classical problems and that leads us to believe the particle "is in two classical states," and then we have to try to grapple with what that even means. More accurately, the particle is in one quantum state ("it's state"), and that state is mathematically equivalent to two classical-looking states. This is similar in nature to the famous particle/wave duality. It doesn't say that a particle is sometimes a classical particle and sometimes a classical wave. It says that a particle is *always* a quantum wave, and in the extremes that quantum wave can act either like a classical particle or act like a classical wave, depending on the circumstances. But never once did it stop behaving as a quantum wave. Trying to think about when the particle stopped being a classical wave and started being a classical particle only leads to confusion. [Share](https://physics.stackexchange.com/a/757759 "Short permalink to this answer") Cite [Improve this answer](https://physics.stackexchange.com/posts/757759/edit) Follow answered Apr 1, 2023 at 14:38 [](https://physics.stackexchange.com/users/47472/cort-ammon) [Cort Ammon](https://physics.stackexchange.com/users/47472/cort-ammon) 54\.8k66 gold badges104104 silver badges178178 bronze badges \$\\endgroup\$ [Add a comment](https://physics.stackexchange.com/questions/757650/what-actually-is-superposition "Use comments to ask for more information or suggest improvements. Avoid comments like “+1” or “thanks”.") \| 0 \$\\begingroup\$ There is a difference between a classical state and a quantum state. In classical mechanics, you specify the state of a particle by giving its position and momentum. Each of these is a vector. The set of all positions is a vector space. Likewise the set of all momenta. The important property of vectors is that you can break a vector into components that add up to the vector. You can do this many different ways. You can calculate the potential energy of a particle from its vector position. Or you can choose a way to break it into components, calculate the potential energies for each component, and add the energies together. In quantum mechanics, you specify a wave function. The wave function tells you everything about the position and momentum. It doesn't tell you as much as the classical state. And this is correct because an electron or photon doesn't have a position or momentum like a classical point particle does. It is spread out over a range of positions and a range of momenta. Spread out means something entirely different than in a classical world. It is not like a classical gas, where some part of it is in each position. The electron has two properties that would contradict each other in any classical universe. It is something like a classical point and something like a classical wave, but not really either one. An electron is an indivisible point with no internal parts. In the double slit experiment, when it hits the screen it hits one atom. Never two or more. It can hit a single target far smaller than an electron. Nobody has found targets so small that an electron can't hit just one. If it has a size, so far it is too small to measure. While in flight, it is spread out in a non classical way. A part of the electron in one place never repels a part in another. There are no parts. If you block the slits with detectors to how it was going to go through both, it only hits one detector. If you unblock them, you must conclude it goes through both like a wave to account for the pattern on the screen. For lack of a good way to say it, the wave function describes the "amount of presence" the point particle has at each position. From the wave function, you can calculate the probability that it will hit different points on the screen. It is more likely to hit where it has more "presence". In quantum mechanics, you often start with a particle in one state, do something to the particle, and find the new state of the particle. Doing something is mathematically represented by an operator, a function that that takes an initial state as input and produces a final state as output. For example in the double slit experiment, it is possible to find an operator that takes the wave function at the back of chamber and produces the wave function at the screen. For the double slit, this function cannot do what a similar classical function might do. It cannot take a single position and velocity that a particle might have at the back of the chamber to a single position and velocity at the screen. It needs the whole wave to produce the outcome at the screen. All the positions in the back where the point like electron has some "presence" affect each point where it has "presence" at the screen. The wave function is a vector. It can be broken into components in many different ways. You can choose how to break it into components, do something to each each component to get new states, and add the new states. This will add to the same state from doing something to the total wave function. When people say a particle is in a superposition of states, they mean the state of a particle is the sum of those states. For example, one way to break the wave function into components is known as the position basis. Each component describes a particle as totally present at one point and \$0\$ presence everywhere else. The wave function