The Fock space is a special kind of Hilbert space. It is the Hilbert space of a free field theory or, equivalently, an infinite-dimensional harmonic oscillator

The Fock space is a special kind of Hilbert space. It is the Hilbert space of a free field theory or, equivalently, an infinite-dimensional harmonic oscillator

The Hilbert space of all realistic systems is infinite-dimensional; the fock space of quantum field theory include states which have many particles in it. however, this does not mean the properties of the system is simple to work out by following every particles in that state.

The Hilbert space of all realistic systems is infinite-dimensional;

Where does the wave function of the universe live? Please describe its home

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Where does the wave function of the universe live? Please describe its home. I think this is the Hilbert space of the universe. (Greater or lesser, depending on which church you belong to.) Or maybe it is the Fock space of the universe, or some still bigger, yet more complicated stringy thingy. I will leave it to you whether you want to describe the observable universe, the total universe, or even the multiverse. Please give a reasonably accurate and succinct mathematical description, including at least dimensionality. Thank you. quantum-field-theory universe wavefunction hilbert-space multiverse share|improve this question asked Apr 10 '12 at 12:54 Jim Graber 2,7431026

2   I dont know why this is but the funny formulation of this question makes me chuckle somehow ;-). Stephen Hawking writes a bit about this issue in his "Great Design" book. He considers some kind of a path integral over the whole multiverse (each path corresponds to the evolution of one of the 10^**** individual members). The evolution of our own universe from the beginning to the present would then be described by the most probable (or classical limit) path. –  Dilaton Apr 10 '12 at 13:21 @Dilaton: I guess the book was grand design.. –  Vineet Menon Apr 10 '12 at 15:27 @VineetMenon Whoops yes, you are right. The typo is probably due to the fact that I thinkd this design is great, ha ha :-) –  Dilaton Apr 10 '12 at 15:32 I don't think there is any consensus about what a wavefunction of the universe means, let alone how to formulate it. Maybe someone in theoreticalphysics.stackexchange.com could comment. –  John Rennie Apr 10 '12 at 17:55 add a comment |

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The wave function lives in the quantum café, see the segment from 3:40 or so to the end of http://www.youtube.com/watch?v=unJ2ajHH-94 More seriously, a wave function is a more special name of the "state vector" which is the element of the Hilbert space H  , a complex vector space with an inner product. The Hilbert space of all realistic systems is infinite-dimensional; for an infinite dimension, one can't really say whether the basis is countable or as large as a continuum because these two bases are actually fully equivalent. Finite-dimensional Hilbert spaces are only used as simplified toy models for some aspects of some physical systems. But they're still very important in theory and practice because realistic situations are often composed of similar small Hilbert spaces by taking tensor products. The two-dimensional Hilbert spaces (e.g. spin-up vs spin-down) seem very simple but they're already very rich and are used as tools to teach quantum mechanics. Quantum computing usually takes place in Hilbert spaces for N  qubits which is 2 N   -dimensional, also finite-dimensional. The remaining infinitely many states of a real physical system are assumed to be inaccessible so we may "truncate" the Hilbert space. But note that systems as simple as en electron orbiting a proton or a harmonic oscillator already have an infinite-dimensional Hilbert space. The Fock space is a special kind of Hilbert space. It is the Hilbert space of a free field theory or, equivalently, an infinite-dimensional harmonic oscillator. One usually defines the free - bilinear - Hamiltonian on the Fock space, too. If we don't say that there's a Hamiltonian, the identity of the Fock space is actually meaningless because all infinite-dimensional Hilbert spaces are isomorphic or "unitary equivalent" to each other. So the Fock space isn't really "something completely different" (or larger) than the Hilbert space; it's a special case of it. The same thing holds for the Hilbert spaces associated with any theory you can think of (describing the world around us or describing a fictitious or hypothetical world), whether it's the Standard Model, the Minimal Supersymmetric Standard Model, or – the most comprehensive theory – String Theory. All these theories, much like any other theories respecting the postulates of quantum mechanics, have their own Hilbert space and all these infinite-dimensional spaces in string theory or a simple infinite-dimensional harmonic oscillator or even a simple Hydrogen atom are actually isomorphic to each other. The theories only differ by different Hamiltonians – or other dynamical laws that describe the evolution in time. Also, one should mention that the actual state of the physical system isn't given by all the information included in an element of the Hilbert space. The phase and the absolute normalization – i.e. the full multiplicative factor that may be complex – is unphysical. So the space of inequivalent "pure states" is actually the quotient H/C ∗   . Aside from "wave functions" i.e. pure states that are elements of the Hilbert space, up to a normalization, one may also describe a physical system by a more general "density matrix" which lives in the space of Hermitian matrices ρ  . For pure states, ρ=|ψ⟩⟨ψ|  and the phase cancels. However, there are also more general mixed states that are superpositions of similar terms. share|improve this answer edited Apr 11 '12 at 8:40 answered Apr 11 '12 at 6:43 Luboš Motl 108k7155312

