In literature on an introduction to quantum mechanics which I am working through, there is a section which explains that a vector has different representations based on the basis you choose and then relates this to quantum mechanics. It does this by making giving the following explanation:

Imagine an ordinary vector $A$ in two dimensions. How would you describe this vector to someone? The most convenient way is to set up cartesian coordinates axes, $x$ and $y$, and specify the components of $A$: $A_x = \hat{i} \cdot A$, $A_y = \hat{j}\cdot A$. Of course you could have a different set of axis $x'$ and $y'$ and the components would then be: $A_x' = \hat{i}'\cdot A$, $A_y = \hat{j}'\cdot A$. But it's the same vector.

The same is true for the state of a system in quantum mechanics. It is represented by a vector, $\lvert\mathcal{S(t)}\rangle$, that lives "out there in Hilbert space," but we can express it with respect to any number of different bases. The wave function $\Psi(x,t)$ is actually the coefficient in the expansion of $\mathcal{S(t)}$ in the basis of position eigenfunctions: $$\Psi(x,t) = \langle x | \mathcal{S(t)} \rangle$$ (with $|x\rangle$ standing for the eigenfunction of $\hat{x}$ with eigenvalue $x$), whereas the momentum space wavefunction $\Phi(p,t)$ is the expansion of $| \mathcal{S} \rangle$ in the basis of momentum eigenfunctions: $$\Phi(p,t) = \langle p | \mathcal{S}(t) \rangle$$ (with $|p \rangle$ standing for the eigenfunction of $\hat{p}$ with eigenvalue $p$).

It then states that $\Psi$ and $\Phi$ contain the same information and describe the same state.


In the description, when it is stated that "$A$: $A_x = \hat{i} \cdot A$, $A_y = \hat{j}\cdot A$" and "$A_x' = \hat{i}'\cdot A$, $A_y = \hat{j}'\cdot A$" for different basis. In this description $A$ is the same but it's representation is different when working out components.

Why is it then that when describing the components of a system, $| \mathcal{S} \rangle$, in quantum mechanics, $| \mathcal{S} \rangle$, that the representation of $|\mathcal{S} \rangle$ does not change when working out the components in different basis (specifically the basis consisting of position operator eigenvectors and momentum operators eigenvectors), as can be seen by the following

$$\Psi(x,t) = \langle x | \mathcal{S(t)} \rangle = \int^{\infty}_{-\infty}\delta(x-y)\Psi(y,t)dy$$ and

$$\Phi(p,t) = \langle p | \mathcal{S(t)} \rangle= \int^{\infty}_{-\infty}\frac{e^{\frac{-ipy}{\hbar}}}{\sqrt{2 \pi \hbar}}\Psi(y,t)dy$$

where we have basis vectors (eigenvetors) $\delta(x-y)$ and $\frac{1}{\sqrt{2 \pi \hbar}}e^{\frac{ipy}{\hbar}}$ respectively.

Why is the vector $|\mathcal{S} \rangle$ represented as $\Psi$ for both bases when working out the components for the quantum mechanics case above?


3 Answers 3


Let me rephrase those precise equations in the language of finite-dimensional linear algebra. You have a vector $A$ and two bases $\beta=\{e_i\}_i$ and $\beta'=\{e_i'\}_i$. This means you can write the components of $A$ with respect to $\beta$ as $$ A_i=e_i·A=\sum_j\delta_{ij}e_j·A $$ and the components with respect to $\beta'$ as $$ A_i'=e_i'·A=\sum_j(e_i'·e_j)e_j·A. $$ Here the $\delta_{ij}=e_i·e_j$ is the direct equivalent of the $\delta(x-y)=⟨x|y⟩$, and the $e_i'·e_j$ is the direct equivalent of $e^{-ipx/\hbar}/\sqrt{2\pi\hbar}=⟨x|p⟩$.

Simply put, the presence of $\Psi(y,t)$ on the right-hand side of both expressions is simply because you're doing the inner products themselves in the position representation. You could equally write $$\Psi(x,t) = \langle x | \mathcal{S(t)} \rangle = \int^{\infty}_{-\infty} \frac{e^{\frac{+iqx}{\hbar}}}{\sqrt{2 \pi \hbar}} \Phi(q,t)dq$$ and $$\Phi(p,t) = \langle p | \mathcal{S(t)} \rangle= \int^{\infty}_{-\infty}\delta(p-q)\Phi(q,t)dq,$$ or any combination of the two representations for the inner products.

