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The idea that the state of a system does not depend on the basis that we choose to represent it in, has always puzzled me. Physically I can imagine that the basis ought to just yield an equivalent representation of the system, but I cannot convince myself of this independence. I apologize for the naive character of this post.

For example, in most QM books one frequently comes across statements such as: "The state of the system is represented by a vector (or ket) $|\psi\rangle$ in the corresponding Hilbert space", or that a "vector is not the same thing as the set of its components in a basis". Physically this means that e.g. the momentum $\mathbf{p}$ of a system should not depend on how we choose to orient the axes of our basis.

It is a very abstract idea that the state is just a vector lying in the entire Hilbert space (spanned by the set of all the common eigenvectors of the observables describing the system). How can I write or speak of this $|\psi\rangle$ without having a basis to represent it in. As soon as I want to say anything about this $\psi$, e.g. its norm, I will need its components in some basis, to compute $\sqrt{\langle \psi|\psi \rangle}.$

  1. So what do I know about $|\psi\rangle$ without a chosen basis? How can I express that knowledge (of $\psi$) without a basis?
  2. Is this independence better illustrated when one considers the fact that the set of eigenvalues for any chosen observable of the system, are the same regardless of the chosen basis?
  3. Why it seems so difficult to imagine vector spaces, or vectors lying in abstract high-dimensional spaces ($|\psi\rangle \in \mathcal{H}$), without a basis? In what sense do we mean that a vector is more than just its components $(\langle v_1|\psi\rangle, \langle v_2|\psi\rangle, \dots)$?
  4. But as soon as I want to compute overlaps such as $\langle v_1|\psi\rangle$ or norms $||\psi||$ I need the components of $|\psi\rangle$ in some basis. So how can I convince myself that no matter what basis I choose, this abstract $|\psi\rangle \in \mathcal{H}$ will not depend on it?
  5. Finally, how should I interpret $|\psi\rangle \in \mathcal{H}$ without necessarily having an observable in mind? (i.e. the general statement that the state of the system lies in the Hilbert space).
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  • $\begingroup$ There is nothing abstract about "linearity". You can measure it and I think therein lies your reap problem: you haven't spent enough time (did you spend any?) in the lab and you are mistaking physics for math. Nobody can convince you of any of the above mathematically because it isn't derived from mathematics. It's derived from measurements. $\endgroup$ – CuriousOne Feb 5 '16 at 14:32
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    $\begingroup$ Draw an arrow on a sheet of paper. It's just an arrow. There it is. Now draw two perpendicular lines and label them $x$ and $y$. Now you can express your arrow as components in terms of $x$ and $y$. Now draw two more perpendicular lines and label the $p$ and $q$. You can also express your arrow as components in terms of $p$ and $q$. That is all there is to it. If you have vector, you can either think about it by itself, or you can communicate about it in terms of components. $\endgroup$ – DanielSank Feb 5 '16 at 15:46
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    $\begingroup$ Frankly I don't see why you should even talk about quantum mechanics and infinite-dimensional Hilbert spaces. You should apply all your questions to the vectors on 2d plane. Try to turn your visual imagination off as you've never experienced them other than as a pair of numbers. $\endgroup$ – OON Feb 5 '16 at 15:50
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The state is a vector in the Hilbert space of the Hamiltonian, which gives it a natural basis in terms of the eigenvectors; distinct eigenvalues then exist in distinct (orthogonal) subspaces - for degenerate values the subspaces are larger, but they are still distinct from all others. Clearly this situation gives many advantages in analysis.

However, this representation, though natural, and dependent only upon the spectral decomposition of the Hamiltonian, is not unique. The experimentalist will prefer to establish a basis which corresponds to his experimental setup! Chosing a different basis will change all of the coordinates, but does not change the states.

To make this clear recall that the state vectors of our Hilbert space can also be viewed a as rays, which are similar to the geometric rays of Euclidean geometry. Imagine the ray first, then superimpose a grid upon it - as you rotate the grid about the origin of the ray the intersections of the grid with the ray define the coordinates for that basis. The coordinates change, but the ray doesn't change. For ordinary geometry the relationships between two rays (angles, projections) don't change, so it is clear that some relationships between the coordinates are fixed - these are properties of a metric space.

Our Hilbert space is a normed space - distances don't mean anything, but a normalized state vector always has a length of 1, even when you change the basis; nor does the length change under unitary operators - hence their name.

All of this becomes clear in a good linear algebra course.

