# Position and Momentum operators in Heisenberg and Schrödinger picture

The position operator $$\hat{x}$$ has eigenstates

$$\hat{x}|x\rangle=x|x \rangle.$$

Usually in the Schrödinger picture the operators are time independent and the states carry the time dependence. However in the equation above it seems like $$\hat{x}$$ and also $$|x\rangle$$ are time independent? What kind of picture is used? The Schrödinger or the Heisenberg picture? Is the time dependence in the states above hidden and I should write

$$\hat{x}|x(t)\rangle=x(t)|x(t)\rangle~?$$

Edit:

The question arose when I looked at the path integral formulation of QM in Peskin. They use the generalized position states and operators denoted by $$q$$. On page 280 they write the completness relation $$\left(\prod_i\int dq_k^i\right) |q_k\rangle\langle q_k|$$, where $$k$$ is a time index. I thought $$|q_k\rangle=|q,t_k\rangle$$ and therefore the states are in Schrödinger picture? They always use $$\hat{q}|q_k\rangle=q_k|q_k\rangle$$.

• Use \rangle for kets. $|x\rangle$ vs $|x>$. I have edited it this time. Apr 5, 2022 at 23:48
• Sorry. Will do next time. Thanks for editing!
– user255856
Apr 5, 2022 at 23:53

In the Schrodinger picture, the components of the state vector changes with time. This is your wave function for position space:

$$|\Psi(t)\rangle=\int\psi(x,t)|x\rangle\,\text dx$$

Where the evolution in time follows the Schrodinger equation.

In the Heisenberg picture the operators, and hence the eigenstates, vary with time. The state vector remains constant, but the components still change since the basis vectors are changing.

• Thanks for the answer. My original question arose when I looked at the path integral formulation of QM in [Peskin][1]. On page 280 they write the completness relation $\prod_i\int d q_k^i|q_k\rangle\langle{q_k}|$, where $k$ is a time index? I thought $\langle{q_k}|=\langle{q,t_k}|$ and therefore the states are in Schrödinger picture? [1]: books.google.com/…
– user255856
Apr 6, 2022 at 0:16
• The $q$'s are generalized position operators
– user255856
Apr 6, 2022 at 0:20
• @jojo123456 If you have a new question you should make a new post :) Apr 6, 2022 at 2:55
• I mean it's not a new question. I just wanted to explain where my confusion is coming from...
– user255856
Apr 6, 2022 at 3:32
• @jojo123456 Ah ok sorry for the confusion. Thanks :) Apr 6, 2022 at 4:44

That's an interesting quirk of notation. The eigenstate $$|x\rangle$$ is defined by how the operator $$\hat{x}$$ acts on that eigenstate. Therefore, in the Schrodinger picture, where $$\hat{x}$$ is time-independent, the eigenstate of $$\hat{x}$$ called $$|x\rangle$$ is time independent. In the Heisenberg picture, the eigenstate is time dependent.

The time dependence of the state in the Schrodinger picture comes from the time dependence of the full state, $$|\Psi\rangle$$. If you expand the state in the position basis, $$$$|\Psi(t) \rangle = \int dx \psi(x, t) | x \rangle$$$$ the point is that the expansion coefficients $$\psi(x, t)$$ change in time, not the position basis states $$|x \rangle$$.

• Great minds think alike, haha Apr 6, 2022 at 0:00
• @BioPhysicist Amazing that we wrote the same equation with almost identical notation independently! Apr 6, 2022 at 0:01
• Almost makes you think there's some truth to this science thing after all :D Apr 6, 2022 at 0:06