for this is a delta function. The uncertainty principal says that such a particle can have any momentum at all. Another way is the momentum basis. A component is a wave function that describes an electron with a single momentum. The uncertainty principle says that such a particle can be at any position at all. When an electron is in a superposition of states from the position basis, it simply means it is in the state you get by adding all those components. One of the problems with quantum mechanics is that the idea of operators taking the particle from one state to another only works up to a certain point. In the double slit experiment, that point is where the electron hits the screen. When that happens, the electron hits an atom and enters a new state. Theory does not predict which atom will be hit. It does not predict the new state of the electron. It just gives probabilities. This is called the collapse of the wave function. This is not a shortcoming of theory. Theory matches reality. There is no way to predict which atom will be hit. It is random. [Share](https://physics.stackexchange.com/a/757713 "Short permalink to this answer") Cite [Improve this answer](https://physics.stackexchange.com/posts/757713/edit) Follow [edited Apr 1, 2023 at 5:49](https://physics.stackexchange.com/posts/757713/revisions "show all edits to this post") answered Apr 1, 2023 at 5:19 [](https://physics.stackexchange.com/users/37364/mmesser314) [mmesser314](https://physics.stackexchange.com/users/37364/mmesser314) 52\.6k66 gold badges7272 silver badges193193 bronze badges \$\\endgroup\$ 1 - \$\\begingroup\$ A wave function doesn't tell us anything about the system. It tells us something about all possible measurement outcomes that we can get on the ensemble of the system. Electrons are also not point-like. They are not even things. Quanta are irreversible energy, momentum, angular momentum and charge exchanges. \$\\endgroup\$ FlatterMann – [FlatterMann](https://physics.stackexchange.com/users/346466/flattermann "3,243 reputation") 2023-04-01 10:11:57 +00:00 Commented Apr 1, 2023 at 10:11 [Add a comment](https://physics.stackexchange.com/questions/757650/what-actually-is-superposition "Use comments to ask for more information or suggest improvements. Avoid comments like “+1” or “thanks”.") \| \-1 \$\\begingroup\$ A quantum mechanical system is in a known state immediately after an observation or when it's in its ground state. Why is that? Because "a measurement" removes a known amount of energy from the system and in the ground state the system is in the lowest possible energetic state. Either way we can "know" how much energy the system has left. Because of energy conservation the energy of the system can't change, unless we remove or add energy without making a measurement. This can happen if we let the system interact with another system that is not a "measurement system". In practice this "other system" is often just the physical vacuum and one of its fields like the electromagnetic field. Imagine an atom that is in an excited state. If we just let it "sit there", it will, eventually, emit light, which means it will send out a quantum of energy. If we don't detect this quantum, then we don't know any longer if the atom is still in the excited state or if it has returned into its ground state. In general we can't know the state with certainty even in these most simple scenarios (like a single atom in vacuum) and we have to consider all possible states that it could be in. This is why we have to allow for a "superposition". A new measurement will again remove a known quantity of energy and only then we can know with perfect certainty what state the system is in. At least that is the trivializing physical picture. There are a number of mathematical complications that are being covered by the theory, but at the end of the day "knowing" in physics means knowing the possible energy exchanges that can happen between systems. [Share](https://physics.stackexchange.com/a/757656 "Short permalink to this answer") Cite [Improve this answer](https://physics.stackexchange.com/posts/757656/edit) Follow [edited Mar 31, 2023 at 17:59](https://physics.stackexchange.com/posts/757656/revisions "show all edits to this post") answered Mar 31, 2023 at 17:52 [](https://physics.stackexchange.com/users/346466/flattermann) [FlatterMann](https://physics.stackexchange.com/users/346466/flattermann) 3,24399 silver badges2121 bronze badges \$\\endgroup\$ [Add a comment](https://physics.stackexchange.com/questions/757650/what-actually-is-superposition "Use comments to ask for more information or suggest improvements. Avoid comments like “+1” or “thanks”.") \| \-1 \$\\begingroup\$ In quantum mechanics, the state of a system satisfies an equation of motion such as the Schrodinger equation. If you have two solutions of the Schrodinger equation for a system and you add them together the sum is also a solution. One component of the sum may look a bit like a particle in position \$x\$ while another looks like a particle in position \$y\$. So then the state superficially looks like a particle is in two places at once. There is a controversy over what this means that is commonly described as being about the interpretation of quantum theory. If you dim a laser beam enough then a suitably sensitive detector will