  • $\begingroup$ @EmilioPisanty Does $\langle x| p \rangle = \frac{1}{\sqrt{2 \pi \hbar}}e^{\frac{ipx}{\hbar}}$ follow from $\langle x|p \rangle = \int^{\infty}_{-\infty}\delta(x-y)e^{\frac{ipy}{\hbar}}dy = \frac{1}{\sqrt{2 \pi \hbar}}e^{\frac{ipx}{\hbar}}$? $\endgroup$
    – user114445
    Apr 15, 2016 at 13:13
  • $\begingroup$ @LucioD That's a circular argument. The formula for $⟨x|p⟩$ follows from a number of different things, depending on what you want to think of as fundamental. That's for a separate question, though. $\endgroup$ Apr 15, 2016 at 14:01
  • $\begingroup$ Okay but is the equation $\langle x| p \rangle = \frac{1}{\sqrt{2 \pi \hbar}}\int_{- \infty}^{\infty} \delta(x-y)e^{\frac{ipy}{\hbar}}dy = \frac{1}{\sqrt{2 \pi \hbar} }e^{\frac{ipx}{\hbar}}$ correct? Also, I'm not following why it is a circular argument, I'm simply taking the inner product ($\langle f , g \rangle := \int_{-\infty}^{\infty} f^* g dx$) of the eigenvector $\langle x| = \delta(x-y)$ and $| p \rangle = \frac{1}{\sqrt{2 \pi \hbar} }e^{\frac{ipx}{\hbar}}$. $\endgroup$
    – user114445
    Apr 17, 2016 at 16:43
  • $\begingroup$ Because it assumes the position-space wavefunction of momentum states, $⟨y|p⟩=e^{ipy/\hbar}/\sqrt{2\pi\hbar}$, for the inner product. (Note also that $⟨x|=\delta(x-y)$ and $|p⟩=e^{ipx/\hbar}/\sqrt{2\pi\hbar}$ are incorrect and inconsistent.) I'm not particularly interested in explaining further - if you still have a question, you can ask it separately. $\endgroup$ Apr 17, 2016 at 20:01
  • $\begingroup$ @EmilioPisanty To confirm what you stated in your answer, which references finite dimensional linear algebra. If you want to express $\Psi$ in terms of the basis of Hamiltonian eigenfunctions i.e. $|n \rangle.$ Then $\Psi(x,t) = \sum_nc_ne^{\frac{-iE_n t}{\hbar}}\psi_n(x)$, does this correspond to $\langle x | n \rangle = c_n e^{\frac{-iE_n t}{\hbar}}$ and $\psi_n(x) = e_j \cdot A$? $\endgroup$
    – user101311
    Apr 26, 2016 at 15:20

Why is the vector |S⟩ represented as Ψ for both bases when working out the components for the quantum mechanics case above?

The first of the final two equations is simply an expression for the sifting property of the delta 'function'.

$$f(x) = \int dx' f(x')\delta(x - x') $$

Let's back up just a bit and write the state (ket) as a weighted 'sum' of position basis kets

$$|\mathcal{S}(t)\rangle = \int dx'\: |x'\rangle\langle x' | \mathcal{S}(t)\rangle = \int dx'\: |x'\rangle \Psi(x',t)$$

Here, $\Psi(x',t) = \langle x' | \mathcal{S}(t)\rangle$ is, for a specific $x'$, a complex number, the 'amount' of $|\mathcal{S}(t)\rangle$ in the $|x'\rangle$ 'direction'.

Weighting the basis ket $|x'\rangle$ by this number and then 'summing' over all these components, we get back the ket $|\mathcal{S}(t)\rangle$ (since the position basis is complete).

Now, contract both sides with the position basis bras $\langle x |$ to find

$$\langle x|\mathcal{S}(t)\rangle = \Psi(x,t) = \int dx'\: \langle x|x'\rangle \Psi(x',t) = \int dx'\: \delta(x - x')\Psi(x',t)$$

But we could have chosen the momentum basis kets $|p\rangle$ instead and then, rerunning the above, end up with

$$\langle p|\mathcal{S}(t)\rangle = \Phi(p,t) = \int dp'\: \langle p|p'\rangle \Phi(p't) = \int dp'\: \delta(p - p')\Phi(p',t)$$

However, these formal expressions don't actually help us find the $\Psi(x,t)$ and $\Phi(p,t)$. But, assuming we have solved for the components of $|\mathcal{S}(t)\rangle$ on one or the other basis, we can find the components on the other basis.

It follows, from the commutation relation for the position and momentum operators, that

$$\langle x | p\rangle = \frac{e^{i\frac{p}{\hbar}x}}{\sqrt{2\pi\hbar}}$$


$$\langle p | x\rangle = \frac{e^{-i\frac{p}{\hbar}x}}{\sqrt{2\pi\hbar}}$$

Then, starting with

$$|\mathcal{S}(t)\rangle = \int dx'\: |x'\rangle\langle x' | \mathcal{S}(t)\rangle = \int dx'\: |x'\rangle \Psi(x',t)$$

and then contracting with the momentum basis eigenbras, find

$$\langle p|\mathcal{S}(t)\rangle = \Phi(p,t) = \int dx'\: \langle p|x'\rangle \Psi(x',t) = \int dx'\: \frac{e^{-i\frac{p}{\hbar}x'}}{\sqrt{2\pi\hbar}}\Psi(x',t)$$

which is the 2nd of your final two equations expressing the relationship of between $\langle p|\mathcal{S}(t)\rangle$ and $\langle x|\mathcal{S}(t)\rangle$

  • $\begingroup$ This is a very good answer. One question, is the $\langle x| x' \rangle = \delta (x-x')$ defined in this way for the reasons that there exists a type of orthonormality when treating it as a scalar product (which it is not, since the position basis is not square integrable) and thus it is analogous in a way to the finite dimensional case of the Kronecker delta? Or is it derived from something else? $\endgroup$
    – user100411
    May 23, 2016 at 11:07

Take the inverse Fourier transform of your last equation, $\psi(x,t)=\frac{1}{\sqrt{2\pi\hbar}}\int e^{i p x/\hbar}\phi(p,t) dp$, to see that the "coefficients" of $\psi(x,t)$ are not the same in the two representations: in the $x$ representation, the coefficients are $\psi$ and in the $p$ representation, the coefficients are $\phi$.


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