I am having a lot of trouble with this one universal infinite dimensional Hilbert space. if the dimensionality of the universe doesn't tell us, how do we know if there is one spatial dimension or three or seventy seven? Also how do we know how many particles are in our universe? TIA –  Jim Graber Apr 11 '12 at 12:25 2   The Hilbert space is a mathematical construction that has essentially nothing to do with real space. So saying that the Hilbert space is infinite-dimensional does not imply anything about the spatial dimensionality of the universe. That is a separate issue. –  David Z♦ Apr 11 '12 at 19:40 Dear @Jim, I agree with David. In QFT, you may try to determine the spacetime dimensions by isolating one-particle states and finding that the Hilbert space of one-particle states has a simple basis diffeomorphic to R d−1   , the spatial momentum, or in an equivalent way. But the Hilbert space is much greater than the regular space. Every basis vector of the Hilbert space in a basis corresponds to one mutually exclusive state in which the whole physical system may be. There are usually infinitely many. –  Luboš Motl Apr 12 '12 at 12:17 So the choice of a greater or lesser Hilbert space only makes a difference for finite-dimensional Hilbert spaces? Because for an infinite Hilbert space the two are the same? Or at least isomorphic? –  Jim Graber Apr 14 '12 at 10:33 The youtube video you linked to is dead. Maybe the wave function is vacationing at the Hilbert Hotel? .... I'll see myself out –  David H Sep 7 '13 at 3:54 add a comment |

