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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 ...


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 ...


23

Welcome to Stack Exchange! I do not know much about quantum control theory, but I can give you a simple example from regular quantum mechanics: that of a particle in a box. This is one of the simplest systems one can study in QM, but even here an infinite dimensional space shows up. Indexing the energy eigenstates by $n$ so that $$H|n\rangle=E_n|n\rangle$$ ...


21

In Bra-Ket notation $| x \rangle$ is a vector and $\langle x |$ is a covector. Formally, a covector is a map that goes from $V \rightarrow \mathbb{R}$, i.e. it “eats a vector” and gives you a number. When we write $$ \langle x | x \rangle $$ we are saying that the covector $\langle x |$ is acting on the vector $|x\rangle$, so $\langle x | x\rangle$ is a ...


19

Bra-ket notation is useful because it lets you get rid of excess superscripts and subscripts. For example, in conventional vector notation you might call the unit vectors in 3D space $\vec{e}_x$, $\vec{e}_y$, and $\vec{e}_z$. There's something like "$\vec{e}$" that you "hang an index on", in order to specify which basis vector you mean. But in bra-ket ...


16

I feel like all the answers here are missing the point. The cat is not both alive and dead at the same time. That would be, as you put it, ludicrous. The truth is that the cat is in a superposition state of the states "alive" and "dead". The problem is that there is no way to make sense of this statement without studying the underlying mathematics. Humans ...


13

It's not so bad if you just set all the constants to one, $m = \omega = \hbar = 1$. You can always recover these later by dimensional analysis. Classically, we have the Hamiltonian $$H = \frac{p^2 + x^2}{2}.$$ Even without knowing quantum mechanics, you can think of phase space as the complex plane, with $x$ as the real axis and $p$ as the imaginary axis. ...


13

Formally self-adjoint, but unbounded, operators can easily take a normalizable state (i.e a state in the Hilbert space) and make it unnormalizable and therefore no longer in the space. This can lead to all sorts of aparent paradoxes. For example consider the operator $p^4= \partial_x^4$ on infinite square well $[0,1]$. Let it act on the wavefunction $$ \...


12

In a vector space over the field of complex numbers the notion of complex conjugation is basis dependent. You might say a vector is "real" if its components in some basis are real numbers, but if you change to another basis and the matrix expressing the new basis in terms of the old has complex entries then the "real" vector will have complex components in ...


11

It's just notation. $H_1$ acts on Hilbert space ${\mathcal H}_1$, $H_2$ acts on Hilbert space ${\mathcal H}_2$. By $ |\psi_1,\psi_2\rangle$ they have an implicit tensor product $$ |\psi_1,\psi_2\rangle\stackrel{\rm def}{=} |\psi_1\rangle\otimes |\psi_2\rangle\in {\mathcal H}_1\otimes {\mathcal H}_2. $$ Then physiscists, usually without saying so, extend $...


10

By the definition of a Hilbert space, any element of a Hilbert space can be represented as a sum $\sum_n c_n \lvert \psi_n\rangle$ with the sequence $(c_n)_{n\in\mathbb{N}}$ being square summable and the $\lvert \psi_n\rangle$ being normalized and orthogonal. So if $\lambda$ is small enough that the sequence $(\lambda^n)_{n\in\mathbb{N}}$ is square summable, ...


10

This is a nice exercise! It can be completely solved with relatively elementary mathematical techniques. Let us start by assuming $\hbar:=1$, defining the formally selfadjoint differential operator over smooth functions $$D := \frac{1}{2}(XP +PX) = -i \left(x \frac{d}{dx}+\frac{1}{2}I\right)\:,$$ and proving that it admits a unique selfadjoint extension ...


10

If $H(t)$ is selfadjoint and bounded (thus everywhere defined), the theory is standard and quite easy to handle using the uniform operator topology. You can find all required proofs in the first or second volume of Reed and Simon's textbook on mathematical methods. If the operators $H(t)$ are unbounded, the theory is much more difficult also in view of ...


10

Forgive the length. I find Schrödinger's cat is much easier to make sense of as a journey through QM, rather than just a few equations that someone says "solves your problems." Schrödinger's cat was definitely meant to be taken seriously, in that it was intended to be a serious challenge to naively applying the Copenhagen interpretation to macroscopic ...


10

If $A:V\to V$ is a linear operator on a vector space $V$ with basis $|m\rangle$, Dirac defined the action $\langle n|A$ of $A$ on the bra vector $\langle n|$ (an element of the dual space $V^*$)as a new bra vector that is evaluated on a ket $|m\rangle$ to give the same answer as the evaluation of $\langle n|$ on $A|m\rangle$. In other words he set $$ (\...


