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 $...


9

From $\mathbf{L}=\mathbf{Q}\times \mathbf{P}$ we have $L_z=Q_xP_y-Q_yP_x$. Then, introduce the following new operators (assuming units of $\hbar=1$): \begin{align} q_1=\frac{Q_x+P_y}{\sqrt{2}},\\ q_2=\frac{Q_x-P_y}{\sqrt{2}},\\ p_1=\frac{P_x-Q_y}{\sqrt{2}},\\ p_2=\frac{P_x+Q_y}{\sqrt{2}}. \end{align} It is immediate to check that $$ [q_1,q_2]=[p_1,p_2]=0,\...


7

Expanding, we have: $$Ae^{ikx}+Be^{-ikx}=(A+B)\sin(kx)+i(A-B)\cos(kx)$$ Setting $A'=A+B$ and $B'=i(A-B)$ we have that $$Ae^{ikx}+Be^{-ikx}=A'\sin(kx)+B'\cos(kx)$$ so the two solutions are equivalent. It really doesn't matter in the end.


7

Simply, it is just a matter of definitions. Being $$ a=\frac{\hbar^24\pi\epsilon_0}{me^2}, $$ a constant generally called Bohr's radius, you can invert this formula to obtain $e^2/4\pi\epsilon_0$ instead. Then, put it in your equation and you are done.


6

A basic difference between quantum mechanics and classical mechanics is that the potentials do not act on masses in quantum mechanics. Instead they are part of the differential equation that has to be solved to give the wavefunction for the system under consideration. In the case of a single atom, lets take the hydrogen atom, the differential equation is ...


6

It would be helpful for you to define what $H_1$ and $H_2$ are in your exercise. My guess is that your source is being sloppy in their notation for how they are defining tensor products of operators. Probably, they are working on some tensored Hilbert space $\mathcal{H}_A \otimes \mathcal{H}_{B}$ and they mean something like $$ H_1 = h_1 \otimes I_{B} \\ H_2 ...


5

You're right, both are solutions to the Schrodinger equation for this potential. They are equivalent representations using Euler's Formula. The preference for complex exponentials vs sinusoids comes down to mathematical convenience. They are generally interchangeable. As a rule of thumb, if the wave function is 0 outside of a bounded region of space, you ...


5

"Differential" may be a bit of a red herring, as the generic $\hat T$ operator may have strings of x s and $\partial_x$s. There is no good reason to hyperfocus on derivatives, which, in the coordinate representation you are using, correspond to the momentum operator $\hat p$, $$ \hat p = \int dx |x\rangle \frac {\hbar} {i} \partial _x \langle x|, $$ so $$ \...


4

What you've written from your text/exercise is an abuse of notation, but it is standard. The composite state $|\psi \alpha\rangle$ is the tensor product of the states $|\psi\rangle \in \mathcal H_1$ and $|\alpha\rangle \in \mathcal H_2$, which is sometimes written $|\psi\alpha\rangle \equiv |\psi\rangle \otimes |\alpha\rangle$. A very typical example of ...


3

It's true that because of QM you can't think of the electrons in the atoms as having precise positions. It's not just position that is affected by QM but all "observable" phenomena. You are right that a theory that explains forces as a function of position is therefore likely to run into problems. But there are quantum theories of how forces like ...


2

If positive real numbers are forwards numbers and negative real numbers are backwards numbers then imaginary numbers are sideways numbers. In terms of angles, positive real numbers could be thought of as having an angle of 0°, negative numbers have an angle of 180° and sideways or imaginary numbers are at ±90°. This is useful in electrical engineering when ...


2

Complex numbers do two obvious things. If you think of them as 2D vectors on a plane, starting at your arbitrary point (0,0), then adding complex numbers is vector addition. And if you think of them as angles off an arbitrary polar-coordinate angle (0,1), then when you multiply two of them you get the sum of the angles (and the product of the magnitudes). ...


2

I assume that in transformation you restrict yourself to unitary transformations, such that probability is maintained $|| \hat{T} |\psi\rangle || = 1$. I will further assume that you limit the question to forms of transformation that are position-based somehow, so the notion of "Kernel" $\hat{K}(x,x')$ makes sense (so for example rotations in spin-space are ...


2

In addition to the comments mentioned, when you solve for 2 electron problems in quantum mechanics, you do include a term of the form $\frac{kq_1q_2}{r_{12}}$ which represent the interaction between the two wavepackets. For more information on how the 2 electron system is solved, see https://en.wikipedia.org/wiki/Two-electron_atom#Schr%C3%B6dinger_equation. ...


2

Because the orthogonality is supplied by the azimuthal direction when the two $z$-components of the angular momentum differ. This is why we usually bundle them together into the spherical harmonics instead of handling them separately.


2

In general the wavefunction depends on all the coordinates. However, there are special states called $s$ states, without any angular momentum, where it depends only on $r$.


2

From my point of view it seems that you are mixing two different cases together: It is correct that your probability density is time-independent if you just choose one solution of the time-dependent Schrödinger equation. But in this case the wavefunction does not cease (I am still not completely sure what you mean by this), since we have just figured out ...