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  • $\begingroup$ I think that the state is a linear combination of vectors, and a Hilbert space should not be "of hamiltonian"... It might not be the solution you are searching for,but maybe the state must be project to some basis then they start to have some meanings $\endgroup$ – cindy Feb 5 '16 at 12:35
  • $\begingroup$ A linear combination of states is another state; in QM this may be a superposition of eigenstates, but it is still a distinct state, even if unknown. I used Hamiltonian in the general sense: the operator & boundary conditions which generated the Hilbert space; this is similar to classical mechanics where the configuration space must be specified in order to write the Lagrangian, and this determines the phase space of the Hamiltonian. $\endgroup$ – Peter Diehr Feb 5 '16 at 13:20
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    $\begingroup$ State vectors are rays, and are normalized to 1 to conserve probability ... this is the most important use of the norm; also projections are taken, a use of the inner product, during measurements. You will find both operations in the foundational axioms. But unlike Eucldean geometry, distance is not a fundamental part of the Hilbert space. Some applications, e.g., measures of fidelity in quantum computing have distance like functions, but they aren't geometric distances. $\endgroup$ – Peter Diehr Feb 6 '16 at 11:35
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    $\begingroup$ Peter, thank you for this intuitive answer. So as soon as we want to characterize our vector in H, some eigenbasis has to be chosen (e.g. energy eigenstates, position eigenvectors or...) relative to which we compre various state vectors that we want to study, but what is important is that the choice will not affect the states thus results, about right? So when one says, we have two rays $|\psi\rangle$ and $|\phi\rangle$ in Hilbert space of a system, we need to decide on some basis if we want to start comparing the states together, right? $\endgroup$ – user929304 Feb 6 '16 at 12:02
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    $\begingroup$ Correct! If you could do everything in the eigenbasis, from beginning, all of the calculations would be at their simplest. $\endgroup$ – Peter Diehr Feb 6 '16 at 13:09
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$\newcommand{\real}{\mathbb R}\newcommand{\field}{\mathbb F}\newcommand{\cx}{\mathbb C}\newcommand{\ip}[2]{\left< #1,#2\right>}$We need to dive into mathematics of vector spaces and inner products in order to understand what a vector means and what is it mean to take a scalar product of two vectors. There is a long post ahead so bear with me even though you may think that the maths is too abstract and has nothing to do with QM. In fact without understanding the abstract concept of vector spaces it is almost impossible to understand QM thoroughly.

Vector Spaces

Let's turn to linear algebra and think for a second what we mean by a vector space $V$ on a field $(F,+ , \cdot , 0 ,1)$. It consists of a set $V$, a field $F$ a scalar multiplication $ \odot : F \times V \to V$, $(\lambda , v) \mapsto \lambda \odot v$ and vector addition $ \oplus : V \times V \to V $, $ (v,w) \mapsto v \oplus w $, which has the following properties:

  • $(V, \oplus, \tilde 0)$ is an abelian group
  • $\forall v,w \in V$ and $\forall \lambda \in F: \; \; \lambda \odot (v \oplus w) = \lambda \odot v \oplus \lambda \odot w$
  • $ \forall v \in V$ and $\forall \lambda, \mu \in F: \; \; (\lambda + \mu) \odot v = \lambda \odot v \oplus \mu \odot v $
  • $\forall v \in V: \; \; 1 \odot v = v$

Note that the summation in the third axiom takes place on $F$ on the left hand side and on $V$ on the right hand side. So a vector space can really be anything, with these properties. Normally people don't distinguish between $\odot$ and $\cdot$ or $\oplus$ and $+$ because they show similar properties. For the sake of being absolutely clear I'll nevertheless make this distinction. If it bothers you replace $\oplus \to +$ and $\odot \to \cdot$ in your head. The elements of the set $V$ is called vectors and I didn't mention a thing called basis to define them.

Basis and Generators

The basis of a vector is then defined to be a set $B \subseteq V$ which has the following two properties:

$$\forall B' \subseteq B\; \text{ with }\; \# B' < \infty: \; \; \bigoplus_{b \in B'} \lambda_b \odot b = \tilde 0 \implies \lambda_b = 0$$

where $\lambda_b \in F$ and $$ \forall v \in V, \; \exists B' \subseteq B \; \text{ with }\; \# B' < \infty:\;\; v = \bigoplus_{b \in B'} \lambda_b \odot b$$

for some $\lambda_b \in F$.

