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I'm currently working through Dirac's book The Principles of Quantum Mechanics. In it, he describes the nature of superpositions and at one point states:

"... if the ket vector corresponding to a state is multiplied by any complex number, not zero, the resulting ket vector will correspond to the same state."

He then goes on to state that,

"Given two states corresponding to the ket vectors $|A\rangle$ and $|B\rangle$, the general state formed by superposing them corresponds to a ket vector $|R\rangle$ which is determined by two complex numbers, namely the coefficients $c_{1}, c_{2}$ [from $c_{1}|A\rangle + c_{2}|B\rangle = |R\rangle$]. If these two coefficients are multiplied by the same factor (itself a complex number), the ket vector $|R\rangle$ will get multiplied by this factor and the corresponding state will be unaltered. Thus only the ratio of the two coefficients is effective in determining the state $|R\rangle$. Hence, this state is determined by one complex number, or by two real parameters. Thus from two given states, a twofold infinity of states may be obtained by superposition."

Correct me if I am wrong, but Dirac seems to be saying that one gets an infinite number of different states of the form $c_{1}|A\rangle + c_{2}|B\rangle$ depending on $c_{1}$ and $c_{2}$ (or really determined by their ratio). But given Dirac's earlier statement, we seem to get an equivalence class of states $\{z|A\rangle : z\in \mathbb{C}, z\neq 0\}$ and $\{z|B\rangle : z\in \mathbb{C}, z\neq 0\}$, so why wouldn't $c_{1}|A\rangle + c_{2}|B\rangle$ and $c_{3}|A\rangle + c_{4}|B\rangle$ always represent the same state no matter what the coefficients are (except 0)?

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Note that defining quantum states as equivalence classes of unit vectors that differ only by a global phase will also lead to problems with composite states and tensor products (where the phase difference between two tensor factors now suddenly matters.) –  Sam Hopkins Dec 24 '12 at 6:31
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After reading the answers is it correct to say that there is no well defined operation which takes two states and outputs the superposition of them? –  Michael Bächtold Dec 24 '12 at 15:20
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To emphasize (perhaps excessively) a point from Robert Israel's answer: Despite the common phrase "superposition of states", one does not really superpose states; one superposes vectors that represent states. One can think of states as equivalence classes of vectors (vectors being equivalent iff they differ by a scalar factor), but superposition operations (with specified coefficients) are not well-defined on the equivalence classes. –  Andreas Blass Dec 24 '12 at 15:28
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@Michael: you're perfectly right. Superposition is not a well-defined operation on quantum states. In other words, superposition does not make sense physically. –  Joël Dec 24 '12 at 21:36
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I urge the people voting to close this question not to do so--it seems like a very reasonable question on a difficult subject for someone not familiar with physics (and there are many research mathematicians not familiar with physics). –  Daniel Litt Dec 24 '12 at 23:12

8 Answers 8

The point is that addition of vectors is not just an operation on the states determined by those vectors. So no, they don't represent the same state. For example, consider a two-dimensional Hilbert space with orthonormal basis $|1\rangle$ and $|2\rangle$. Then $|1\rangle + |2\rangle$ and $|1 \rangle - |2 \rangle$ are orthogonal vectors, representing quite different states.

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Let $ |A \rangle $ and $ |B \rangle $ be two non-zero vectors of a Hilbert space $ \mathcal{H} $ that belong to two different one-dimensional subspaces of $ \mathcal{H} $. According to Dirac, $ |A \rangle $ and $ |B \rangle $ represent two different quantum states.

