I am pretty much confused with this notation I believe. The Heisenberg states are denoted by $\left|x,t\right>$ and the Schrodinger states are given by $\left|x(t)\right>$. It seems like both of these are parameterised with time, but the Schrodinger state is parameterised through the position vector,

But then in the case of Heisenberg picture, is the state parameterised using two variables $x,t$? What is this Linear Vector space

Secondly, in the case of position operator in the Heisenberg picture $\hat X_H(t)$ the eigenvalue equation is given like this,

$$ \hat X_H(t)\left|x,t\right> = x\left|x,t\right> . $$ So does that mean that for every time $t$ of $\hat X_H(t)$ there is a eigenvalue $x$ with the eigenvector $\left|x,t\right>$. So does that mean

$$ \hat X_H(t')\left|x,t\right> = 0 .$$


I) Recall that in the Heisenberg picture, operators [such as e.g. the position operator $\hat{X}(t)$] evolve in time $t$, while states (kets & bras) are independent of time $t$.

In particular, an instantaneous position eigenstate $| x_0,t_0 \rangle_H $ in the Heisenberg picture does not depend of time $t$, cf. Ref. 1. An instantaneous position eigenstate satisfy

$$\tag{1} \hat{X}(t=t_0)| x_0,t_0 \rangle_H~=~x_0| x_0,t_0 \rangle_H, \qquad {}_H\langle x_1,t_0 | x_2,t_0 \rangle_H~=~\delta(x_1-x_2). $$

It is important to stress that there are no requirements to $| x_0,t_0 \rangle_H$ for $t\neq t_0$.

II) One typical application of instantaneous position eigenstates [e.g. in connection with the timeslice procedure for the Feynman path integral, cf. Ref. 1] is to decompose the unit operator ${\bf 1}$ via an integral representation of instantaneous position eigenstates

$$\tag{2} {\bf 1}~=~ \int_{\mathbb{R}} \! d x_0 ~|x_0,t_0 \rangle_{H~H}\langle x_0,t_0 |. $$


  1. J.J. Sakurai, Modern Quantum Mechanics, 1994; Section 2.1 and Section 2.5.
  • $\begingroup$ +1 Great ! This answer is in the direction I wish to see, thanks. However now, if this is the case how do we enumerate the space, it has to be like for each $x$ i have infinite $t$ possibilities. So the vector space of states seems a little weird unlike in the case of Schrodinger states. I know these are non-normalisable states and make no proper mathematical sense, but just being curious how to define it. $\endgroup$ – user35952 May 7 '14 at 14:37
  • $\begingroup$ I updated the answer. $\endgroup$ – Qmechanic May 7 '14 at 15:58
  • $\begingroup$ With this kind of a completeness relation, does it mean for every time slice I have complete set of orthonormal vectors, i.e. $$ \mathcal {\textbf 1} = \int dx_0\left| x_0,t_0\right>_{HH}\left< x_0,t_0\right|\qquad \forall t_0\in \mathbb R$$ $\endgroup$ – user35952 May 8 '14 at 6:34
  • $\begingroup$ Yes, eq. (2) holds for any $t_0$. $\endgroup$ – Qmechanic May 8 '14 at 6:57

I think your confusion stems from the difference between the two pictures. For the Schrodinger picture the states $\left|{x(t)}\right>$ can evolve in time, while the operators are fixed. However for the Heisenberg picture, the states are fixed, and do not evolve with time, but rather the operators evolve with time.

Schrodinger picture: Given some base ket, $\left|x\right>$, we can evolve it in time to get $\left|x(t)\right> = U(t)\left|x\right>$, where $U(t) = e^{-iHt / \hbar}$ is the time evolution operator. The eigenvalue equation is now:

$$ X \left|x(t)\right> = x(t)\left|x(t)\right> $$

Heisenberg picture: Now the state is fixed $\left|x\right>$, but we evolve the operator. $X(t) = U^\dagger(t) X U(t)$, where $U(t)$ is the the same time evolution operator as before. The eigenvalue equation now reads:

$$ X(t) \left|x\right> = x(t)\left|x\right> $$

EDIT: So to answer the first two questions, the state is labeled as $\left|x,t\right>$ to indicate it is a fixed state at some point in time $t$, it is still part of the Hilbert space. For the next two questions, for every time $t$ of $X(t)$, there is an eigenvalue $x(t)$. I don't know what time you are referring to by $t'$, but in general no, is is not zero.

  • $\begingroup$ I am sorry, but I don't understand how this answers my question. I understand all that you have said. If the state does not have time dependence what is the reason behind choosing the notation $\left|x,t\right>$. Does it mean to say that the every state itself has time parameterisation by itself unlike in the Schrodinger case ? This also doesn't answer the last equation in the question. $\endgroup$ – user35952 May 7 '14 at 13:26
  • $\begingroup$ I answered the questions more specifically, but am a little unsure about some of the notation you used. $\endgroup$ – peanut_butter May 7 '14 at 13:42

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