Quantum states are unit vectors... with respect to which norm?

16

4

The most general definition of a quantum state I found is (rephrasing the definition from Wikipedia)

Quantum states are represented by a ray in a finite- or infinite-dimensional Hilbert space over the complex numbers.

Moreover, we know that in order to have a useful representation we need to ensure that the vector representing the quantum state is a unit vector.

But in the definition above, they don't precise the norm (or the scalar product) associated with the Hilbert space considered. At first glance I though that the norm was not really important, but I realised yesterday that the norm was everywhere chosen to be the Euclidian norm (2-norm). Even the bra-ket notation seems to be made specifically for the euclidian norm.

My question: Why is the Euclidian norm used everywhere? Why not using an other norm? Does the Euclidian norm has useful properties that can be used in quantum mechanics that others don't?

Adrien Suau

Posted 2018-07-13T11:43:57.257

Reputation: 3 670

2Actually I just wanted to add a comment but I don't have the reputation for it: note that, as you write in your question - quantum states are rays in the Hilbert space. This means that they are not normalized, but rather that all vectors in the Hilbert space that point in the same direction are equivalent. It is more convenient to work with normalized states but the physics is actually hidden in the overlap of the states with each other. It is for this reason that there is no norm present in the definition of a state. – Omri Har-Shemesh – 2018-07-14T07:09:08.080

Answers

6

Born's rule states that $|\psi(x)|^2 = P(x)$ which is the probability of finding the quantum system in the state $|x\rangle$ after a measurement. We need the sum (or integral!) over all $x$ to be 1:

\begin{align} \sum_x P_x &= \sum_x |\psi_x|^2 = 1,\\ \int P(x)dx &= \int |\psi(x)|^2 dx= 1. \end{align}

Neither of these are valid norms because they are not homogenous. You can make them homogenous simply by doing the square root:

\begin{align} \sqrt{\sum_x |\psi_x|^2} = 1,\\ \sqrt{\int |\psi(x)|^2dx} = 1. \end{align}

and you may recognize this as the Euclidean norm and a generalization of the Euclidean norm to a non-discrete domain. We could also use a different norm:

\begin{align} \sqrt{\sum_x \psi_x A \psi_x^*} = 1,\\ \sqrt{\int \psi(x)A\psi^*(x)} = 1, \end{align}

for some positive definite matrix/function A.


However a $p$-norm with $p>2$ would not be as useful because for example:

\begin{align} \sqrt[5]{\sum_x |\psi_x|^5} \\ \end{align}

does not have to be 1.

In this way the Euclidean norm is special because 2 is the power in Born's rule, which is one of the postulates of quantum mechanics.

user1271772

Posted 2018-07-13T11:43:57.257

Reputation: 8 162

This answer is related to my comment on @DaftWullie's one. So the euclidian norm is used because the postulate of measurement tells us that it's the only $p$-norm that is valid?

– Adrien Suau – 2018-07-13T13:31:47.270

2It is the only p-norm that is meaningful. We want the sum of probabilities to be 1 (which is a law of mathematics) and probabilities are defined by the square of the wavefunction (which is a postulate of quantum mechanics called Born's rule). – user1271772 – 2018-07-13T13:33:19.273

@Nelimee: Thanks for your message on Chat. I can't reply because I'm banned from chat for 2 more days. The reason for the first answer was because I read your questions "Why is the Euclidian norm used everywhere? Why not using an other norm?" and immediately considered a case where a valid norm is not the Euclidean norm but a different 2-norm, which is a 2-norm on a non-discrete set of variables. I thought this was enough to explain that the Euclidean norm is not the only valid norm, and why the Euclidean norm is used when it is. But when I noticed daftwullie got the upvote and I didn't, I – user1271772 – 2018-07-13T13:42:52.423

wrote the second answer (this one), which is even better. – user1271772 – 2018-07-13T13:43:04.827

2so your answer is "because of Born's rule"? Doesn't that just move the question to "why does Born's rule use the power of 2?"? – DaftWullie – 2018-07-13T14:14:03.483