2

You can understand the relationship between these two expressions when using a common trick performed in quantum mechanical calculations, which is multiplication by one. The unit matrix in quantum mechanics can be expressed as the sum of the outer product over a complete basis, i.e. $ 1 = \sum_n \vert n \rangle \langle n \vert$, where the states $ \vert n \...


2

The basic idea behind the factorization is to replace a 2nd-order differential equation by a pair of first order ones. It was made popular in physics by the work of Hull and Infeld: Infeld, Leopold, and T. E. Hull. "The factorization method." Reviews of modern Physics 23.1 (1951): 21 although in fact earlier examples, such as the factorization of the ...


1

No. Since the square well is finite the boundary conditions do not give $E_n=\frac{n^2 \pi^2\hbar^2}{2ma^2}$ since the wavefunction need not be $0$ at the edges of the well. Instead, the possible energies are found numerically by solving a transcendental equation, as given later in your linked page, which amounts to finding the intersection points of two ...


1

I found in the work of Groenwold, 1946, a closed form for the solution of \begin{equation} k_{mn}(\zeta_1,\zeta_2)=\frac{1}{\sqrt{\pi 2^{m+n} m!n!}}\int_{\mathbb R} \mathrm d \zeta \; e^{-\zeta^2} H_{m}(\zeta+\zeta_1) H_{n}(\zeta+\zeta_2) \end{equation} can be written in terms of the associated Laguerre polynomials by considering \begin{equation*} \sum_{mn=0}...


1

By experimenting in Mathematica I found that $$\int_{-\infty}^\infty d\zeta e^{-\zeta^2}H_n(\zeta+\zeta_1)H_{n-1}(\zeta+\zeta_2)= \sqrt{\pi}2^n(n-1)!\zeta_1L_{n-1}^1(-2\zeta_1\zeta_2)$$ holds for $n=1,...,6$ so I conjecture that it holds for higher values of $n$ as well. I didn't try to prove it. It seemed reasonable to guess that if Laguerre polynomials ...


1

Dhruv Maroo, Quantum particles are indistinguishable. You cannot "label" the electrons. So, a state in which two electrons are exchanged is the same state as the original one. The probability of an electron to be found very far away from the nucleus is very low. In order to calculate the probability for an electron to "jump" from an atom to another atom you ...


1

This wavefunction is one over 3D space, given its dependence on $x,\,y$ and $z$. Hence, your normalization should be done over all space: $$ 1=\iiint_D\Psi^*\Psi\,\mathrm{d}V=\iiint_D\Psi^*\Psi\,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z $$ where $D\subset\mathbb{R}^3$. Additionally, if your function is separable, then $\Psi(\mathbf{x})=\psi(x)\phi(y)\xi(z)$ and ...


1

The Schrödinger equation gives a standing wave solution at a hard boundary, just like a water wave equation does. A second quantum phenomenon is that the wave exponentially penetrates - tunnels into - the barrier.


1

You have $$\bar\Psi = \begin{pmatrix} \cos(x) & 0 \\ 0 & -i \sin(x) \end{pmatrix}.$$ Then, $$ \bar\Psi\bar\Psi= \begin{pmatrix} \cos^2(x) & 0 \\ 0 & -\sin^2(x) \end{pmatrix}. $$ This yields $$ \bar\Psi\bar\Psi\begin{pmatrix}1 \\ -1\end{pmatrix}= \begin{pmatrix} \cos^2(x) & 0 \\ 0 & -\sin^2(x) \end{pmatrix}\begin{pmatrix}1 \\ -1\...


1

The assumption is an example of the Born Rule, which was formulated by Max Born in 1926. More generally the rule states that if you make a measurement of an observable property of a particle associated with some operator O, say, then the result must be one of the eigenvalues of O, and the wave function of the particle immediately after the measurement must ...


1

It is a postulate: the probability of outcome $E_i$ is given by the inner product of the eigenfunction associated with $E_i$ with the initial state. In Dirac notation $\vert \langle \psi_i\vert \Psi(t)\vert^2$. If more than one eigenfunction is associated with the eigenvalue $E_i$ then one must sum over these eigenfunctions: $$ \sum_{n} \vert \langle \...


1

$$\Psi(x)=Ae^{ikx}+Be^{-ikx}=Acos(kx)+Aisin(kx)+Bcos(kx)-Bisin(kx)=Ccos(kx)+Disin(kx)$$ You have to impose continuity of the solution at the border of the hole: $$\Psi(a)=Ccos(ka)+Disin(ka)=0$$ $$\Psi(-a)=Ccos(ka)-Disin(ka)=0$$ which are true if you put $D=0$ or $C=0$. In the first case: $$0=cos(k^*a)$$ $$\Psi_{1}(x)=Ccos(k^*x)$$ In the second case: $$0=...


1

Sorry for a long story, which only address the title of your question (and not the inner questions). I remember the first time I was introduced to complex numbers at school. The teacher (of mathematics, not physics) was explaining us how to solve quadratic equations (a.x^2+b.x+c=0). After giving us the method, he ended up with the well known solution for ...


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