77

The state \begin{equation} |\Psi \rangle = \frac{1}{\sqrt{2}}\left(|\psi_1\rangle +|\psi_2\rangle \right) \end{equation} is a pure state. Meaning, there's not a 50% chance the system is in the state $|\psi_1\rangle$ and a 50% it is in the state $|\psi_2\rangle$. There is a 0% chance that the system is in either of those states, and a 100% chance the system ...


59

The Bloch sphere is beautifully minimalist. Conventionally, a qubit has four real parameters; $a e^{i\chi} |0\rangle + b e^{i\phi} |1\rangle.$ However, some quick insight reveals that the $a$-vs-$b$ tradeoff only has one degree of freedom due to the normalization $a^2 + b^2 = 1$ and some more careful insight reveals that, in the way we construct expectation ...


51

I don't know of any books which use this language exclusively, but the basic idea is pretty straightforward: All Hilbert spaces are isomorphic (if their dimensions match). This would present conceptual problems in quantum mechanics if we ever talked about the Hilbert space alone; how could we distinguish them? But it's OK because we are actually ...


51

Well, the Dirac delta function $\delta(x)$ is a distribution, also known as a generalized function. One can e.g. represent $\delta(x)$ as a limit of a rectangular peak with unit area, width $\epsilon$, and height $1/\epsilon$; i.e. $$\tag{1} \delta(x) ~=~ \lim_{\epsilon\to 0^+}\delta_{\epsilon}(x), $$ $$\tag{2} \delta_{\epsilon}(x)~:=~\frac{1}{\epsilon}...


50

The assumption that a measurable property exists whether or not we measure it is inconsistent with the experimental facts. Here's a relatively simple example. Suppose we have four observables, $A,B,C,D$, each of which has two possible outcomes. (For example, these could be single-photon polarization observables.) For mathematical convenience, label the two ...


43

They are not the same. One is the outer product and one is the inner product. In the finite-dimensional real-number case for example, if $$|x\rangle = \begin{bmatrix}1\\2\\3\end{bmatrix},$$ then $$|x\rangle\langle x| = \begin{bmatrix}1\\2\\3\end{bmatrix} \begin{bmatrix}1&2&3\end{bmatrix} = \begin{bmatrix}1&2&3\\2&4&6\\3&6&9\...


41

Before reading this answer (and to those who are downvoting), I am addressing if the cat is both alive and dead. I don't think the question is asking for a complete explanation of the Schrodinger's cat experiment, nor is it asking how this links to all of the deeper mysteries of quantum mechanics and how we should think of them. Therefore, while there is ...


37

I) Well, one can identify a complex-valued observable with a normal operator $$\tag{1} A^{\dagger}A~=~AA^{\dagger}.$$ A version$^1$ of the spectral theorem states that an operator $A$ is orthonormally diagonalizable iff $A$ is a normal operator. Thus normal operators are the only kind of operators that we can consistently extract measurements [i.e. ...


36

You need nothing more than your understanding of $$ \int_{-\infty}^\infty f(x)\delta(x-a)dx=f(a) $$ Just treat one of the delta functions as $f(x)\equiv\delta(x-\lambda)$ in your problem. So it would be something like this: $$ \int\delta(x-\lambda)\delta(x-\lambda)dx=\int f(x)\delta(x-\lambda)dx=f(\lambda)=\delta(\lambda-\lambda) $$ So there you go.


36

The canonical commutation relations are not well-defined on finite-dimensional Hilbert spaces. The canonical prescription is $$ [x,p] = \mathrm{i}\hbar\mathbf{1}$$ and, recalling that the trace of a commutator must vanish, but the trace of the identity is the dimension of the space if it is finite-dimensional, we conclude that we have a space for which the ...


34

A mixed state is mathematically represented by a bounded, positive trace-class operator with unit trace $\rho : \cal H \to \cal H$. Here $\cal H$ denotes the complex Hilbert space of the system (it may be nonseparable). The set of mixed states $S(\cal H)$ is a convex body in the complex linear space of trace class operators $B_1(\cal H)$ which is a two-side ...


34

There are many different problems with interactions; or, rather, many manifestations of the same problem. For example, interactions are always non-linear in the equations of motion, e.g., $$ (\partial^2+m^2)\phi=\lambda\phi^3 $$ or similar equations for QED. As the operators are distributions, their products are ill-defined, and there is no rigorous method ...


29

Note: There is a short summary at the bottom. This is actually also described in Nielsen&Chuang: You don't learn about general measurements, because they are completely equivalent to projective measurements + unitary time evolution + ancillary systems, all of which is described in your usual QM formalism. The Measurement Postulate Let's start from ...


28

One problem with the given $3\times 3$ matrix example is that the eigenspaces are not orthogonal. Thus it doesn't make sense to say that one has with 100% certainty measured the system to be in some eigenspace but not in the others, because there may be a non-zero overlap to a different eigenspace. One may prove$^{1}$ that an operator is Hermitian if and ...