sometimes go off and sometimes remain dark, so the light you can detect comes in chunks called photons. If you shine the laser beam on a suitably narrow pair of slits in an opaque screen and measure the number of photons arriving at each point on a screen on the other side you will see a pattern of light and dark bars when enough photons have arrived at the detectors. If you introduce more slits the arrangement of light and dark bars will change. Some places that were previously light will become dark and vice versa. This effect is called quantum interference. These patterns continue to be present if you dim the laser beam so that less than one photon at present in the experiment at a time on average. Now consider a point on the screen that would be light with two slits and dark with four slits. Something has to be coming through the additional slits that prevents a photon from arriving at a dark point on the screen. So unless we're going to give up on explanation entirely we have to conclude that the solutions described by the Schrodinger equation represent something real. That real thing is blocked by opaque material and can pass through an empty slit. And if we do other experiments with lenses or filters or mirrors or whatever that thing acts like a photon. It even acts like a photon if you put a detector that absorbs photons in front of the slit: it acts like the slit is blocked and the pattern on the screen changes accordingly. So then this "dark" photon is interacting with something that looks like the detector we put in front of the slit. So there has to be "dark" detector we don't see that goes off. And that dark detector is connected to a dark computer that records whether the detector goes off and so on. So there are whole dark universes full of detectors and computers and people setting those things up. This is commonly called the many worlds interpretation of quantum mechanics (MWI). For reasons that are unclear this is controversial but the MWI is just a straightforward implication of the theory and it is the only existing explanation of experiments like the one I described above and many others. For a longer version of this argument see "The Fabric of Reality" by David Deutsch, Chapter 2. When you observe an experimental result, you don't force the system to decide what state is in. Rather, after the experiment there is one version of you for each of the possible measurement results. This looks random because because before the experiment there is no single fact of the matter about what state you will be in and so there is no way to predict what state you will be in. You don't see the other versions of the system you're measuring because copying information out of a system prevents interference, which is how different versions of a system interact, see <https://arxiv.org/abs/1111.2189> [Share](https://physics.stackexchange.com/a/757732 "Short permalink to this answer") Cite [Improve this answer](https://physics.stackexchange.com/posts/757732/edit) Follow answered Apr 1, 2023 at 9:28 [](https://physics.stackexchange.com/users/28512/alanf) [alanf](https://physics.stackexchange.com/users/28512/alanf) 12\.4k11 gold badge1717 silver badges3636 bronze badges \$\\endgroup\$ 11 - \$\\begingroup\$ MWI is completely false. All you have to do to know that is to read Everett's thesis. His second sentence is already false and his argument never recovers from that mistake. \$\\endgroup\$ FlatterMann – [FlatterMann](https://physics.stackexchange.com/users/346466/flattermann "3,243 reputation") 2023-04-01 10:14:59 +00:00 Commented Apr 1, 2023 at 10:14 - \$\\begingroup\$ @FlatterMann What is happening in reality to produce the results of single particle interference experiments? \$\\endgroup\$ alanf – [alanf](https://physics.stackexchange.com/users/28512/alanf "12,361 reputation") 2023-04-01 15:17:43 +00:00 Commented Apr 1, 2023 at 15:17 - \$\\begingroup\$ There is no such thing as "single particle interference". That is merely a very poor mental model. In general, even in classical physics "interference" is not actually a physical interaction. It's a property of linear systems that are free of self-interaction that allows for wave-like solutions to traverse the bulk without being scattered. \$\\endgroup\$ FlatterMann – [FlatterMann](https://physics.stackexchange.com/users/346466/flattermann "3,243 reputation") 2023-04-01 16:35:46 +00:00 Commented Apr 1, 2023 at 16:35 - \$\\begingroup\$ @FlatterMann You haven't provided an explanation or a link to an explanation. \$\\endgroup\$ alanf – [alanf](https://physics.stackexchange.com/users/28512/alanf "12,361 reputation") 2023-04-01 20:41:26 +00:00 Commented Apr 1, 2023 at 20:41 - \$\\begingroup\$ I simply tell people to read the second sentence in Everett's thesis. I leave it up to them to compare it carefully both to the theory of quantum mechanics and all known quantum mechanical experiments. That's work that everybody has to do on their own to "understand" what is actually going on here. I can't create that understanding in people's minds. I do admit that one has to think very carefully about all of this... way more so than is usually required in the rushed QM 101 lectures that are the starting and the endpoint for most physicists as far as QM is concerned. \$\\endgroup\$ FlatterMann – [FlatterMann](https://physics.stackexchange.com/users/346466/flattermann "3,243 reputation") 2023-04-01 20:45:16 +00:00 Commented Apr 1, 2023 at 20:45 \| [Show **6** more comments](https://physics.stackexchange.com/questions/757650/what-actually-is-superposition "Expand to show all comments on this post") Start asking to get answers Find the answer to your question by asking. 