Now, consider two non-trivial superpositions of $ |A \rangle $ and $ |B \rangle $: $$ |R_{1} \rangle := a_{1} |A \rangle + b_{1} |B \rangle \quad \& \quad |R_{2} \rangle := a_{2} |A \rangle + b_{2} |B \rangle, $$ where non-trivial means that $ (a_{1},b_{1}),(a_{2},b_{2}) \in \mathbb{C}^{2} \setminus \lbrace (0,0) \rbrace $. If $ |R_{1} \rangle $ and $ |R_{2} \rangle $ are to represent the same quantum state, then they must lie in the same one-dimensional subspace of $ \mathcal{H} $, i.e., they must be non-zero scalar multiples of each other. Knowing this, write $ |R_{2} \rangle = \lambda |R_{1} \rangle $, where $ \lambda \in \mathbb{C}^{\times} $. As $ \lbrace |A \rangle,|B \rangle \rbrace $ is a linearly independent subset of $ \mathcal{H} $, it follows that the condition $ (a_{2},b_{2}) = \lambda (a_{1},b_{1}) $ must be met. This condition does not hold for all choices of $ (a_{1},b_{1}) $ and $ (a_{2},b_{2}) $ in $ \mathbb{C}^{2} \setminus \lbrace (0,0) \rbrace $. Therefore, in general, $ |R_{1} \rangle $ does not represent the same quantum state as $ |R_{2} \rangle $.

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So, would it be correct to say that even though $a_{1}|A\rangle$ is the same state as $a_{2}|A\rangle$ and $b_{1}|B\rangle$ is the same state as $b_{2}|B\rangle$, $|R_{1}\rangle$ and $|R_{2}\rangle$ represent two different ways of superposing these states, so that the probabilities of being in the state represented by $|A\rangle$ and $|B\rangle$ are different for each? –  Ryan Dec 24 '12 at 18:15
    
Hi Ryan. We always have $ a_{1} |A \rangle \sim a_{2} |A \rangle $ and $ b_{1} |B \rangle \sim b_{2} |B \rangle $, yet we may still end up with $ |R_{1} \rangle \nsim |R_{2} \rangle $. Superposition of vectors is not a quantum-state-preserving binary operation on $ \mathcal{H} \setminus \lbrace 0_{\mathcal{H}} \rbrace $, unless $ |A \rangle \sim |B \rangle $. You can prepare a myriad of quantum states from just two distinct quantum states. This principle is important in quantum computing, in which one can produce infinitely many states from a fixed basis of a finite-dimensional Hilbert space. –  Leonard Dec 24 '12 at 21:22
    
Hence, you are right. As Professor Andreas Blass has mentioned in his comment below the wording of your question, superposition is a binary operation on non-zero vectors, not on states, which are equivalence classes of non-zero vectors. –  Leonard Dec 24 '12 at 21:32

As a linear algebraist, the following interpretation is the one that helped me the most: the set of (non-necessarily pure) quantum states is in a one-to-one correspondence with trace-1 symmetric positive-semidefinite matrices; pure states are the ones that correspond to rank-1 matrices $\mid \! u \rangle \langle u \! \mid$, while non-pure states correspond to their convex combinations.

So instead of thinking about vectors and their "formal convex combinations", you just have to think about symmetric rank-1 matrices and their convex combinations in the usual sense. This interpretation will also come in useful later on in your study of QM.

EDIT: Fixed two mistakes pointed out in the comments.

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Not just any matrices, though, but the positive (semi-definite) ones. –  Branimir Ćaćić Dec 24 '12 at 10:55
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You can't add trace one matrices and get a trace one matrix though. I guess you are adding and renormalising ? Forming a convex combination as Konrad points out. –  Michael Murray Dec 24 '12 at 11:23
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Both your observations are correct, I was sloppy above. One needs indeed to consider positive semidefinite trace-1 matrices, which are the convex hull of their rank-1 subset. –  Federico Poloni Dec 24 '12 at 11:36
    
Well, I could as well delete this answer now, Konrad has written these concepts up in a much better way –  Federico Poloni Dec 24 '12 at 11:42
    
A convex combination of two different density matrices produces a classical mixture, not a quantum superposition, even if the two matrices you began with represent pure states. –  Andreas Blass Aug 8 '13 at 2:10

Try not to lose sight of the physics (the ultimate goal, really) in this mathematical jargon:

