I Momentum/Position space wave function

WeiShan Ng
Messages
36
Reaction score
2
These are from Griffith's:
Momentum space wave function ##\Phi(p,t)## is the Fourier transform of ##\Psi(x,t)##
$$\Phi(p,t)=\frac{1}{\sqrt{2\pi \hbar}} \int_{-\infty}^{\infty} e^{-ipx/\hbar} \Psi(x,t) \, dx$$
Position space wave function ##\Psi(x,t)## is the inverse transform of ##\Phi(p,t)##
$$\Psi(x,t)=\frac{1}{\sqrt{2\pi \hbar}} \int_{-\infty}^{\infty} e^{ipx/\hbar} \Phi(x,t) \, dp$$
And ##|\Phi(p,t)|^2 = |c(p)|^2## is the probability of getting one of the eigenvalue of the momentum operator.
Momentum eigenfunctions are ##f_p(x) = (1/\sqrt{2\pi\hbar}) exp(ipx/\hbar)##
$$c(p) = \langle f_p|\Psi \rangle = \frac{1}{\sqrt{2\pi\hbar}} \int_{-\infty}^{\infty} e^{-ipx/\hbar} \Psi(x,t) \, dx$$
while ##|\Psi(y,t)|^2 = |c(y)|^2## is the probability of getting one of the eigenvalue of the position operator.
Position eigenfunctions are ##g_y(x) = \delta(x-y)##
$$c(y)=\langle g_y|\Psi\rangle = \int_{-\infty}^{\infty} \delta(x-y) \Psi(x,t) \, dx = \Psi(y,t)$$
My lecture note says that
Physical duality of ##\Psi## and ##\Phi## specify the same state of the system and we can compute one from another


I am having quite a confusion over here...Does the ##\Psi## in the expression ##\langle f_p|\Psi \rangle## equals to ##\Psi(x,t)##? I understand it as ##\Psi(x,t)## being the component of the position basis to form ##\Psi##, so ##\Psi## is a state vector and ##\Psi(x,t)## is the "coefficients"?
And when it says ##\Psi## and ##\Phi## both specifying the same state of the system, should they be ##\Psi(x,t)## and ##\Phi(p,t)## (the coefficients) instead? If so we will have
$$\begin{align*} \Psi &= \int c(p) f_p \, dx =\int \left[ \int \frac{1}{\sqrt{2\pi\hbar}} e^{-ipx'/ \hbar} \Psi(x',t) \, dx' \right] \frac{1}{\sqrt{2\pi\hbar}} e^{ipx/\hbar} \, dx \\
&= \int c(y) g_y \, dx = \int \Psi(y,t) \delta(x-y) dx = \Psi(y,t) \end{align*}$$
And if I use the Fourier transform of ##\delta(x)##
$$\delta(x) = \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{ikx} \, dk$$
I get
$$\frac{1}{2\pi\hbar} \int e^{ipx/\hbar} \, dx = \delta(p) $$
which means the first line will be
$$\Psi = \int e^{-ipx'/\hbar} \Psi(x',t) \, dx' \delta(p) = \int \Psi(x',t) dx'$$
So I get ##\int \Psi(x',t) \, dx## and ##\Psi(y,t)=\Psi(x,t)## both equal to ##\Psi##?
 
Physics news on Phys.org
WeiShan Ng said:
I am having quite a confusion over here..

Write the momentum eigen vector \hat{P} | p \rangle = p | p \rangle in the coordinate space as \langle x | p \rangle = (2\pi \hbar)^{-1/2} e^{i p \cdot x / \hbar} . Now, any vector |\Psi \rangle can be expanded in an arbitrary orthonormal basis \{| \alpha \rangle\} according to |\Psi \rangle = \int d \alpha \ | \alpha \rangle \langle \alpha | \Psi \rangle . The component of the vector |\Psi \rangle along the “x-direction” in the coordinate space, i.e., the wavefunction \Psi (x) is calculated from \Psi (x) \equiv \langle x | \Psi \rangle = \int d \alpha \ \langle x | \alpha \rangle \langle \alpha | \Psi \rangle . Or \Psi (x) = \int d \alpha \ \langle x | \alpha \rangle \Psi (\alpha) . \ \ \ \ \ \ \ \ \ \ \ \ \ (1) If the \langle x | \alpha \rangle is a Kernel of a Fourier transform, we usually write \tilde{\Psi}(\alpha) or \Phi (\alpha) instead of \Psi (\alpha) on the RHS of (1). This is the case when you take \alpha = p.
 