@DaftWullie: My answer is not so much: "because of Born's rule". What is wrong with the norm in my answer with 0 votes, which gives a norm that is not a Euclidean norm? What is wrong with the square of the Euclidean norm, or the Euclidean norm to the power 1000 ? As long as the sum (or integral!) of the probabilities = 1, we are good, no? Why do we need the square root? Also, not everyone believes in Born's rule, there are people that have explored $|\psi(x)|^{(2+\epsilon)}$ and $|\psi(x)|^2 + \epsilon|\psi(x)|^4$ and things like that, because Born's rule can only be tested to some finite – user1271772 – 2018-07-13T14:19:43.547

precision, and also because quantum mechanics at present is missing something (without any improvement, it is not compatible with general relativity). So Born's rule is maybe not so fundamental, and the Euclidean norm is not so fundamental, but the fundamental thing (in my mind at least) is that the sum of the probabilities = 1. – user1271772 – 2018-07-13T14:21:31.587

So if you're using one of your variant norms, how do you define the inner product, $|\langle \psi|\phi\rangle|$, remembering that the case of $\psi=\phi$ must give the length of the vector (under your norm)? Now, what does that mean for matrix multiplication (which is fundamentally founded on taking the inner product). What are the matrices that are length preserving? I don't know the answer, but I suspect it comes out rather similarly to what I mentioned in my answer. – DaftWullie – 2018-07-13T14:26:32.740

In my answer with 0 votes, I gave an inner product as $\int \psi(x)\psi^*(x) dx$, which (if x was discrete) would be the square of the Euclidean norm, isn't it? – user1271772 – 2018-07-13T14:28:52.000

I'm a bit too rusty with continuous variables. I'm specifically talking about the discrete case. – DaftWullie – 2018-07-13T14:36:42.470

@DaftWullie: I'm quite rusty on it too, so it took me a long time to remember, but I think Euclidean norm (or 2-norm) is NOT necessary, since there is also $\int \psi(x)\psi^*(x)dx=1$ which is also a commonly used norm. You must remember this one from your first QM class right? – user1271772 – 2018-07-13T17:48:24.420

That is the normalisation of the state, yes. I’m not so sure that’s the same thing as the norm. – DaftWullie – 2018-07-13T18:03:13.437

Remove the "= 1" part and isn't that the norm ?? – user1271772 – 2018-07-13T18:27:25.067

The argument in the answer seems very weak. I would argue conversely that Born's rule uses the square BECAUSE this way the probability is preserved. – Norbert Schuch – 2018-07-13T20:46:19.297

1Seems like a "what came first, the chicken or the egg?" case. – user1271772 – 2018-07-13T20:58:53.620

The norm is a very precisely defined concept, and it’s the formal mathematics of that that i’ve Never studied in continuous variables. But colloquially you could specify it as “the number that you have to divide a state by to make it normalised”. Just because you can take arbitrary powers of the number 1 and get back 1 does not let you arbitrarily redefine the norm. – DaftWullie – 2018-07-14T06:00:56.213

@user1271772 That's not at all the way you present it. And it is not true either. It is about "What are valid chicken-egg pairs", and the answer is "unitary evolution and the 2-norm" and "stochastic matrices and the 1-norm". There is a nice essay of Scott Aaronson on that. – Norbert Schuch – 2018-07-14T08:13:14.337

It doesn't have to be the 2-norm Norbert. – user1271772 – 2018-07-14T13:59:11.080

8

Some terminology seems a little bit jumbled here. Quantum states are represented (within a finite dimensional Hilbert space) by complex vectors of length 1, where length is measured by the Euclidean norm. They are not unitary, because unitary is a classification of a matrix, not a vector.

Quantum states are changed/evolved according to some matrix. Given that quantum states have length 1, it turns out to be necessary and sufficient that the maps of pure states to pure states are described by unitary matrices. These are the only matrices that preserve the (Euclidean) norm.