This addition-operation is literally superposition, so let's understand what's going on in such terms. The coefficients represent the probability that your object (described by the superposed vector) is actually in one of those state-vectors or the other. That being said, note that if your object is only in $|A\rangle$, then the wavefunction is literally $1|A\rangle$, because you're 100% there... it doesn't make sense for it to only be $\frac{1}{3}|A\rangle$, but regardless that's fine because we normalize the wavefunction, so any coefficient will disappear (i.e. become "1")... so this pertains to Dirac's first statement. Now For $|R\rangle$ there is still normalization, that being $\sqrt{|c_1|^2+|c_2|^2}=1$. But we have freedom here in this condition, so the coefficients themselves don't just pass to 1. In particular, they are important numbers: such a pair $(c_1,c_2)$ is telling you with what likelihood your object lies in $|A\rangle$ versus $|B\rangle$, in such a way that the object is definitely somewhere (that normalization above). Now we see clearly that varying those coefficients are going to change your state.

(If you want the pure math explanation, see the other posts; I just kept with the flow of the title of the post).

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But if states are rays in a Hilbert space then you have the problem that rays cannot be added. I think that is Ryan's concern. –  Michael Murray Dec 24 '12 at 11:22
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Am I not clearing that up?: States are not rays, they just 'correspond' to them. The ray $|A\rangle$ can be added to itself without affecting the state, because on the state-level, "adding" is superposition (and superposing an object with itself gives you the same object). –  Chris Gerig Dec 24 '12 at 11:57
    
Chris Gerig, see my question to Leonard above. –  Ryan Dec 24 '12 at 18:17
    
The answer is yes, precisely for the reason explained here (physics-wise). –  Chris Gerig Dec 24 '12 at 20:29

I like the perspective that the set of states is precisely the set of positive trace class operators $M$ of trace one. A state is called pure if $M=pr^{\perp}_{U}$ is the orthogonal projection onto a one-dimensional subspace $U$.

So, every non-zero vector $\psi$ defines a pure state. Since the orthogonal projection onto $\mathbb{C}\psi$ is the same operator as the orthogonal projection onto $\mathbb{C}x\psi$, for every $x\in \mathbb{C}^\times$, there is no confusion about equiavlence classes.

Concerning the superposition of states, one shows:

  • The set of states is not a vector space, rather it is a convex space. This explains (a) that in a linear combination of vectors only the ratio of the coefficients counts, and it explains (b) that convex combinations $$ aM_1 + (1-a)M_2 $$ of states is the only allowed operation on the space of states.

  • Every state is a convex combination of pure states.

  • A state is pure if and only if can be written as a convex combination of other states only in a trivial way.

This point of view is described in the book

  • L. A. Takhtajan, "Quantum mechanics for mathematicians", Graduate Studies in Mathematics Vol. 95, AMS, 2008
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This answer doesn't explicitly distinguish between two notions of superposition which come into play here, namely addition of vectors and addition of trace class operators, which could be confusing to the uninitiated. –  Qiaochu Yuan Dec 24 '12 at 11:51
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@Qiaochu Yuan: I see. Maybe it helps to emphasize that there is no addition at all on the set of states? It's just a convex set, not a vector space. –  Konrad Waldorf Dec 24 '12 at 12:40
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@Konrad: yes, it helps. It helps even so much that I think you should incorporate this in your answer. –  Joël Dec 24 '12 at 21:35
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@Joël: Good, I've tried to include it. –  Konrad Waldorf Dec 25 '12 at 13:44
    
Definition: "A convex space (also called barycentric algebra and other terms, invented independently many times) is a set equipped with a notion of taking weighted averages, or convex-linear combinations, of its elements. Do not confuse this with an (abstract) convex set , which a special kind of convex space, also defined below." || mathoverflow.net/a/117138/238 –  Tom LaGatta Aug 8 '13 at 5:11

While all the relevant information needed to answer this question is already available in the various given answers and comments, I think that it is worth clarifying some things. The various precise definitions have already been given, so I'll be brief with that.

