Description
Pandas version checks
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I have checked that this issue has not already been reported.
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I have confirmed this bug exists on the latest version of pandas.
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I have confirmed this bug exists on the main branch of pandas.
Reproducible Example
import pandas as pd
date = pd.Timestamp("2025-01-01")
df = pd.DataFrame(columns=["date"], index=["a", "b", "c"])
df["date"] = date
print(df["date"].dtype) # Output: datetime64[s] Expected: datetime64[ns]
print(df.to_json()) # Output: {"date":{"a":1696,"b":1696,"c":1696}}
# Expected: {"date":{"a":1735689600000,"b":1735689600000,"c":1735689600000}}
Issue Description
When assigning a pd.Timestamp to a column in a DataFrame, the resulting dtype of the column is not as expected, and the output of to_json() is incorrect.
Expected Behavior
The dtype of the date column should default to datetime64[ns] after assignment.
The output of df.to_json() should correctly represent the timestamp in milliseconds since the epoch.
Installed Versions
INSTALLED VERSIONS
commit : 0691c5c
python : 3.12.4
python-bits : 64
OS : Darwin
OS-release : 24.4.0
Version : Darwin Kernel Version 24.4.0: Fri Apr 11 18:33:47 PDT 2025; root:xnu-11417.101.15~117/RELEASE_ARM64_T6000
machine : arm64
processor : arm
byteorder : little
LC_ALL : None
LANG : en_US.UTF-8
LOCALE : en_US.UTF-8
pandas : 2.2.3
numpy : 2.2.4
pytz : 2025.2
dateutil : 2.9.0.post0
pip : 25.0.1
Cython : None
sphinx : None
IPython : None
adbc-driver-postgresql: None
adbc-driver-sqlite : None
bs4 : None
blosc : None
bottleneck : None
dataframe-api-compat : None
fastparquet : None
fsspec : None
html5lib : None
hypothesis : None
gcsfs : None
jinja2 : 3.1.6
lxml.etree : None
matplotlib : None
numba : None
numexpr : None
odfpy : None
openpyxl : None
pandas_gbq : None
psycopg2 : None
pymysql : None
pyarrow : None
pyreadstat : None
pytest : None
python-calamine : None
pyxlsb : None
s3fs : None
scipy : None
sqlalchemy : None
tables : None
tabulate : None
xarray : None
xlrd : None
xlsxwriter : None
zstandard : None
tzdata : 2025.2
qtpy : None
pyqt5 : None
None