5 votes
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What is the distinction between a ket and a state in quantum mechanics?

I think all other answers so far do not address the problem directly. The definition of the OP regarding kets is a quite common, though not the most used one, see e.g. references 1 and 2. So I don't ...
Tobias Fünke's user avatar
5 votes

What is the distinction between a ket and a state in quantum mechanics?

This is really two questions in one, namely what is a ket and what is a state. Both questions are bound to cause some degree of controversy between physicists of different mathematical pedigrees. I ...
J. Murray's user avatar
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4 votes
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Antisymmetry and coordinate invariance

How can we prove that $\phi(y_1,y_2) = -\phi(y_2,y_1)$? You cannot. Consider $$f_1(y_1,y_2) = \frac{1}{2}(y_1 + y_2),\qquad f_2(y_1,y_2) = \frac{1}{2}(y_1 - y_2).$$ Then, let us take $x_2=0$. In ...
mdi's user avatar
  • 353
3 votes

What is the distinction between a ket and a state in quantum mechanics?

A ket is an element of a Hilbert space. A ket may also be called a state vector. An operator acting on a ket produces a ket. Suppose we have a two-dimensional Hilbert space and a Hermitian operator $\...
Andrew Steane's user avatar
3 votes

Does the inner product of wavefunctions really have units?

You forgot that the wave functions need to satisfy the normalization condition: $$\int dx\, \overline{\psi(x)} \, \psi(x) = 1$$ $$\int dx\, \overline{\phi(x)} \, \phi(x) = 1$$ Hence the wavefunctions $...
Thomas Fritsch's user avatar
3 votes

What is the distinction between a ket and a state in quantum mechanics?

A ket is an element of the Hilbert space, $| \psi \rangle \in \mathcal{H}$, i.e., it is an element (a vector) of the complex inner-product space. I think this is well-accepted terminology (introduced ...
Ben H's user avatar
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3 votes
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Wave packet for particle in a potential

Apart from the time development of a quantum state in the Schrodinger picture, the potential only effects the eigenvalues and eigenstates of the Hamiltonian, i.e. the eigenstates and energy levels ...
Albertus Magnus's user avatar
3 votes
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Anderson localization in the continuum case

Yes, Anderson localization persists to a continuum model. The statement is actually much stronger than this. For any value of $V_0$, all states will be localized in 1d (and, marginally, in 2d as well)....
Rococo's user avatar
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3 votes

Is each INDIVIDUAL photon a PHYSICAL wave?

Firstly, we can only say anything about the quantum world in as far as we can make observation of it. For that reason, the underlying mechanisms that cannot be directly observed always rely on how we ...
flippiefanus's user avatar
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3 votes
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What is the difference between $(\mathcal{H}\setminus \{ 0\})/\mathbb{C}^*$ and $\mathcal{H}_1/U(1)$?

The text mentions two definitions of the space of physical states. One is $\mathcal{H} \setminus \{0\}$ modulo $\mathbb{C}^*$, and the other is the unit sphere in $\mathcal{H}$ modulo $U(1)$. They are ...
Javier's user avatar
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2 votes

Is there a potential energy for every wave function?

Hoping I understand correctly, you are given one wavefunction over a time interval and a space interval, and the mass. Then, by Schrödinger's equation we have the answer: $\frac{ i\hbar\frac{\partial}{...
Rol's user avatar
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2 votes

Interpretation of the vector space in notation bra ket

The bra, ket notation for quantum states is a way of expressing the state in the abstract. The quantum state ket $|\alpha\rangle$ is a vector in Hilbert space. Here the concept of vector is to be ...
Albertus Magnus's user avatar
2 votes
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Interpretation of the vector space in notation bra ket

You can think of $\psi(\vec x, t)$ as being a vector in an infinite dimensional complex vector space (a Hilbert space), where $|\psi(\vec x, t)|^2$ is the probability that the particle will be found ...
gandalf61's user avatar
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2 votes
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Why does the curvature change with respect to the sign of the wave function?

Note the curvature is varying based on the sign of the second derivative. $\kappa=\frac{y''}{(1+y'^2)^{3/2}}$ The sign of the curvature is the sign of the second derivative. So if you have $\frac{-\...
R. Romero's user avatar
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2 votes
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What does $f(x)$ satisfies the given equation means?

Now the solution just requires one to prove that $\partial^2/\partial x^2=\partial^2/\partial (-x)^2$. Yes, if you can "just" prove the above-quoted equality you are basically "done.&...
hft's user avatar
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2 votes

What does $f(x)$ satisfies the given equation means?

As you note, the form of the equation doesn't change, which implies that $\psi(-x)$ is just as much a solution as $\psi(x)$ is. Similarly, one can take any linear combination $\phi(x)=c_1\psi(x)\pm ...
Albertus Magnus's user avatar
2 votes

Are EM waves telling us the probability of finding a photon?

Light is not a wave, that was only true in classical physics. It can not be represented (as you claim) by a sine wave with discrete values of amplitude. It can be represented as a wave functional, ...
Jos Bergervoet's user avatar
1 vote

What is the distinction between a ket and a state in quantum mechanics?

Besides the correct introduction to the bra-ket notation done by @AndrewSteane, a cleaner mathematical notation may help to understand the situation and how manipulations go. Kets are elements of the ...
GiorgioP-DoomsdayClockIsAt-90's user avatar
1 vote

What will be wave function after application of operator?

No, the state vector after a position measurement is not given by $X |\psi\rangle$. Suppose your detector can decide whether the position of the particle is inside oder outside a certain (space) ...
Hyperon's user avatar
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1 vote

"Derivation" of group velocity and "sharply" peaked wave functions

The point is of course, to talk about the group velocity of a free particle with a somewhat well defined momentum. If you use a Dirac delta function then you throw any concept of "group" or ...
Albertus Magnus's user avatar
1 vote
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Wave packet at $t<0$

In the Heisenberg picture, the time evolutions of the position operator $X(t)$ and the momentum operator $P(t)$ of a free particle with mass $m$ are given by $$X(t)=X(0)+\frac{P(0) }{m}t, \quad P(t)=P(...
Hyperon's user avatar
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1 vote
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Boundaries of finite potential well

You are correct that this definition of finite potential well is not well-defined. In the best traditions of quantum mechanics boundary condition should be omitted, because particle state at these ...
Agnius Vasiliauskas's user avatar
1 vote

Justification of discarding the backward wave in step potential scattering

The general solution to Schrödinger's time-independent equation (for energy $E$) is $$\psi(x)=\begin{cases} A_1 e^{-i k_1 x} + B_1 e^{i k_1 x} & \text{for } x < 0 \\ A_2 e^{-i k_2 x} + B_2 e^{i ...
Thomas Fritsch's user avatar
1 vote

Justification of discarding the backward wave in step potential scattering

Typically, we don't set $A_2=0$ - we set $B_2=0$. Our ability to do this is essentially a boundary condition. Because the Schrodinger equation is a 2nd order linear differential equation, there will ...
J. Murray's user avatar
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1 vote
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Unbound States of the 1D Finite Potential Well

The correct ansatz for a simultaneous eigenfunction of the Hamilton operator and the paritity operator with energy $E>V_0$ and even parity is given by $$\psi_E^+(x) = \begin{cases} C \cos kx +D \...
Hyperon's user avatar
  • 4,848
1 vote

Wave functions and Schwartz spaces (QM)

It is possible for a function to go to zero at infinity without "getting all flat". For example, imagine a function which oscillates in such a way that its amplitude goes to zero at infinity,...
Albertus Magnus's user avatar

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