  • There are two kinds of operations that can be performed on states: "mixtures" and "superpositions".
  • Mixtures take two or more pure states and produce a mixed state. If states are represented by positive, trace-class, unit trace operators or as positive, normalized linear functionals on the relevant algebra of bounded operators, mixtures correspond precisely to convex linear combinations of states, in the most obvious unambiguous way.
  • Superpositions take two or more pure states and produce a pure state. Super positions are most easily performed given individual state vectors, say $|\psi\rangle$ and $|\phi\rangle$, each of which is a representative from the ray corresponding to the pure state $\rho_\psi = |\psi\rangle\langle\psi|$ or $\rho_\phi = |\phi\rangle\langle\phi|$. Given any complex numbers $c_1$ and $c_2$, the new pure state $\rho_\chi = |\chi\rangle\langle\chi|$, where $|\chi\rangle = c_1|\psi\rangle + c_2 |\phi\rangle$ is a representative from the ray of the new superposition state.
  • Of course, one could imagine a superposition operation that acts directly on the states, rather than their representative state vectors, something like $\rho_\chi = S(c_1,c_2;\rho_\psi,\rho_\phi)$. But then, there is essentially an implicit choice of representative state vector ($\rho_\psi \to |\psi\rangle$, etc.) involved in such a definition. The operation $S'(c_1,c_2;-) = S(c_1\lambda_1,c_2\lambda_2;-)$, where $\lambda_1$ and $\lambda_2$ are any non-zero complex numbers is an equally good choice. So, in this sense, neither $S$ nor $S'$ is "the superposition" operation.

Physicists usually do not worry about such subtleties because they work with explicit state vector representatives for all states they care to discuss.

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Of course you can regard projectivisation of corresponding Hilbert space, where you have Hamiltonian as quadratic function $\frac{(x,Hx)}{(x,x)}$ and Kahler structure which gives really classical picture of quantum dynamics of states (without phase). Moreover, states modulo phase are the same from their expectation values on all observables. So, question is - why you care about superposition? In quantum mechanics it's usefull to take some basic states and regard any other as superposition of given, and what you can't see on projectivisation is rule for changing superposition expansion from one basis to another. So, state as itself object is phase-independent, but globally you need linear structure and this can be pointed by rule 'two systems product <-> tensor product of spaces of states'.

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I think the answers given by Konrad Waldorf and Federico Poloni are fantastic. However, if you're just starting to learn quantum mechanics—especially from an older book, like the one you're using—you may want to think only about pure states (this can be done without loss of generality). In that case, this answer may be helpful for you.


In my opinion, thinking about linear combinations of vectors as representing "superpositions" of the corresponding states is extremely misleading. It might suggest, for example, that if a photon in the state represented by $|H\rangle$ is guarnateed to pass through a horizontal polarizer, and a photon in the state represented by $|L\rangle$ is guaranteed to pass through a left circular polarizer, then a photon in the "superposition" state $|H\rangle + \sqrt{2}|L\rangle$ should have at least some chance of passing through a horizontal polarizer. However, by choosing the vectors $|H\rangle$ and $|L\rangle$ appropriately, you can set things up so that a photon in the state represented by $|H\rangle + \sqrt{2}|L\rangle$ will never pass through a horizontal polarizer.

Even when the concept of superposition isn't actively harmful, I find it totally unhelpful, and I would urge you to forget about it entirely. If you absolutely must have it, however, you may read on for a description of the only situation I know of in which "superposition" makes any kind of sense.


Suppose you have a quantum system with state space $\mathcal{H}$. You can think of each orthonormal basis for $\mathcal{H}$ as an abstract description of an experiment that could be done on the system; the basis vectors represent the possible outcomes of the experiment.

Say $|v_1\rangle, \ldots, |v_n\rangle$ is an orthonormal basis for $\mathcal{H}$, and $$|\psi\rangle = c_1|v_1\rangle + \ldots + c_n|v_n\rangle$$ is a unit vector representing the state of the system. If you do the experiment described by the basis $|v_1\rangle, \ldots, |v_n\rangle$, you have probability $|c_k|^2$ of getting the outcome represented by $|v_k\rangle$.