A set $U \subseteq V$ with the first property is called a linear independent set and a set $T \subseteq V$ with the second property is called the generator of the vector space. You have namely

$$B \text{ is a basis} :\iff B \text{ is a generator of the vector space and linear independent}$$

In a linear algebra class it is shown that all the bases of $V$ has the same cardinality. We define there for the dimension of $V$ to be $\mathrm{dim}(V)=\#B$ for $B$ being a basis of $V$.

Representation of a vector as tuple

If your vector space is finite dimensional ie if $\mathrm{dim}(V)< \infty$ then you can simplify the above conditions as:

$$\bigoplus_{b\in B} \lambda_b \odot b = \tilde 0 \; \text{and} \; \forall v\in V: \; \; v= \bigoplus_{b\in B} \lambda_b \odot b$$

In a linear algebra course it is taught that every vector space (finite or infinite) $V$ has a basis and for a chosen basis the scalars $\lambda_b$ is unique. A basis is called an ordered basis if you order the elements of your basis in a tuple. If your vector space is finite dimensional, and $B$ is an order basis you can define a vector space isomorphism $\iota_B$

$$\iota_B : V \to F^n, \; v = \bigoplus_{b \in B} \lambda_b \odot b \mapsto (\lambda_{b_1}, \dots, \lambda_{b_n}) $$

where $b_i \in B$ $\forall i = 1 \dots n$ and $n = \#B$. There you can see the components of the vector $b$ as the numbers $\lambda_{b_i} \in F$. Note that even though the representation of a vector $v \in V $ as a n-tuple is unique for a given basis, there is no unique representation for the vector $v \in V$ for all bases. As an example will consider the set $V:=\real^2$ as a $F:=\real$ vector space, where $\oplus$ and $\odot$ are defined in the following fashion. I'll denote the elements of $V$ with square brackets and their representations with normal brackets to avoid confusion. $$[a,b] \oplus [c,d] = [a+c, b+d]$$ and $$\lambda \odot [a,b] = [\lambda \cdot a, \lambda \cdot b]$$ for all $a,b,c,d \in F$ and $[a,b],[c,d] \in V$. I leave you to check the vector space axioms. Note that whilst writing $[a,b] \in \real^2$ I'm not referring to any basis. It is merely a definition of being in $\real^2$ that let's me write $[a,b]$.

Now let $B=\{b_1,b_2\}$ with $b_1 = [1,0]$ and $b_2=[1,0]$. Note that $\lambda \odot b_1 \oplus \mu \odot b_2 = [0,0] \implies \mu= \lambda =0 $ so $B$ is linear independent. Furthermore $[a,b] = a \odot b_1 + b \odot b_2$, $\forall [a,b] \in V $ as you can easily check. Thus the isomorphism $\iota_B$ well-defined and you get

$$\iota_B([a,b]) = (a,b) \;\; \forall [a,b] \in V$$

However for another Basis $C=\{c_1, c_2\}$ with $c_1 =[-1,0]$ and $c_2=[0,2]$ you get:

$$\iota_C([a,b])=(-a,b/2) \;\; \forall [a,b] \in V$$

as you can easily check.

Note that in this example your vectors are $[a,b]\in \real^2$, which exist as an element of $\real^2$ independent of any basis.

Another example of a $n$-dimensional vector space could be the solutions of a homogeneous linear differential equation of degree $n$. You should play with it choose some basis and represent them as tuples. Note that in this case your vectors are functions, which solve that particular differential equation.

In the case of a infinite dimensional vector space things get a little bit difficult but to understand the basis concept finite dimensional vector spaces are the best way to go.

Inner/Scalar product

Let $V$ be a vector space, and $\field = \{\real, \cx\}$ be either real or complex numbers. Furthermore I'll now replace $\otimes \to +$ and $\odot \to \cdot$ to make my point more clear. A scalar product is a function $\ip \cdot \cdot: V \times V \to \field$, $(v,w) \mapsto \ip vw$ with following properties:

  • $\forall v_1,v_2,w \in V, \; \forall \lambda \in \field: \;\; \ip{v_1 + \lambda v_2} w = \ip{v_1} w + \lambda^* \ip{v_2}w $
  • $\forall v,w_1,w_2 \in V, \; \forall \lambda \in \field: \;\; \ip{v}{ w_1 + \lambda w_2} = \ip{v}{ w_1} +\lambda \ip{v}{w_2}$
  • $\forall v,w \in \field: \;\;\ip vw = \ip wv^*$
  • $\forall v \in V\setminus\{0\}:\;\; \ip vv \in \real_{>0}$

Again note that $\ip \cdot\cdot$ could be any function with these properties. Furthermore I didn't need a basis of $V$ to define the scalar product, so it cannot possibly be dependent of the basis chosen. A vector space with an scalar product is called an inner product space.