10

Yes, you just have to check that $e^{-i\beta P/\hbar}|q\rangle$ is an eigenket of $Q$ (uppercase represents operators) with eigenvalue $q+\beta$. We evaluate $$ Q \big ( e^{-i\beta P/\hbar}|q\rangle \big ) $$ by using the first identity (which is actually a consequence of the canonical commutation relation): $$ Q e^{-i\beta P/\hbar} = e^{-i\beta P/\hbar}Q + ...


9

The time dependent Schroedinger equation looks like this: $$ i\hbar \frac{\partial \Psi}{\partial t} = H \Psi = \left ( -\frac{\hbar^2 }{2 m}\frac{\partial^2}{\partial x^2} + V(x,t) \right ) \Psi(x,t) ,$$ you attempt a solution via separation of variables: $\Psi(x,t) = \psi(x) T(t)$, plug it in. If the potential $V$ is time independent such that $V(x,t) = ...


9

The statement by yuggib is correct. To put it in perspective, I'll start with a completely general formulation, and then I'll show how vector-states and density operators fit into that picture. I won't try to be mathematically rigorous here, but I'll try to give an overview with enough keywords and references to enable further study. State = normalized ...


9

I am in the middle of writing some notes on quantum mechanics for my brother, a mathematician. I'm trying to introduce as little notation as possible, and so I am not using bra-ket notation. The advantages I've found for not using bra-ket notation are Don't have to explicitly introduce the notation. Explaining what a bra is does take a little time. Makes ...


9

Schrodinger's cat is an exercise in seeing how nonsensical the Copenhagen interpretation is, so answers that attempt to clarify it in terms of CI are not very helpful, in my opinion. As a framework for this answer, I'll repeat a point I make frequently: QM describes not the probabilistic evolution of a single deterministic state, but rather the ...


9

In order to calculate the matrix elements of an operator, you need to know how the operator acts on the basis states. This is usually something that you're given (because it's almost always how an operator is defined in the first place - by its action on basis states). So, if you want to calculate the matrix elements, you need at least that much information (...


9

Two complex topological vector spaces $X$ and $Y$ are said to be in duality if there is a sesquilinear map $$b:X\times Y\to \mathbb{C}\; .$$ The idea is that, given such map, a dual action of $X$ on $Y$ and viceversa is defined: for any $x\in X$, $$b(x,\cdot): Y\to \mathbb{C}$$ is a linear functional. There are two natural spaces in duality with a given ...


9

In QM while doing these manipulations you always assume that the wave function is square integrable i.e. in $L^2(\mathbb R)$ space. Now mathematically speaking functions in the $L^2(\mathbb R)$ space can be wild. So rather than working with $L^2(\mathbb R)$ functions, we usually restrict ourselves to Schwartz functions $\mathcal S(\mathbb R)$, which are ...


8

$U(H)$ is a connected and continuous-path connected topological group with respect to the strong operator topology, which is the most natural one for physical applications. This topology permits to extend some features of Lie groups to the case of an "infinite-dimensional" group where the differentiable structure is not easy to define (there are different ...


8

I think your formula is confused. The wavefunction is $$ \psi(x) = \langle x\vert \psi\rangle = \int \delta(x-x') \psi(x')\,dx' $$ where $\delta(x-x')= \langle x'\vert x\rangle$ is the wavefuction of the position eigenfunction $\vert x\rangle$ in the position eigenfunction basis. This not what you have written with the "$x$" operator. For the ocillator ...


8

This operator is an element of the $\mathfrak{su}(1,1)$ Lie algebra. This Lie algebra is spanned by $\{\hat K_0,\hat K_\pm\}$ where these are conveniently realized (for your question) in terms of harmonic oscillator operators: $$ \hat K_0=\frac{1}{4}\left(\hat a^\dagger \hat a+\hat a\hat a^\dagger\right) \, ,\quad \hat K_+=\frac{1}{2}a^\dagger a^\dagger\, ,\...


7

Restricting to finite-dimensional Hilbert spaces, you have a complete basis of orthonormal eigenkets if and only if the operator is normal, meaning that it commutes with its adjoint, $[A, A^\dagger] = 0$. In particular, Hermitian operators are normal since they are equal to their adjoints, anti-Hermitian operators are normal, and unitary operators are ...


7

In general $\langle A\vert B\rangle$ is a scalar product whereas $\vert A\rangle\langle B\vert$ is an operator. To this this last suppose $\vert \psi\rangle$ is an arbitrary vector in your space. Then $$ \vert A\rangle\langle B\vert \psi\rangle $$ is a vector proportional to $\vert A\rangle$ since $\langle B\vert \psi\rangle \in \mathbb{C}$ is a (complex) ...


7

It is not true. Take for example $$ B = 2\pi \mathrm{i} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} , $$ then $\exp(B) = 1$ and $[A,\exp(B)] = 0$ for all $A$. However, clearly, $[A,B] \neq 0$ for some $A$.


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