Last edited:
  • Like
Likes WeiShan Ng and dextercioby
I'm still trying to get my head around this, not sure if I understood it correctly... When we write ##|\Psi\rangle##, it means we haven't specify any particular basis set to represent the state vector, when we write ##\Psi(x)##, it means we are writing the component of ##|\Psi\rangle## along an element of a basis set that uses variable ##x##, like ##\{1/x,1/x^2,\dots\}##?

And in
samalkhaiat said:
$$Ψ(x)≡⟨x|Ψ⟩=∫dα ⟨x|α⟩⟨α|Ψ⟩.$$
for the ##∫dα ⟨x|α⟩⟨α|Ψ⟩##, does it mean we are finding ##|\Psi\rangle## in the direction of ##|x\rangle## then represent this using a set of basis vectors ##\{|\alpha\rangle\}##, i.e. we perform a basis tranformation??
 
WeiShan Ng said:
I'm still trying to get my head around this, not sure if I understood it correctly... When we write ##|\Psi\rangle##, it means we haven't specify any particular basis set to represent the state vector, when we write ##\Psi(x)##, it means we are writing the component of ##|\Psi\rangle## along an element of a basis set that uses variable ##x##, like ##\{1/x,1/x^2,\dots\}##?
Make the following correspondence with Linear Algebra |\Psi \rangle \to \vec{V} , \ \ \ \mbox{Abstract Vector},| \alpha \rangle \to \hat{e}_{i} , \ \ \ \mbox{Orthogonal unit vectors},\langle \alpha | \Psi \rangle \to \hat{e}_{i}\cdot \vec{V} = V_{i} , \ \ \mbox{component in i-direction}, \int d \alpha \to \sum_{i}. Now the expansion | \Psi \rangle = \int d \alpha \ | \alpha \rangle \langle \alpha | \Psi \rangle , will correspond to \vec{V} = \sum_{i} \hat{e}_{i} \left( \hat{e}_{i} \cdot \vec{V}\right) = \sum_{i} \hat{e}_{i}V_{i}. Do you recognise this equation?

we perform a basis tranformation??
Yes, it is simply a linear transformation relating the components of the vector in two different “coordinate systems”. That is the component of the vector |\Psi \rangle “along” the “unit” vector |x\rangle (i.e., the number \Psi (x) \equiv \langle x | \Psi \rangle) is related to its component “along” the “unit” vector |\alpha \rangle (i.e., the number \Psi (\alpha) = \langle \alpha | \Psi \rangle) by the transformation “matrix” \langle x | \alpha \rangle \equiv M(x, \alpha). So \langle x | \Psi \rangle = \int d \alpha \langle x | \alpha \rangle \langle \alpha | \Psi \rangle is same as \Psi (x) = \int d \alpha \ M( x , \alpha) \Psi (\alpha ) . This corresponds to the familiar linear (matrix) transformations in vector algebra V^{'}_{i} = \sum_{j} M_{ij} V_{j}

Remember | \Psi \rangle is an abstract vector (just like the vector \vec{V} in ordinary vector algebra) and \langle \beta | \Phi \rangle = \Phi ( \beta ) is a complex number representing the component of the vector |\Phi \rangle in the basis | \beta \rangle (just like the real number V_{i} which represents the component of \vec{V} along the unit vector \hat{e}_{i}).
 
Last edited:
  • Like
Likes MichPod and WeiShan Ng
Not an expert in QM. AFAIK, Schrödinger's equation is quite different from the classical wave equation. The former is an equation for the dynamics of the state of a (quantum?) system, the latter is an equation for the dynamics of a (classical) degree of freedom. As a matter of fact, Schrödinger's equation is first order in time derivatives, while the classical wave equation is second order. But, AFAIK, Schrödinger's equation is a wave equation; only its interpretation makes it non-classical...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. Towards the end of the first lecture for the Qiskit Global Summer School 2025, Foundations of Quantum Mechanics, Olivia Lanes (Global Lead, Content and Education IBM) stated... Source: https://www.physicsforums.com/insights/quantum-entanglement-is-a-kinematic-fact-not-a-dynamical-effect/ by @RUTA
Is it possible, and fruitful, to use certain conceptual and technical tools from effective field theory (coarse-graining/integrating-out, power-counting, matching, RG) to think about the relationship between the fundamental (quantum) and the emergent (classical), both to account for the quasi-autonomy of the classical level and to quantify residual quantum corrections? By “emergent,” I mean the following: after integrating out fast/irrelevant quantum degrees of freedom (high-energy modes...
Back
Top