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There is a difference between a classical state and a quantum state. In classical mechanics, you specify the state of a particle by giving its position and momentum. Each of these is a vector. The set of all positions is a vector space. Likewise the set of all momenta. The important property of vectors is that you can break a vector into components that add up to the vector. You can do this many different ways. You can calculate the potential energy of a particle from its vector position. Or you can choose a way to break it into components, calculate the potential energies for each component, and add the energies together. In quantum mechanics, you specify a wave function. The wave function tells you everything about the position and momentum. It doesn't tell you as much as the classical state. And this is correct because an electron or photon doesn't have a position or momentum like a classical point particle does. It is spread out over a range of positions and a range of momenta. Spread out means something entirely different than in a classical world. It is not like a classical gas, where some part of it is in each position. The electron has two properties that would contradict each other in any classical universe. It is something like a classical point and something like a classical wave, but not really either one. An electron is an indivisible point with no internal parts. In the double slit experiment, when it hits the screen it hits one atom. Never two or more. It can hit a single target far smaller than an electron. Nobody has found targets so small that an electron can't hit just one. If it has a size, so far it is too small to measure. While in flight, it is spread out in a non classical way. A part of the electron in one place never repels a part in another. There are no parts. If you block the slits with detectors to how it was going to go through both, it only hits one detector. If you unblock them, you must conclude it goes through both like a wave to account for the pattern on the screen. For lack of a good way to say it, the wave function describes the "amount of presence" the point particle has at each position. From the wave function, you can calculate the probability that it will hit different points on the screen. It is more likely to hit where it has more "presence". In quantum mechanics, you often start with a particle in one state, do something to the particle, and find the new state of the particle. Doing something is mathematically represented by an operator, a function that that takes an initial state as input and produces a final state as output. For example in the double slit experiment, it is possible to find an operator that takes the wave function at the back of chamber and produces the wave function at the screen. For the double slit, this function cannot do what a similar classical function might do. It cannot take a single position and velocity that a particle might have at the back of the chamber to a single position and velocity at the screen. It needs the whole wave to produce the outcome at the screen. All the positions in the back where the point like electron has some "presence" affect each point where it has "presence" at the screen. The wave function is a vector. It can be broken into components in many different ways. You can choose how to break it into components, do something to each each component to get new states, and add the new states. This will add to the same state from doing something to the total wave function. When people say a particle is in a superposition of states, they mean the state of a particle is the sum of those states. For example, one way to break the wave function into components is known as the position basis. Each component describes a particle as totally present at one point and \$0\$ presence everywhere else. The wave function for this is a delta function. The uncertainty principal says that such a particle can have any momentum at all. Another way is the momentum basis. A component is a wave function that describes an electron with a single momentum. The uncertainty principle says that such a particle can be at any position at all. When an electron is in a superposition of states from the position basis, it simply means it is in the state you get by adding all those components. One of the problems with quantum mechanics is that the idea of operators taking the particle from one state to another only works up to a certain point. In the double slit experiment, that point is where the electron hits the screen. When that happens, the electron hits an atom and enters a new state. Theory does not predict which atom will be hit. It does not predict the new state of the electron. It just gives probabilities. This is called the collapse of the wave function. This is not a shortcoming of theory. Theory matches reality. There is no way to predict which atom will be hit. It is random.