# Tag Info

0

If you read the next paragraph, it seems that Dirac means that the eigenvalues are non-degenerate, i.e. for a set of simultaneous eigenvalues $\xi_1^\prime, \xi_2^\prime\dots$ there is exactly one corresponding eigenbra. There are no two distinct bras which are eigenbras of all the operators $\xi_1, \xi_2\dots$ and have the same eigenvalues for all of them. ...

0

If the electron is confined to the $x-y$-plane, it's $z$-position is fixed, i.e. certain, and hence the $z$-momentum infinitely uncertain by the uncertainty relation. That paragraph is trying to say that, if the magnitude of $\vec L$ is larger than $L_z$, then $\vec L$ is not fixed, and if angular momentum is not fixed, i.e. conserved, then the motion does ...

0

If the particle is known to be in the $xy$-plane, then $\Delta z = 0$, and so (by the uncertainty principle) $\Delta p_z = \infty$. Roughly speaking, this means that there's a good chance that the particle's momentum would be greater than "escape momentum", and thus that it could escape from the hydrogen atom. (I'm kind of dubious about this argument ...

0

$\newcommand{\ket}[1]{\lvert #1 \rangle}$The Schrödinger equation does not say what you claim it does. The time-independent Schrödinger equation is an eigenvalue equation for the Hamiltonian operator $$H \ket{\psi_E} = E \ket{\psi_E}$$ where solving for $\ket{\psi_E}$ for a concrete $H$ gives us the occuring eigenstates of $H$. It does not say that these ...

2

$\newcommand{\ket}[1]{\left| #1 \right>}$If your state is in an eigenstate of the energy operator then the answer is that you'll get the same value for the energy every time you measure the particle's energy. That is the reason why the energy eigenstates are also called stationary states. On the other hand you can also have a superposition of energy ...

1

No, that is not quite correct. The Schrödinger equivalence says that if the system has a definite energy, then this energy can only be an eigenvalue of the system's hamiltonian $\hat H$. There is no requirement, however, for the system to have a definite energy; if the energy is undefined then an energy measurement may return different (eigen)values on ...

1

Note that you only can identify the numbers with the operators when you hold a plane wave function. For an arbitrary wave functions, you can't identify the operator $(\hbar/i)(\partial / \partial x)$ with some number $p$, but you can hold the operator to have some significant meaning related to this numbers. If you apply  \langle P\rangle = \int dx \, ...

7

Not equal, but equivalent, in the sense that they have the same effect on the wavefunction in question. More precisely, the book is using a slight abuse of terminology. Taking momentum as an example, it's not really the case that the dynamical quantity of momentum is equivalent to the operator $\frac{\hbar}{i}\frac{\partial}{\partial x}$, because a number ...

2

As argued by von Neumann, the measuring process has many properties that resemble those found in the theory of operator algebras. For instance, if you have an instrument, you can measure something, say the length of a table, to get a certain value $x$ within experimental errors. What you can now do is relabel the ticks of your instrument according to a ...

8

The reason operators correspond to measured values has to do with what happens when you connect a measurement apparatus to the system under observation. Suppose the Hamiltonian of the system by itself is $H_S$ and the Hamiltonian of the measurement apparatus by itself is $H_M$. When $M$ is physically connected to $S$, we get a additional "interaction" term ...

0

You will have to settle for momentum since speed is not a quantum mechanical observable (because $\dot{x}$ is not a classical observable on the phase space, which are functions of $x$ and $p$, but a function of a classical trajectory $(x(t),p(t))$, so canonical quantization does not produce a "speed" observable). The probability for a certain momentum, ...

Top 50 recent answers are included