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Multiple choice
For a single particle of mass
m
m
m
in one dimension under potential
V
(
x
,
t
)
V(x,t)
V
(
x
,
t
)
, the Hamiltonian operator
H
^
\hat{H}
H
^
is built from the classical energy
E
=
p
2
/
(
2
m
)
+
V
E = p^2/(2m) + V
E
=
p
2
/
(
2
m
)
+
V
by replacing momentum
p
p
p
with the operator
p
^
=
−
i
ℏ
d
/
d
x
\hat{p} = -i\hbar\,d/dx
p
^
=
−
i
ℏ
d
/
d
x
. What explicit form results?
H
^
=
+
(
ℏ
2
/
2
m
)
d
2
/
d
x
2
+
V
(
x
,
t
)
\hat{H} = +(\hbar^2/2m)\, d^2/dx^2 + V(x,t)
H
^
=
+
(
ℏ
2
/2
m
)
d
2
/
d
x
2
+
V
(
x
,
t
)
H
^
=
−
(
ℏ
2
/
2
m
)
d
2
/
d
x
2
+
V
(
x
,
t
)
\hat{H} = -(\hbar^2/2m)\, d^2/dx^2 + V(x,t)
H
^
=
−
(
ℏ
2
/2
m
)
d
2
/
d
x
2
+
V
(
x
,
t
)
H
^
=
−
i
ℏ
d
/
d
t
+
V
(
x
,
t
)
\hat{H} = -i\hbar\, d/dt + V(x,t)
H
^
=
−
i
ℏ
d
/
d
t
+
V
(
x
,
t
)
H
^
=
−
(
ℏ
2
/
2
m
)
d
/
d
x
+
V
(
x
,
t
)
\hat{H} = -(\hbar^2/2m)\, d/dx + V(x,t)
H
^
=
−
(
ℏ
2
/2
m
)
d
/
d
x
+
V
(
x
,
t
)
Check answer