It is certainly a valid question "could we use a different ($p$) norm for our quantum states?" If you then classify the operations that map normalised states to normalised states, they are incredibly limited. If $p\neq 2$, the only valid operations are permutation matrices (with different phases on each element). Physics would be a whole lot more boring.

A good way to get a feel for this is to try drawing a 2D set of axes. Draw on it the shapes corresponding to the set of points of length 1 under different $p$-norms. $p=2$ gives you the circle, $p=1$ gives you a diamond, and $p\rightarrow\infty$ gives a square. What operations can you do that map the shape onto itself? For the circle, it's any rotation. For anything else, it's just rotations by multiples of $\pi/2$. The following comes from Wikipedia:

enter image description here

If you want more details, you might want to look here.

DaftWullie

Posted 2018-07-13T11:43:57.257

Reputation: 35 722

Thanks for the terminology precisions! You are right, I misused the terms. – Adrien Suau – 2018-07-13T13:16:48.093

However the question is fine as long as you replace "unitary" by "unit vector" – user1271772 – 2018-07-13T13:24:17.137

But this answer does not answer why we use the euclidian norm. I understood that the other norms are not convenient, but we don't really have the control on what is "convenient" within the physics laws and what is not, do we? – Adrien Suau – 2018-07-13T13:24:56.107

@Nelimee It's not inconvenient. It's that lots of operations do not exist if you don't use the 2-norm. Operations such as the square-root of not, which we can go out, do an experiment, and observe. So that excludes everything except the 2-norm – DaftWullie – 2018-07-13T13:26:29.170

So the use of the 2-norm in the theoretical background of quantum computing is like measurement: "we don't really know why, but nature tells us so with the outputs of some experiments so it must be true"? – Adrien Suau – 2018-07-13T13:27:55.250

1as with all physics! All theories are that, theories which best fit the available data. – DaftWullie – 2018-07-13T13:29:50.683

Yes, I probably have not done enough physics to realise that, and since my first introduction to QC, the "postulates" are annoying me :/ – Adrien Suau – 2018-07-13T13:33:45.520

5

More mathematically, because $\mathbb{R}^n$ with an $L^p$ norm is a Hilbert space only for $p=2$.

Federico Poloni

Posted 2018-07-13T11:43:57.257

Reputation: 159

I have upvoted your answer (which is a great first answer to the QCSE!), but does it have to be a 2-norm? You are saying that 1-norm and 3-norm are invalid, but what about the norm in my answer, which is the square of the 2-norm? – user1271772 – 2018-07-13T17:50:23.953

3

@user1271772 Thanks! If I understand correctly, the function that you suggest isn't even a vector norm because it's not homogeneous.

– Federico Poloni – 2018-07-13T18:00:50.387

2Anyway, what you suggest is true: one can construct a Hilbert space structure with a norm different than the $L^2$ norm (although not with the $L^p$ norm in place of the 2-norm). The simplest example is: choose any positive definite matrix $A$ and take the norm $|x|_A:=\sqrt{x^*Ax}$. – Federico Poloni – 2018-07-13T18:03:09.847

it is positive homogenous with $k=2$, why does it have to be with $k=1$ ?

– user1271772 – 2018-07-14T07:44:51.497

@user1271772 $k=1$ is a requirement in the definition. One of the axioms of vector norms is 2. p(av) = |a| p(v) (being absolutely homogeneous or absolutely scalable) (check, for a quick reference, that Wikipedia page I linked above). Of course, that's just a tautological argument "because it's defined that way", and I understand that a physicist may want a more physical reason. – Federico Poloni – 2018-07-14T07:56:55.370

why does it need to have k=1 ? – user1271772 – 2018-07-14T13:55:50.087

4

An elegant argument can be derived by asking which theories can we build which are described by vectors $\vec v = (v_1,\dots,v_N)$, where the allowed transformations are linear maps $\vec v\to L\vec v$, probabilities are given by some norm, and probabilities must be preserved by those maps.

It turns out that there are basically only three options:

  1. Deterministic theories. Then we don't need those vectors, since we are always in one specific state, i.e. the vectors are $(0,1,0,0,0)$ and the like, and the $L$'s are only permutations.