28

Our physics prof once put it informally that way: A state is a set of variables describing a system which does not include anything about its history. The set of variables (position, velocity vector) describes the state of a point mass in classical mechanics, while the path how the point mass got from point $A$ to point $B$ is not a state.


27

This question was 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 ...


27

If $\cal H$ is a complex Hilbert space, and $A :D(A) \to \cal H$ is linear with $D(A)\subset \cal H$ dense subspace, there is a unique operator, the adjoint $A^\dagger$ of $A$ satisfying (this is its definition) $$\langle A^\dagger \psi| \phi \rangle = \langle \psi | A \phi \rangle\quad \forall \phi \in D(A)\:,\forall \psi \in D(A^\dagger)$$ with: $$D(A^\...


27

Hilbert spaces of infinite dimension are necessary, in the minimal case, to describe the non-relativistic quantum mechanics of a massive particle with at least a single real degree of freedom, and they are needed to allow the theory to describe, in general, states with arbitrarily high level of detail and at arbitrarily far-away positions. However, for any ...


27

The definition of a state of a system, in physics, strongly depends on the area of physics one is dealing with and it comes as one of the initial definitions once such underlying theory has to be set up. In particular one has: classical mechanics: a state of a system is a point $m\in TQ$ (or equivalently $T^*Q)$ in the tangent bundle of the configuration ...


27

This is definitely not a dumb question. If we work in a (linear) Hilbert space, then our inner product $\langle \cdot,\cdot \rangle$ induces the usual natural flat metric (given by $d(\psi,\phi) = || \psi - \phi ||$). However, often we take the viewpoint that our states are elements of projective Hilbert space $\mathbb CP^n$. Then it is more natural to ...


26

First, a historical subtlety: Schrödinger has actually stolen the idea of the cat from Einstein. Second, both men – Einstein and Schrödinger – used the thought experiment to "explain" a point that was wrong. They thought it was absurd for quantum mechanics to say that the state $a|{\rm alive}\rangle+b|{\rm dead}\rangle$ was possible in Nature (it was ...


26

There are at least three notions of basis depending on the mathematical structure you are considering. I will quickly discuss three cases relevant in physics (topological vector spaces are relevant too, but I will not consider them for the shake of brevity). (1) Pure algebraic structure (i.e. vector space structure over the field $\mathbb K=$ $\mathbb R$ ...


25

UPDATE: the insights and conceptual meanings advocated below in this answer are detailed and technically developed, for example, in these wonderful articles: Fuchs; Peres - Quantum Mechanics Needs No Interpretation, Physics Today (2000), vol. 53, issue 3, p. 70. Englert - On Quantum Theory, Eur. Phys. J. D (2013) 67: 238. Duvenhage - The Nature of ...


25

The Hilbert space ${\cal H}$ of the one-dimensional harmonic oscillator in the position representation is the set $L^2(\mathbb{R})={\cal L}^2(\mathbb{R})/{\cal N}$ (of equivalence classes) of square integrable functions $\psi:\mathbb{R}\to\mathbb{C}$ on the real line. The equivalence relation is modulo measurable functions that vanish a.e. The Dirac delta ...


25

Basically the answer is yes, the cat is both dead and alive. People used to discuss this sort of thing in terms of the Copenhagen interpretation (CI) and the Many-Worlds interpretation (MWI), but those discussions tend not to be satisfying, because both CI and MWI are designed so that in almost all real-world measurements, they give the same predictions. A ...


24

I understand the definition of a Hilbert space. But I do not understand why non-commutativity compels us to use Hilbert spaces. It doesn't, but that's not what Scrinzi is saying. The reason is doesn't is because we could work, for example, in Wigner quasiprobability representation: $$\rho\mapsto W(x,p) = \frac{1}{\pi\hbar}\int_{-\infty}^\infty\langle x+x'|\...


24

This question gets to the heart of what makes quantum mechanical amplitudes different from classical probabilities. It is true that if you make measurements in the basis of states $\{\lvert 00 \rangle ,\lvert 11 \rangle\}$ then the two states have the same measurement statistics, and so cannot be distinguished. The interesting thing is that it is possible to ...


23

Here's a mathematician's stab at an answer. The tl;dr version: "symmetric" and "self-adjoint" are the same thing for bounded operators, whilst they are the same thing for unbounded operators only insofar as the Aharonov--Bohm effect doesn't exist! A densely defined operator $A$ on your Hilbert space $H$ is symmetric if for any vectors $\xi$ and $\eta$ in ...


23

That the eigenfunctions of the free Hamiltonian $H\propto p^2$ are not actually normalizable due to its completely continuous spectrum and therefore cannot be actual quantum states is well-known, although rarely suitably emphasized. (See e.g. Why are eigenfunctions which correspond to discrete/continuous eigenvalue spectra guaranteed to be normalizable/non-...


23

The relation $$ \langle Ay | x \rangle = \langle y | A x \rangle \text{ for all } x \in \text{Domain of } A\tag1$$ makes no reference to any basis at all, so it is indeed basis-independent. In fact, this definition, which seems pretty strange when you first meet it, arises precisely out of a desire to make things basis-independent. The particular ...


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