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The word "space" in mathematics is not the same ontological object as physical space. It is sort of equivalent to asking, in classical physics, where is the space of all velocity vectors located? Its not that they are actually somewhere "out there", they are just mathematical abstractions from which useful information can be extracted and inferences can be made unto measurable quantities which are analogous to the abstraction. A vector space (Hilbert, Fock, and many other variations of vector spaces exist!) is a mathematical object which makes convenient many computations one can do with sets of numbers(vectors, matrices, tensors, etc...) endowed with human-invented algebraic properties (closure, commutativity, associativity, etc...). Quantum mechanics makes use of linear algebra almost solely for the fact that one can extract 'spectrum' of eigenvalues which are analogous to the measurable discrete quantities one finds when dealing with such objects. 一叶斋 一叶障目，一叶知秋 博问 闪存 联系 订阅 管理 三维旋转：旋转矩阵，欧拉角，四元数 原文见我的博客主站，欢迎大家过去评论。 如何描述三维空间中刚体的旋转，是个有趣的问题。具体地说，就是刚体上的任意一个点P(x, y, z)围绕过原点的轴(i, j, k)旋转θ，求旋转后的点P\'(x\', y\', z\')。 旋转矩阵 旋转矩阵乘以点P的齐次坐标，得到旋转后的点P'，因此旋转矩阵可以描述旋转， ⎡ ⎣ ⎢ ⎢ ⎢ x ′ y ′ z ′ 1 ⎤ ⎦ ⎥ ⎥ ⎥ =R⋅⎡ ⎣ ⎢ ⎢ ⎢ xyz1 ⎤ ⎦ ⎥ ⎥ ⎥   绕x，y，或z轴旋转θ的矩阵为： R x (θ)=⎡ ⎣ ⎢ 100 0cosθsinθ 0−sinθcosθ ⎤ ⎦ ⎥   R y (θ)=⎡ ⎣ ⎢ cosθ0sinθ 010 −sinθ0cosθ ⎤ ⎦ ⎥   R z (θ)=⎡ ⎣ ⎢ cosθsinθ0 −sinθcosθ0 001 ⎤ ⎦ ⎥   所以，绕任意轴旋转的矩阵为 R x (−p)⋅R y (−q)⋅R z (θ)⋅R y (q)⋅R x (p)  这表示： 1. 绕x轴旋转角度p使指定的旋转轴在xz平面上 2. 绕y轴旋转角度q使指定的旋转轴与z轴重合 3. 绕z轴旋转角度θ 4. 绕y轴旋转角度-q 5. 绕x轴旋转角度-p 其中，p和q的值需要用i,j,k计算出来。 欧拉角 欧拉角也可以描述三维刚体旋转，它将刚体绕过原点的轴(i,j,k)旋转θ，分解成三步（蓝色是起始坐标系，而红色的是旋转之后的坐标系。）。   1. 绕z轴旋转α，使x轴与N轴重合，N轴是旋转前后两个坐标系x-y平面的交线 2. 绕x轴（也就是N轴）旋转β，使z轴与旋转后的z轴重合 3. 绕z轴旋转γ，使坐标系与旋转后的完全重合 按照旋转轴的顺序，该组欧拉角被称为是“zxz顺规”的。对于顺规的次序，学术界没有明确的约定。 欧拉角的旋转矩阵为： R z (α)⋅R x (β)⋅R z (γ)  在旋转矩阵一节中，最先进行的旋转其矩阵在最右侧，说明该矩阵最先与点的齐次坐标相乘，旋转矩阵按照旋转的次序从右向左排列。而在欧拉角中，最先进行的旋转其旋转矩阵在最左边。这是因为，**对于前者（旋转矩阵），我们始终是以绝对参考系为参照来的，对于后者（欧拉角），我们每一次旋转的刻画都是基于刚体的坐标系。**比如，在欧拉角中的第2步，绕x轴旋转β，这里的x轴实际上是N轴了（而不是蓝色的x轴）。 为什么旋转参考系的不同会导致旋转矩阵次序的差异呢？细想一下便知，旋转矩阵左乘叠加用以描述三维变换效果的叠加，这本身就是基于绝对坐标系的，所以旋转矩阵一节没有疑问；而对于欧拉角一节的这种旋转方式，这样考虑： 1. 如果有一个“影子坐标系3”与原坐标系重合，然后首先进行了第3步（绕z轴旋转γ）； 2. 然后有一个“影子坐标系2”也与原坐标系重合，然后与“影子坐标系3”一起（视作同一个刚体）进行了第二步； 3. 最后一个“影子坐标系1”，与前两个坐标系一起进行了第一步。 此时，考察“影子坐标系”1和2，他们就分别落在了欧拉角旋转的两个“快照”上，而“影子坐标系3”就落在旋转后的位置上（红色的）。而在上述过程中，“影子坐标系3”就是相对于绝对坐标系依次进行了第三步，第二步，和第一步。所以欧拉角的旋转矩阵写成那样，也是行得通的。 这个想法，我猜在很多第一人称游戏中，已经得到了广泛应用了。这样，玩家对人物的控制就可以绕开人物的实时状态（位置，角度等）直接对人物的模型矩阵产生影响。 万向节死锁是欧拉角的一个弊端，这是一个直观的例子。 四元数 四元数是今天的主角，它能够很方便的刻画刚体绕任意轴的旋转。四元数是一种高阶复数，四元数q表示为： q=(x,y,z,w)=xi+yj+zk+w  其中，i，j，k满足： i 2 =j 2 =k 2 =−1  ij=k,jk=i,ki=j  由于i，j，k的性质和笛卡尔坐标系三个轴叉乘的性质很像，所以可以将四元数写成一个向量和一个实数组合的形式： q=(v ⃗ +w)=((x,y,z),w)  可以推导出四元数的一些运算性质，包括： * 四元数乘法 q1q2=(v 1  → ×v 2  → +w 1 v 2  → +w 2 v 1  → ,w 1 w 2 −v 1  → ⋅v 2  → )  * 共轭四元数 q ∗ =(−v ⃗ ,w)  * 四元数的平方模 N(q)=N(v ⃗ )+w 2   * 四元数的逆 q −1 =q ∗ N(q)   四元数可以看做是向量和实数的一种更加一般的形式，向量可以视作为实部为0的四元数，而实数可以是作为虚部为0的四元数。上述四元数的运算性质也是实数或向量的运算性质的更一般的形式。 四元数可用来刻画三维空间中的旋转，绕单位向量(x,y,z)表示的轴旋转θ，可令： q=((x,y,z)sinθ2 ,cosθ2 )  刚体坐标系中的点p(P,0)（写成四元数的形式），旋转后的坐标p'为： p ′ =qpq −1   接下来我们来证明这一点。 首先，我们证明 qpq −1 =(sq)p(sq) −1   其中s为实数。显然 (sq)p(sq) −1 =sqpq −1 s −1 =sqp −1   此时，我们可以将q看做是单位矩阵，因为如果q不是单位矩阵，我们就可以乘以一个常数s将其化为单位矩阵。 然后，我们证明qpq^{-1}和p的模长相等 下面将q视为单位四元数： q −1 =q ∗   四元数q的标量: S(q)=(q+q ∗ )/2  那么： 2S(qpq −1 )=2S(qpq ∗ )=qpq ∗ +(qpq ∗ ) ∗ =qpq ∗ +qp ∗ q ∗ =q(p+p ∗ )q ∗ =q2S(p)q ∗ =2S(p)  最后，我们证明 p ′ =qpq ∗   如图所示，u为旋转轴，旋转角度为σ，向量v旋转到w处。旋转到σ/2处为k（图中未标出）。 下面也用相同的字母指代四元数，如u就表示向量u的四元数形式((ux,uy,uz),0)。 首先，令u方向上的单位向量为u（为了方便，命名不变，后面的u都是指旋转轴方向的单位四元数），那么根据q的定义，参见四元数乘法法则： q=(u ⃗ sinθ2 ,cosθ2 )=(v ⃗ ×k ⃗ ,v ⃗ ⋅k ⃗ )=(v ⃗ ,0)(−k ⃗ ,0)=kv ∗   现在令 w=qvq ∗   如果能证明w与v的夹角是σ，那么就说明w确实是v旋转σ得到的，整个命题就得证了。 注意v，k和w都是实部为0的单位四元数，表示单位向量，我们有： wk ∗ =(qvq −1 )k ∗ =qvq ∗ k ∗ =qvvk ∗ k ∗ =q  所以 wk ∗ =kv ∗   上面的式子拆分成实部和虚部，虚部表明w与-k的平面和k与-v的平面重合，实部表明w和-k之间的夹角与k和-v之间的夹角相等，都是π-σ/2。这就说明了w与v的夹角是σ，原命题就得证了。