  2. Classical probabilistic theories. Here, we use the $1$-norm and stochastic maps. The $v_i$ are probabilities.

  3. Quantum mechanics. Here, we use the $2$-norm and unitary transformations. The $v_i$ are amplitudes.

These are the only possibilities. For other norms no interesting transformations exist.

If you want a more detailed and nice explanation of this, Scott Aaronson's "Quantum Computing since Democritus" has a Lecture on this, as well as a paper.

Norbert Schuch

Posted 2018-07-13T11:43:57.257

Reputation: 3 740

2

The other answers addressed why $p=2$ in terms of which $L^p$ space to use, but not the weighting.

You could put in a Hermitian positive definite matrix $M_{ij}$ so that that the inner product is $\sum x_i^* M_{ij} y_j$. But that doesn't gain you much. This is because you might as well change variables. For ease, consider the case when $M$ is diagonal. with the diagonal case that would be interpreting $M_{ii} \mid x_i \mid^2$ as a probability instead of $\mid x_i \mid^2$. $M_{ii}>0$ so why not just change variables to $\tilde{x}_i = \sqrt{M_{ii}} x_i$. You can think of this as $L^2$ functions on the space of $n$ points where each point is weighted by $M_{ii}$.

For the continuous 1 variable case, yes you could use $L^2 (\mathbb{R} , w(x) dx)$ as well. $w(x)$ just reweights the lengths. That's still a perfectly good Hilbert space. But the problem is that translation $x \to x+a$ was supposed to be a symmetry and $w(x)$ breaks that. So might as well not use $w(x)$. For some purposes, that symmetry is not present, so you do have a $w(x) \neq 1$.

In some cases it is useful not to move to standard form. It shuffles around how you do some calculations. For example, if you're doing some numerics, then you can reduce your errors by this sort of reshuffling to avoid really small or large numbers that your machine finds difficult.

A tricky thing is to make sure you keep track of when you changed your variables and when you didn't. You don't want to get confused between changing to the standard inner product doing some unitary and changing variables back vs trying to do that in one step. You are likely to drop factors of $\sqrt{M_{ii}}$ etc by mistake, so be careful.

AHusain

Posted 2018-07-13T11:43:57.257

Reputation: 3 383

-1

The Euclidean norm on an $n$-dimensional space, as defined here, is not the only norm used for quantum states.

A quantum state doesn't have to be defined on an n-dimensional Hilbert space, for example the quantum states for a 1D harmonic oscillator are functions $\psi_i(x)$ whose ortho-normality is defined by:

$$ \int \psi_i(x)\psi_j^*(x)dx. $$

If $i=j$ we get:

$$ \int |\psi(x)|^2dx = \int P(x)dx = 1, $$

because the total probability must be 1.
If $i\ne j$, we get 0, meaning that the functions are orthogonal.

The Euclidean norm, as defined in the link I gave, is more for quantum states on discrete variables where $n$ is some countable number. In the above case, $n$ (which is the number of possible values that $x$ can be) is uncountable, so the norm doesn't fit into the definition given for a Euclidean norm on an $n$-dimensional pace.

We could also apply a square root operator to the above norm, and still we'd have the required property that $\int P(x)dx=1$, and the Euclidean norm can then be thought of as a special case of this norm though, for the case where $x$ can only be chosen from some countable number of values. The reason why we use the above norm in quantum mechanics is because it guarantees that the probability function $P(x)$ integrates to 1, which is a mathematical law based on the definition of probability. If you had some other norm which can guarantee that all laws of probability theory are satisfied, you would be able to use that norm too.

user1271772

Posted 2018-07-13T11:43:57.257

Reputation: 8 162

@Nelimee: I can't reply to your chat message "I did not get the point of your answer with 0 votes" because I'm banned from chat for 2 more days, but which part of this answer do you not get? – user1271772 – 2018-07-13T13:45:18.543

@Nelimee ? I'm now at -1 so would appreciate knowing which part was unclear – user1271772 – 2018-07-13T23:08:54.630

