New answers tagged hilbert-space
4
Rigged Hilbert spaces have no special relationship to quantum field theory. We can talk about more general elements of "the" Hilbert space such as wave functions looking like distributions. They don't have a finite norm but they're still useful to talk about.
Truly physical states that may be realized in practice are normalizable - their norm may be chosen ...
2
The dimension of the Hilbert space of a free particle is countable. To see this, simply note that
The Hilbert space of a free particle in three dimensions is $L^2(\mathbb{R}^3)$.
The dimension theorem guarantees that any two bases of of a vector space have the same cardinality, which allows us to define the dimension of a vector space as the cardinality ...
1
Q1: That looks like a typo. If it is supposed to be an eigenvalue equation (it probably is), then it is indeed $|\psi_n\rangle$ on both sides.
Q2: A state is described by a vector in Hilbert space. This is an abstract mathematical object, but it is often convenient to choose a basis and use the components in that basis to describe the state. If you have a ...
2
There is no necessity to start from the Gaussian measure on $\mathbb{C}^{2N}$. Any $U(N)$ invariant measure would result the same $U(N)$ invariant measure of the Grassmannian. However, this is the standard choice for two reasons.
1) When GL(2) is factorized out, the resulting measure on GL(2) is the "Ginibre measure" , Please see for example: ...
3
I) First note that there is a group action $\rho: GL(2,\mathbb{C})\times u(2) \to u(2)$ given by
$$\tag{A} g\quad \mapsto\quad\rho(g)\sigma~:= ~g\sigma g^{\dagger},
\qquad g\in GL(2,\mathbb{C}),\qquad\sigma\in u(2). $$
In detail, the Lie group
$$\tag{B} GL(2,\mathbb{C})~:=~ \{ g\in {\rm Mat}_{2\times 2}(\mathbb{C}) \mid \det(g)\neq 0\}$$
acts ...
0
In the comments Alfred raises the notion of the dual space. In fact, if you try to read Dirac's principles of QM, you will find that he starts with dual space.
In Dirac notation $|z\rangle$ is an element of an abstract vector space $\mathcal{H}$. Then, there is a notion of dual space: the dual space $\mathcal{H}^*$ is the space of all (continious) linear ...
2
Without loss of generality, let's take the $|\lambda_i\rangle$ to be orthonormal. Notice that, by the spectral theorem, the hamiltonian can be written as follows:
$$
H = \sum_i \lambda_i P_i, \qquad P_i = |\lambda_i\rangle\langle \lambda_i|
$$
Each operator $P_i$ is a projectors onto the subspace spanned by $|\lambda_i\rangle$. Notice, in particular, ...
7
Starting with:
$$U(t,t_i) = e^{\frac{-i}{\hbar }H(t-t_i)}$$
If $t_i=0$:
$$U(t,0) = e^{\frac{-i}{\hbar }Ht}$$
Using the identity: $\sum\limits_i \left|\lambda_i\right>\left<\lambda_i\right|=\mathbb{I}$
$$U(t,0) = \sum\limits_i e^{\frac{-i}{\hbar }Ht}\left|\lambda_i\right>\left<\lambda_i\right|$$
Since the exponential of an operator is (by ...
1
From Wiki:
For a finite-dimensional vector space, using a fixed orthonormal
basis, the inner product can be written as a matrix multiplication of
a row vector with a column vector:
$ \langle A | B \rangle = A_1^* B_1 + A_2^* B_2 + \cdots + A_N^* B_N = \begin{pmatrix} A_1^* & A_2^* & \cdots & A_N^* \end{pmatrix} \begin{pmatrix} B_1 \\ ...
2
Although I think user Siva's argument is nice for intuition, I feel that the key mathetmatical point is being obscured; you just need to be careful about what you mean by the "size" of a vector space.
The dimension theorem for vector spaces tells us that any two bases for a vector space must have the same cardinality. This allows us to define the dimension ...
6
This question first posed to me by a friend of mine. For the subtleties involved, I love this question. :-)
The "flaw" is that you're not counting the dimension carefully. As other answers have pointed out, $\delta$-functions are not valid $\mathcal{L}^2(\mathbb{R})$ functions, so we need to define a kosher function which gives the $\delta$-function as a ...
6
The Hilbert space ${\cal H}$ of the one-dimensional harmonic oscillator in the position representation is the set $L^{2}(\mathbb{R})$ of square integrable functions $\psi:\mathbb{R}\to\mathbb{C}$ on the real line. The Dirac delta distribution $\delta(x-x_{0})$ is not a function and it is not square integrable. See also this Phys.SE post.
2
If you are acquainted to matrices, then $|\psi\rangle$ is very much like column vector, and $\langle\psi|$ is similar to row vector. Operators correspond to square matrices. The conjugate transpose $^\dagger$ is similar to the matrix transpose $^\mathrm{T}$ in the sense that it turns columns to rows and vice versa.
Then (neither of your 1-3 lines), if we ...
2
To formalize the comments as an answer:
The difference between requiring
$$(\alpha u,v)=\alpha(u,v)\quad\text{ (mathematician's definition)}$$
and
$$\langle u, \alpha v\rangle=\alpha\langle u,v\rangle\qquad\quad\,\,\text{ (physicist's definition)}$$
is purely one of convention, and the two definitions are equivalent as $(u,v)=\langle v,u\rangle$. There's no ...
2
I will try to cover all your questions, but I really recommend a book as the topic is too wide spread for such a "short" answer.
You can think of a Hilbert space as an infinite dimensional vector space, that has similar properties as one knows from $\mathbb{R}^3$:
The elements are vectors and will be denoted by $\phi$
You can measure angles and lengths ...
4
I'll try to answer your questions while explaining as much basic quantum mechanics as possible, so that you'll be able to fill in the blanks with only linear algebra.
Hilbert space- As a novice to QM i am very sad that in none of the books i have read i found the reasons for using Hilbert space H at first place followed by a full geometrical ...
6
The usual convention is that $\langle 0 \vert 0 \rangle$ is indeed $1$. This is true because we are free to pick any overall normalization for $\lvert 0 \rangle$ given its definition of "the eigenvector of $\hat{a}$ with eigenvalue $0$." Scaling an eigenvector keeps it an eigenvector with the same eigenvalue. Thus we declare it to have a normalization that ...
-2
We have to be careful with the bra-ket formalism and its meaning. Unlike $|x_1>$, I am not sure that the notation $|x_1 x_2>$ where $x_1$ and $x_2$ are positional coordinates makes any sense. In literature [1] the notation $|ab>$ designates the Slater determinant or Hartree-Fock state, i.e.:
$$|ab>=c_a^{\dagger}c_b^{\dagger}|0>=\phi_a(x_1) ...
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