# Tagged Questions

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678 views

### Normalization of basis vectors with a continuous index?

I have an infinite basis which associates with each point, $x$, on the $x$-axis, a basis vector $|x\rangle$ such that the matrix of $|x\rangle$ is full of zeroes and a one by the $x^{\mathrm{th}}$ ...
822 views

### Normalizing Propagators (Path Integrals)

In the context of quantum mechanics via path integrals the normalization of the propagator as $$\left | \int d x K(x,t;x_0,t_0) \right |^2 ~=~ 1$$ is incorrect. But why? It gives the correct pre-...
971 views

### Who is doing the normalization of wave function in the time evolution of wave function?

In the Schrodinger equation, at any given time $t$ we should jointly add another sub equation, like $$||\psi_t(x)|| = 1$$ where $\psi_t(x) = \Psi(x,t)$, and then try to solve the two equations ...
171 views

### Physical position eigenfunction normalisation

We know that the Dirac function $$\delta(a)=\lim_{a \rightarrow 0} \delta_{a}(x)$$ can be written as an infinitesimally narrow Gaussian: $$\delta_{a}(x) := \frac{1}{\sqrt{2\pi a^2}}e^{-x^2/2a^2}$$ ...
112 views

### Normalisation of free particle wavefunction

The wavefunction $\Psi(x,t)$ for a free particle is given by $$\Psi(x,t) = A e^{i(kx-\frac{\hbar k}{2m}t)}$$ This wavefunction is non-normalisable. Does this mean that free particles do not exist in ...
302 views

### Green's Functions from Gell-Mann and Low Theorem

What I want to do: $\newcommand{\ket}[1]{\left|#1\right\rangle}$ $\newcommand{\bra}[1]{\left\langle#1\right|}$ $\newcommand{\braket}[1]{\left\langle#1\right\rangle}$ The Gell-Mann Low Theorem tells ...
916 views

### How do we normalize a delta function position space wave function? [duplicate]

I have a position space wavefunction $$\psi(x) = \delta(x-a) + \delta(x+a).$$ Now the question states to compute the following: The Fourier transform of $\psi(x)$. (Which invariably is the momentum ...
100 views

### Free space propagator: reconciling two results

In quantum mechanics, the free space propagator $G(q_f=0,q_i=0;\tau)$ can be easily calculated to be $$\sqrt{\frac{m}{2\pi i \hbar \tau}}$$ by inserting an identity operator. However if we use ...
I have the one dimensional free particle SchrÃ¶dinger equation $$i\hbar \frac{\partial}{\partial t} \Psi (x,t) = -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} \Psi (x,t), \tag{1}$$ with ...