Some people like to think of the state $|\psi\rangle$ as a "superposition" of the possible experimental outcomes represented by the basis vectors $|v_1\rangle, \ldots, |v_n\rangle$. Notice that if you change the coefficients $c_1, \ldots, c_n$, but keep their magnitudes the same, the state $|\psi\rangle$ will change, but the statistics of the experiment described by the basis $|v_1\rangle, \ldots, |v_n\rangle$ will stay the same. In other words, if $$|\psi'\rangle = c'_1|v_1\rangle + \ldots + c'_n|v_n\rangle$$ is a superposition with $|c'_j| = |c_j|$ for all $j$, then the states represented by $|\psi\rangle$ and $|\psi'\rangle$ cannot be distinguished using the experiment described by $|v_1\rangle, \ldots, |v_n\rangle$.

However, consider another orthonormal basis $|w_1\rangle, \ldots, |w_n\rangle$, and write $$|\psi\rangle = d_1|w_1\rangle + \ldots + d_n|w_n\rangle$$ $$|\psi'\rangle = d'_1|w_1\rangle + \ldots + d'_n|w_n\rangle.$$ In general, $|d_j|$ will not be equal to $|d'_j|$ for all $j$. In this case, the states represented by $|\psi\rangle$ and $|\psi'\rangle$ can be distinguished using the experiment described by $|w_1\rangle, \ldots, |w_n\rangle$.


So, if you have a "superposition" of orthonormal basis vectors, changing the coefficients without changing their magnitudes will not change the statistics of the experiment described by the basis you used, but it will generally change the statistics of the experiments described by most other bases. This is why changing the coefficients of a "superposition" generally gives you a representative of a different state.

I say "generally" because there is one exception: if you change your coefficients by multiplying them all by the same number, the statistics of all experiments will remain the same. This is why multiplying a vector by a number gives you a representative of the same state.

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This is just a vamped version of my post. And I completely disagree about "forgetting superposition entirely"... the whole point of these states is probability and superposition. We are trying to understand how to observe the world, and quantum measurements are precisely this. –  Chris Gerig Dec 25 '12 at 8:46
    
As to your photon-polarizer example, I believe it is misleading: Superposition still works, it was just hidden. $|H\rangle$ is a superposition of $|L\rangle$ and $|R\rangle$, and now your two states mix. It's like if I took $A+B$ but forgot to tell you that $B=C-A$. –  Chris Gerig Dec 25 '12 at 8:57
    
@ChrisGerig -- (1) I'd like to understand why you find the idea of superposition important, since I've never been able to get much use out of it. Can you give an example of a situation where you find it particularly useful, or maybe an explanation of why you think it's "the whole point"? (2) Maybe your "B = C - A" comment is a good place to start looking for the disconnect between our points of view. When I read it, my first thought was, "I don't see why it's useful to think of the number 4 as a superposition of the numbers -11 and 7." Am I missing the point somehow? –  Vectornaut Dec 26 '12 at 5:24
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@ChrisGerig -- (3) Just to be clear, I do think that taking convex combinations of density operators is an extremely useful and conceptually important operation. But, as far as I know, summing state vectors has nothing to do with summing the corresponding density operators. Am I mistaken? –  Vectornaut Dec 26 '12 at 5:28
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Summing state vectors is definitely different from summing density operators. Both are important in their own ways. To appreciate the important of superposition of state vectors, one need only open any physics textbook of quantum mechanics. What underlies the importance of superpositions is the Born rule (inner products -> probabilities) which includes in it quantum interference effects. Such effects show up notoriously for particles interacting with barriers or slits, entangled photons in EPR states, non-trivial quantum computer algorithms. Hard to overstate the importance of superposition. –  Igor Khavkine Dec 26 '12 at 6:05

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