As an example take the polynomials of degree $\leq n$, defined on an interval $I\subset\real$. You can easily show that $$V:=\{P:I \to \real \, | \, P \text{ is a polynomial and degree of } P \leq n\}$$ is a vector space. Furthermore we define

$$\forall P,Q \in V\;\; \ip PQ_1 := \int_0^1 P(x) \cdot Q(x) \, \mathrm d x $$

Note that this function is a valid scalar product on $V$. However I can also define $$\forall P= p_n x_n + \dots + p_0 , \,Q = q_n x_n + \dots q_0 \in V \;\; \ip PQ_2 := p_nq_n + \dots + p_0 q_0 $$

which is also a nice definition for a scalar product. Again I'm not referring to any basis as I write $P= p_n x_n + \dots + p_0$. It is just a definition of being in $V$. It is clear that there is no unique scalar product defined on a particular vector space.

Representation of $\ip \cdot \cdot$ wrt a Basis

Now if I choose a ordered basis $B$ of $V$ then I can simplify my life a little bit. Let $v\in V$ with $v= \sum_{i=1}^n v_i b_i$ and $w\in V$ with $w = \sum_{j=0} ^n w_i b_i$. Then I can write:

$$\ip vw = \ip{\sum_{i=1}^n v_i b_i}{\sum_{j=0} ^n w_j b_j} = \sum_{i=1}^n\sum_{j=1}^n v_i^* w_j \ip {b_i} {b_j}$$

You see now that this looks like a matrix product of $\iota_B(v)^\dagger \cdot A \cdot\iota_B(w) $, where $A=(a_{ij}):=\ip{b_i}{b_j}$. Note that this representation of $\ip\cdot \cdot$ depends of the basis chosen. Having chosen a basis however, you can just do the matrix multiplication to get the scalar product.

For the above example with polynomials and the inner product $\ip \cdot \cdot _2$, if you choose the basis to be $b_i = x^i$ then you get for $A = \mathrm{diag}(1, \dots, 1)$

Hilbert Space

A Hilbert space is an inner product space and as a metric space it is complete with the metric induced by the scalar product. This means your norm is defined by $\lVert v \rVert := \sqrt{\ip vv}$ and the metric is defined by $d(v,w):= \lVert v-w \rVert$. For what it means for a space to be complete you can check the Wikipedia article.

Upshot

There is very clear distinction between the vector itself as an element of the set $V$ and its representation as a tuple in $F^n$ with respect to a basis $B$ of $V$. The vector itself exist regardless of any basis, whereas the representation of the vector is basis dependent. The same goes for the scalar product and its representation.

In physics one automatically assumes the standard basis for $\real^n$, which has vectors with zeros everywhere except for one component, that is equal to one, and calculates everything in representation of vectors without specifying any basis, which in turn creates the illusion that vectors are its components and a vector without its a component is unimaginable.

Since a vector space could almost be anything it is really hard to imagine what a vector is without referring to its components. The best way to do that in my opinion is to accept the fact that a vector is just an element the set $V$ and nothing more. For example if you write

$$\left|\psi \right> = c_1 \left|+ \right> + c_2 \left| - \right>$$

you barely say that $\left| \psi \right>$ can be written as a linear combination of $\left| \pm \right>$. So this does not refer to any basis. However the moment you write

$$\left|\psi \right> = c_1 \left|+ \right> + c_2 \left| - \right> = \begin{pmatrix} c_1 \\ c_2 \end{pmatrix}$$

first of all a mathematician dies because $\left|\psi \right> = \begin{pmatrix} c_1 \\ c_2 \end{pmatrix}$ doesn't make any sense as $\left|\psi \right> \in \mathcal H$ and $\begin{pmatrix} c_1 \\ c_2 \end{pmatrix} \in \cx^2$. Regardless of that $\begin{pmatrix} c_1 \\ c_2 \end{pmatrix}$ refers to an ordered basis of $\mathcal H$ namely the basis $B=\{\left|+ \right>, \left|- \right> \}$ and this representation depends on the basis that you've chosen.

