# Momentum/Position space wave function

• I
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##????

samalkhaiat
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 $\alpha = p$.

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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
$$Ψ(x)≡⟨x|Ψ⟩=∫dα ⟨x|α⟩⟨α|Ψ⟩.$$
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??

samalkhaiat
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?
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 $\hat{e}_{i}$).