What you write is just the euclidean norm in infinite dimensions. Your statement "The Euclidean norm on an n-dimensional space, as defined here, is not the only norm used for quantum states." is misleading to the extent of being wrong. – Norbert Schuch – 2018-07-14T08:34:51.357

@Norbert. (1) this is the SQUARE of the euclidean norm. (2) here it is UNCOUNTABLY infinite. It is no longer n-dimensional even for countably infinite n. – user1271772 – 2018-07-14T13:54:26.397

@ (1) That's because you forgot to put the square root. Also, the square root of $1$ is $1$. (2) That's not true. $L^2(\mathbb R^n)$, the space of normlized functions with that norm, is a separable space, i.e. it has a countably infinite basis. – Norbert Schuch – 2018-07-14T14:17:59.243

@NorbertSchuch: (1) I did not forget any square root. https://en.wikipedia.org/wiki/Normalizing_constant (2) $\psi(x)$ for a harmonic oscillator is defined for an uncountably infinite domain for $x$.

– user1271772 – 2018-07-14T18:13:50.400

Doesn't mean it's an uncountably infinite dimensional space. And the square root of 1 is still 1. – Norbert Schuch – 2018-07-14T20:58:58.083

@NorbertSchuch: The square root of 1 is 1, so for unit vectors the norm is the same with or without the sqrt (it's 1 in either case). But "the norm" in general is not the same, so non-unit vectors won't have the same norm. As for the uncountably infinite, it is uncountably infinite dimensional in the sence that a spin-half wavefunction is 2-dimensional ${|0\rangle,|1\rangle}$, a spin-1 wavefunction is 3D: ${|-1\rangle,|0\rangle,|1\rangle}$, and a QHO is countably infinite dimensional. – user1271772 – 2018-07-14T22:48:24.117

As in the discussion you had above with @FedericoPoloni, norms must be homogenous ($||\lambda \vec v|=|\lambda|,|\vec v|$), which is only satisfied with a square root. – Norbert Schuch – 2018-07-14T23:09:36.787

@NorbertSchuch: Yes, but I don't quite see why we need it to satisfy this property though, especially since anything without a norm of 1 (i.e. anything with $\lambda\ne 1$) is not a valid quantum state anyway. Also the QHO wavefunctions live in an uncountably infinite dimensional space. – user1271772 – 2018-07-14T23:11:59.943

That's what things called "norm" have to satisfy. You are free to define whatever you want, but please don't call it a "norm". Next, you will exchange the meaning of "quantum" and "classical". – Norbert Schuch – 2018-07-14T23:14:37.357

@NorbertSchuch: but in that same Wikipedia page there's so many norms that don't satisfy the homogeneity property, one time it even says "it is not a norm in the usual sense because it's not homogenous" – user1271772 – 2018-07-14T23:15:42.677

Which Wikipedia page? – Norbert Schuch – 2018-07-14T23:17:03.897

Wikipedia page for "norm (mathematics)" which was linked by FedericoPoloni – user1271772 – 2018-07-14T23:20:04.120

And where do you see a non-homogenous norm there? There's only some semi-, pre-, quasi-, or whatever-norms. Putting these prefixes means it is not a norm. – Norbert Schuch – 2018-07-14T23:25:02.363

For example "The F-norm described above is not a norm in the usual sense because it lacks the required homogeneity property." – user1271772 – 2018-07-14T23:26:40.113

Not a normal norm either, is it? Not useful for quantum states. – Norbert Schuch – 2018-07-14T23:32:35.710

I am less and less interested in whether or not the norm satisfies some definition of what a "norm" is. The whole point is to make sure that the sum (or integral!) of all possible *probabilities* is not 3 or 0.5, but is 1. This is because probabilities have to add up to 1. So it could be the 2-norm, the square of the 2-norm, the square root of the 2-norm, or some of the norms suggested in other answers, such as $\sqrt{xAx^*}$ for positive-definite A. It can even be something that doesn't satisfy homogeneity! – user1271772 – 2018-07-15T00:42:42.567