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  • $\begingroup$ Please don't use renewcommand, as this is applied to everyone's post and not just yours, see meta.physics.stackexchange.com/q/6951 for more details. $\endgroup$ – Kyle Kanos Feb 5 '16 at 16:10
  • $\begingroup$ Sorry I didn't know that. I changed it now. $\endgroup$ – Gonenc Feb 5 '16 at 16:13
  • $\begingroup$ Thanks a lot for writing this up, this has been most helpful. You showed neatly how the axioms of a vector space, vectors and inner products don't assume any basis, all well, but two subquestions if i may: i) without a basis chosen, how do I go about comparing two vectors $v,w \in V$ with one another? ii) you showed that a function that satisfies the axioms for $\langle,\rangle$ would be a valid inner product, so can I come up with an inner product that does not require a basis for its computation? $\endgroup$ – user929304 Feb 6 '16 at 10:03
  • $\begingroup$ It depends on what you mean by comparing. They are equal if $v-w =0$ and not equal otherwise. You can calculate the distance between them if $V$ is a metric space by $d(v,w)$, where $d(\cdot, \cdot)$ satisfies these axioms but in general there is no way to tell if $v> w$ or $w>v$ since most of the time $V$ is not totally ordered ie. you cannot say $(0,1),(0,1) \in \real^2$ $(0,1)\gtrless (1,0)$. $\endgroup$ – Gonenc Feb 6 '16 at 10:42
  • $\begingroup$ For the second question: Note that I don't need any basis to define $\ip \cdot \cdot$ so by definition the inner product is independent of any basis. Thus you don't need to find and inner product for which the inner product is basis independent. In short every inner product is basis independent. $\endgroup$ – Gonenc Feb 6 '16 at 10:44
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Suppose $\left\{\left|e_i\right\rangle|i\in I\right\}$ is an orthonormal basis of a Hilbert space $\mathcal{H}$, viz. $\left\langle e_i |e_j\right\rangle =\delta_{ij}$. Then the identity operator from $\mathcal{H}$ to $\mathcal{H}$ can be written as an outer product $$\mathbb{I}=\Sigma_{i\in I}\left|e_i\right\rangle\left\langle e_i\right|\quad\left(\ast\right),$$because kets then satisfy $\left|\psi\right\rangle=\Sigma_{i\in I}\left|e_i\right\rangle\left\langle e_i|\psi\right\rangle$, and similarly with bras.

If $\left\{\left|f_j\right\rangle|j\in I\right\}$ is another orthonormal basis, we may write $\left|e_i\right\rangle = \Sigma_{j\in I} U_{ij}\left|f_j\right\rangle$ with $U_{ij}$ a unitary matrix so that $\left|e_i\right\rangle = \Sigma_{k\in I}\delta_{ik}\left|e_k\right\rangle,\,\left|f_j\right\rangle = \Sigma_{l\in I}\delta_{jl}\left|f_l\right\rangle$. Hence $\left\langle e_i\right| = \Sigma_{j\in I}U_{ij}^\ast\left\langle f_j\right|=\Sigma_{j\in I}\left(U^\dagger\right)_{ji}\left\langle f_j\right|$. Finally, $$\Sigma_{i\in I}\left|e_i\right\rangle\left\langle e_i\right|=\Sigma_{i,\,j,\,k\in I}\left(U^\dagger\right)_{ki}U_{ij}\left|f_j\right\rangle\left\langle f_k\right|=\Sigma_{j,\,k\in I}\delta_{kj}\left|f_j\right\rangle\left\langle f_k\right|=\Sigma_{j\in I}\left|f_j\right\rangle\left\langle f_j\right|,$$ so our two "identity operators" match. Thus either can be used to compute a ket as per Eq. $\left(\ast\right)$; the basis choice doesn't matter.

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It's hard to picture the things you want because you're working in an infinite-dimensional space, and I don't know anybody who can intuitively 'see' in that kind of space.

Instead, let's consider a finite-dimensional space: the polarization of a massless spin 1 particle traveling along the $z$ axis. The polarization state is simply described by a unit vector in the $xy$ plane. The 'geometric meaning' of the state is just the direction the arrow is pointing, i.e. the direction of the polarization. The inner product between two states is simply $\cos^{-1}(\theta)$, where $\theta$ is the angle between them; this is clearly basis independent.

Quantum computers scientists have a similar visualization, where they plot the state of qubits on the Bloch sphere. It's not quite as nice as our picture, because it's suited for the polarizations of spin 1/2 particles, so all the angles get doubled. Still, it provides a geometric picture of 'what the state is' without having to choose a basis.

It's much harder to see these things in an infinite-dimensional space, but the math continues to hold.

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protected by Qmechanic Feb 5 '16 at 15:57

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